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Light - Reflection and Mirrors PYQs – Physics, SSC & Railway

106 Light - Reflection and Mirrors (Physics) questions asked in recent SSC and Railway exams, newest first, with exam, date & shift, answers and solutions.

📚 106 questionsSSC 4 · Railway 102
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RRB Group D 2025-26 – 25 August 2026 Shift 2 · Numericals on mirrors

Q1. A small candle is placed 18 cm from a concave mirror with a focal length of 6 cm. What will be the nature and position of the image formed?

  1. (a) Real, 6 cm in front of the mirror
  2. (b) Virtual, 12 cm behind the mirror
  3. (c) Virtual, 9 cm behind the mirror
  4. (d) Real, 9 cm in front of the mirror
Answer & solution

Answer: (d) Real, 9 cm in front of the mirror

Correct Answer:
Correct Answer: Option 4 – Real, 9 cm in front of the mirror
Explanation:
Using the mirror formula 1/v + 1/u = 1/f with the sign convention (distances measured against the direction of incident light are negative): object distance u = −18 cm, focal length f = −6 cm (concave mirror).1/v = 1/f − 1/u = (−1/6) − (−1/18) = −3/18 + 1/18 = −2/18 = −1/9v = −9 cm.Since v is negative, the image formsin front of the mirror(on the same side as the object) — it isreal, located9 cm in front of the mirror.
RRB Group D 2025-26 – 25 August 2026 Shift 3 · Concave mirror - image formation and uses

Q2. A converging mirror has a radius of curvature of 30 cm. If an object is placed on the principal axis at 60 cm from the pole, what is the nature of the image formed?

  1. (a) The image is real and upright on the opposite side
  2. (b) The image is virtual and upright on the same side as the object
  3. (c) The image is at the same position as the object
  4. (d) The image is real and inverted on the same side as the object
Answer & solution

Answer: (d) The image is real and inverted on the same side as the object

Correct Answer:
Correct Answer: Option 4 – The image is real and inverted on the same side as the object
Explanation:
A converging (concave) mirror has focal length f = R/2 = 15 cm, so the centre of curvature C is at 30 cm. The object is placed at 60 cm, which is beyond C (u > 2f). For an object placed beyond C, a concave mirror forms a real, inverted, diminished image between F and C — and being a mirror, the real image forms in front of the mirror, on the same side as the object.
Important Facts:
• Concave mirrors always form real images in front of the mirror (same side as the object), unlike converging lenses where real images form on the opposite side.
• Object beyond C → image between F and C, real, inverted, diminished.
• Object at C → image at C, same size, real, inverted.
RRB Group D 2025-26 – 21 August 2026 Shift 1 · Numericals on mirrors

Q3. A dentist uses a concave mirror of focal length 20 cm to examine a patient's tooth. If the tooth is placed 5 cm from the mirror, what is the magnification produced?

  1. (a) 1.67
  2. (b) 0.5
  3. (c) 0.75
  4. (d) 1.33
Answer & solution

Answer: (d) 1.33

Explanation:
Concave mirror: f = -20 cm, u = -5 cm (object placed within the focal length). Using 1/v + 1/u = 1/f:
1/v = 1/f - 1/u = -1/20 + 1/5 = 3/20, so v = 20/3 cm (positive, virtual image).
Magnification m = -v/u = -(20/3)/(-5) = 4/3 ≈ 1.33 (virtual, erect, magnified — as expected for an object placed between the pole and focus).
Correct Answer: Option 4 – 1.33
RRB Group D 2025-26 – 21 August 2026 Shift 1 · Concave mirror - image formation and uses

Q4. In a concave mirror, a ray parallel to the principal axis is incident. After reflection, the ray passes through:

  1. (a) The pole
  2. (b) The focus
  3. (c) Anywhere along the mirror
  4. (d) The center of curvature
Answer & solution

Answer: (b) The focus

Explanation:
This is a basic property of concave mirrors: any incident ray travelling parallel to the principal axis, after reflection, passes through the principal focus. This is in fact the defining property used to locate the focus of a concave mirror.
Correct Answer: Option 2 – The focus
RRB Group D 2025-26 – 21 August 2026 Shift 2 · Mirror formula and magnification

Q5. A student observes an image that is inverted and smaller than the object when using a mirror. What type of magnification is being observed?

  1. (a) The magnification is greater than one and positive
  2. (b) The magnification is equal to one and positive
  3. (c) The magnification is less than one and negative
  4. (d) The magnification is greater than one and negative
Answer & solution

Answer: (c) The magnification is less than one and negative

Correct Answer:
Correct Answer: Option 3 – The magnification is less than one and negative
Explanation:
An image that is inverted (real) is assigned a negative magnification by sign convention, and since the image is smaller than the object, the magnitude of magnification is less than one. Hence m is negative and |m| < 1, as with a concave mirror when the object is placed beyond the centre of curvature.
RRB Group D 2025-26 – 21 August 2026 Shift 3 · Numericals on mirrors

Q6. A real, inverted image is formed at a distance of 30 cm in front of a concave mirror when an object is placed 60 cm from the mirror. Which statement best describes the focal length of the mirror?

  1. (a) The focal length of the mirror is - 30 cm
  2. (b) The focal length of the mirror is + 10 cm
  3. (c) The focal length of the mirror is - 20 cm
  4. (d) The focal length of the mirror is + 15 cm
Answer & solution

Answer: (c) The focal length of the mirror is - 20 cm

Correct Answer:
Correct Answer: Option 3 – The focal length of the mirror is - 20 cm
Explanation:
Using the mirror formula 1/v + 1/u = 1/f with sign convention (u = −60 cm, v = −30 cm, both real/in front of mirror):
1/f = 1/(−30) + 1/(−60) = −2/60 − 1/60 = −3/60 = −1/20, sof = −20 cm.
Important Facts:
• For concave mirrors, focal length is taken as negative as per the Cartesian sign convention.
RRB Group D 2025-26 – 20 August 2026 Shift 2 · Convex mirror - image formation and uses

Q7. Which scenario best illustrates the use of a convex mirror based on the nature of its image?

  1. (a) A shopkeeper uses it to monitor a wide area inside the store
  2. (b) A scientist uses it to focus sunlight for heating
  3. (c) A student uses it to obtain a real, inverted image on paper
  4. (d) A model uses it as a makeup mirror
Answer & solution

Answer: (a) A shopkeeper uses it to monitor a wide area inside the store

Correct Answer:
Correct Answer: Option 1 – A shopkeeper uses it to monitor a wide area inside the store.
Explanation:
A convex mirror always forms a virtual, erect, and diminished image, and it has a wider field of view than a plane mirror. This makes it ideal for surveillance mirrors in shops and as vehicle rear-view mirrors, where seeing a wide area matters more than image size.
Important Facts:
• Concave mirrors, not convex, are used to focus sunlight (solar cookers) and produce real, inverted images (e.g., shaving/makeup mirrors held close).
RRB Group D 2025-26 – 20 August 2026 Shift 2 · Numericals on mirrors

Q8. A student places a small object 20 cm from a concave mirror of focal length 10 cm. According to the mirror formula, what will be the image distance?

  1. (a) 25 cm behind the mirror
  2. (b) 20 cm in front of the mirror
  3. (c) 10 cm behind the mirror
  4. (d) 15 cm in front of the mirror
Answer & solution

Answer: (b) 20 cm in front of the mirror

Correct Answer:
Correct Answer: Option 2 – 20 cm in front of the mirror
Explanation:
Using the mirror formula 1/v + 1/u = 1/f with sign convention u = −20 cm, f = −10 cm:1/v = 1/f − 1/u = (−1/10) − (−1/20) = −1/20, so v = −20 cm.Since the object is placed at the centre of curvature (2f = 20 cm), the image also forms at 2f, i.e., 20 cm in front of the mirror (real, inverted, same size).
RRB Group D 2025-26 – 18 August 2026 Shift 1 · Concave mirror - image formation and uses

Q9. A car driver uses a rear-view mirror to observe vehicles behind. If the mirror were replaced with a concave mirror of the same size, which of the following is the most likely consequence for the image of a car far behind?

  1. (a) The image would always be virtual, erect, and magnified
  2. (b) The image would always be virtual, inverted, and diminished
  3. (c) The image would always be virtual, erect, and diminished
  4. (d) The image would appear inverted and could be larger or smaller depending on the position
Answer & solution

Answer: (d) The image would appear inverted and could be larger or smaller depending on the position

Concave mirror as a rear-view mirror
Ordinary rear-view mirrors are convex, giving a wider field of view with a virtual, erect, diminished image at all distances — which is why convex mirrors are preferred.
A concave mirror behaves very differently depending on the object's distance relative to its focal length and centre of curvature: it can produce a real, inverted image that is magnified, same-sized, or diminished depending on exact position, and only produces a virtual erect image when the object is very close (within the focal length).
For a car far behind at varying distances, a concave mirror would generally give an inverted image whose size (larger or smaller) changes depending on how far the car is — making it unreliable and inconsistent, unlike the convex mirror's uniform virtual, erect, diminished image.
Correct Answer: The image would appear inverted and could be larger or smaller depending on the position
RRB Group D 2025-26 – 17 August 2026 Shift 2 · Numericals on mirrors

Q10. If an image is formed three times the size of the object and is inverted, what is the possible position of the object with respect to a concave mirror of focal length 10 cm?

  1. (a) Object placed 10 cm from the mirror
  2. (b) Object placed 13.3 cm from the mirror
  3. (c) Object placed 6.6 cm from the mirror
  4. (d) Object placed 20 cm from the mirror
Answer & solution

Answer: (b) Object placed 13.3 cm from the mirror

Concave Mirror: Magnification 3, Inverted, f = 10 cm
An inverted image means it is real, so magnification m = -v/u = -3, giving v = 3u. Using the mirror formula 1/v + 1/u = 1/f with sign convention (f = -10 cm for a concave mirror, u negative):
• 1/(3u) + 1/u = 1/f → 4/(3u) = 1/f → u = 4f/3
• u = 4×(-10)/3 = -13.3 cm, so the object distance is 13.3 cm from the mirror.
• This falls between f (10 cm) and 2f (20 cm), consistent with a real, inverted, magnified image — as expected for concave mirrors in this zone.
Correct Answer: Object placed 13.3 cm from the mirror
RRB Group D 2025-26 – 17 August 2026 Shift 3 · Convex mirror - image formation and uses

Q11. A science fair project uses a convex mirror to monitor a hallway. If a person stands at various distances from the mirror, what remains constant about the images formed, regardless of their position?

  1. (a) The images are always real and inverted
  2. (b) The images are always enlarged and real
  3. (c) The images are always magnified and behind the mirror
  4. (d) The images are always diminished and virtual
Answer & solution

Answer: (d) The images are always diminished and virtual

Images Formed by a Convex Mirror
A convex mirror always diverges reflected rays, so regardless of the object's distance:
• The image formed is alwaysvirtual, erect, and diminished(smaller than the object), located behind the mirror.
• This property makes convex mirrors ideal for surveillance/rear-view use, since they give a wide field of view with a constant type of image.
Correct Answer: The images are always diminished and virtual
RRB Group D 2025-26 – 14 August 2026 Shift 2 · Concave mirror - image formation and uses

Q12. A ray passing through the principal focus of a concave mirror will, after reflection:

  1. (a) emerge parallel to the principal axis
  2. (b) converge at the pole
  3. (c) reflect back through the centre of curvature
  4. (d) diverge away from the mirror
Answer & solution

Answer: (a) emerge parallel to the principal axis

Answer: Option 1 – Emerge parallel to the principal axis.
• A ray through the focus of a concave mirror reflects backparallel to the principal axis(reverse of a parallel ray).
RRB Group D 2025-26 – 14 August 2026 Shift 2 · Numericals on mirrors

Q13. A traffic surveillance camera uses a convex mirror with a radius of curvature of 2 m. Where will the image of a car located 3 m from the mirror appear?

  1. (a) 3 m in front of the mirror
  2. (b) 0.12 m in front of the mirror
  3. (c) 0.75 m behind the mirror
  4. (d) 3 m behind the mirror
Answer & solution

Answer: (c) 0.75 m behind the mirror

Answer: Option 3 – 0.75 m behind the mirror.
• Convex mirror: f = +R/2 = +1 m, u = −3 m ⇒ 1/v = 1/f − 1/u = 1 + 1/3 = 4/3 ⇒ v = +0.75 m (0.75 m behind).
RRB Group D 2025-26 – 13 August 2026 Shift 1 · Numericals on mirrors

Q14. A student places an object 20 cm in front of a concave mirror with a focal length of 10 cm, using the sign convention for spherical mirrors. What is the correct value and sign for the image distance?

  1. (a) -10 cm
  2. (b) +20 cm
  3. (c) +10 cm
  4. (d) -20 cm
Answer & solution

Answer: (d) -20 cm

Answer: Option 4 – −20 cm.
• 1/v + 1/u = 1/f, u = −20, f = −10 ⇒ 1/v = 1/(−10) − 1/(−20) = −1/20 ⇒ v =−20 cm.
RRB Group D 2025-26 – 13 August 2026 Shift 2 · Numericals on mirrors

Q15. A student observes that when an object is placed at 15 cm from a mirror, the image is formed at 30 cm on the same side. What is the focal length of the mirror?

  1. (a) -10 cm
  2. (b) -5 cm
  3. (c) -12 cm
  4. (d) -8 cm
Answer & solution

Answer: (a) -10 cm

Answer: Option 1 – −10 cm.
• Image on same side (real): u = −15, v = −30. 1/f = 1/v + 1/u = −1/30 − 1/15 = −3/30 ⇒ f =−10 cm.
RRB Group D 2025-26 – 13 August 2026 Shift 2 · Numericals on mirrors

Q16. A student places an object at 90 cm from a concave mirror and finds the image at 45 cm in front of the mirror. What will be the distance between the pole and the principal focus of this mirror?

  1. (a) 60 cm
  2. (b) 22.5 cm
  3. (c) 30 cm
  4. (d) 45 cm
Answer & solution

Answer: (c) 30 cm

Answer: Option 3 – 30 cm.
• u = −90, v = −45 ⇒ 1/f = 1/(−45) + 1/(−90) = −3/90 ⇒ f = −30 ⇒ pole–focus distance =30 cm.
RRB Group D 2025-26 – 12 August 2026 Shift 1 · Concave mirror - image formation and uses

Q17. A ray parallel to the principal axis strikes a concave mirror. After reflection, it passes through which point?

  1. (a) The vertex of the mirror
  2. (b) The focus of the mirror
  3. (c) The pole of the mirror
  4. (d) The center of curvature
Answer & solution

Answer: (b) The focus of the mirror

Answer: Option 2 – The focus of the mirror.
• A ray parallel to the principal axis reflects through thefocusof a concave mirror.
RRB Group D 2025-26 – 11 August 2026 Shift 1 · Concave mirror - image formation and uses

Q18. A ray passing through the center of curvature of a concave mirror strikes the mirror. What is the path of the reflected ray?

  1. (a) It becomes parallel to the axis
  2. (b) It passes through the focus
  3. (c) It diverges from the principal axis
  4. (d) It retraces its original path
Answer & solution

Answer: (d) It retraces its original path

Answer: Option 4 – It retraces its original path.
• A ray through thecentre of curvaturehits the mirror normally (along the radius), so itretraces its path.
RRB Group D 2025-26 – 10 August 2026 Shift 1 · Concave mirror - image formation and uses

Q19. An object is placed at the center of curvature of a concave mirror. If the mirror is now partially covered with an opaque material, what change will occur in the image?

  1. (a) The image will remain at the same position but will become less bright
  2. (b) The image will disappear completely
  3. (c) The image will move farther from the mirror and become inverted
  4. (d) The image will shift closer to the mirror and become smaller
Answer & solution

Answer: (a) The image will remain at the same position but will become less bright

Answer: Option 1 — The image remains at the same position but becomes less bright
Covering part of a mirror still forms the full image at the same place, justdimmer→ Option 1.
RRB Group D 2025-26 – 10 August 2026 Shift 3 · Concave mirror - image formation and uses

Q20. Which of the following best describes the image formed by a concave mirror when the object is placed just beyond the focal point but before the center of curvature?

  1. (a) Virtual, inverted, and magnified
  2. (b) Real, upright, and same size
  3. (c) Real, inverted, and magnified
  4. (d) Virtual, upright, and diminished
Answer & solution

Answer: (c) Real, inverted, and magnified

Answer: Option 3 – Real, inverted, and magnified.
• For a concave mirror with the objectbetween F and C, the image isreal, inverted and magnified(formed beyond C).
RRB Group D 2025-26 – 9 August 2026 Shift 2 · Convex mirror - image formation and uses

Q21. A convex mirror has a radius of curvature of 50 cm. A parallel beam of light strikes the mirror. At what distance from the pole does the reflected beam appear to diverge?

  1. (a) 50 cm in front of the mirror
  2. (b) 25 cm in front of the mirror
  3. (c) 25 cm behind the mirror
  4. (d) 10 cm behind the mirror
Answer & solution

Answer: (c) 25 cm behind the mirror

Answer: Option 3 — 25 cm behind the mirror
Convex mirror focus = R/2 = 25 cm, virtual (behind). Reflected parallel beam appears to diverge from25 cm behind→ Option 3.
RRB Group D 2025-26 – 9 August 2026 Shift 3 · Mirror formula and magnification

Q22. When measuring the object distance for a spherical mirror, as per the New Cartesian Sign Convention, which sign is assigned if the object is placed in front of the mirror?

  1. (a) No sign is assigned
  2. (b) Variable sign is assigned
  3. (c) Negative sign is assigned
  4. (d) Positive sign is assigned
Answer & solution

Answer: (c) Negative sign is assigned

Answer: Option 3 — Negative sign is assigned
By the New Cartesian Sign Convention, object distance (in front of the mirror) isnegative→ Option 3.
RRB Group D 2025-26 – 6 August 2026 Shift 2 · Numericals on mirrors

Q23. A student places an object 15 cm in front of a concave mirror of focal length 10 cm. If the image formed is real, what is the magnification produced by the mirror?

  1. (a) + 1
  2. (b) −2
  3. (c) −1
  4. (d) + 2
Answer & solution

Answer: (b) −2

Answer: Option 2 — −2
Concave mirror: 1/v + 1/u = 1/f, u = −15, f = −10. 1/v = −1/10 + 1/15 = (−3 + 2)/30 = −1/30 ⟹ v = −30.
m = −v/u = −(−30)/(−15) = −2 → Option 2.
RRB Group D 2025-26 – 5 August 2026 Shift 1 · Convex mirror - image formation and uses

Q24. A convex mirror has a focal length of 20 cm. If an object is placed 40 cm from the mirror, what will be the nature and relative size of the image?

  1. (a) Real, smaller than the object
  2. (b) Virtual, smaller than the object
  3. (c) Real, larger than the object
  4. (d) Virtual, same size as the object
Answer & solution

Answer: (b) Virtual, smaller than the object

Answer: Option 2 — Virtual, smaller than the object
Aconvex mirroralways forms avirtual, erect, diminishedimage, regardless of object position → Option 2.
RRB Group D 2025-26 – 5 August 2026 Shift 3 · Concave mirror - image formation and uses

Q25. A student places a small candle between the center of curvature and the principal focus of a concave mirror. What type of image will be formed by the mirror according to the ray diagram rules?

  1. (a) Virtual, erect and enlarged image is produced behind the mirror
  2. (b) Real, inverted and enlarged image is produced beyond the center of curvature
  3. (c) Virtual, erect and diminished image is produced between pole and focus
  4. (d) Real, inverted and same size image is produced at the center of curvature
Answer & solution

Answer: (b) Real, inverted and enlarged image is produced beyond the center of curvature

Answer: Option 2 — Real, inverted and enlarged, formed beyond C
For a concave mirror with the objectbetween C and F, the image isreal, inverted, enlarged and beyond C→ Option 2.
RRB Group D 2025-26 – 4 August 2026 Shift 2 · Concave mirror - image formation and uses

Q26. A physics teacher asks students to create a scenario where a concave mirror is used for shaving to get a large, upright image of the face. Where must the face be positioned relative to the mirror for this result?

  1. (a) Beyond the centre of curvature of the mirror
  2. (b) At the focus of the mirror
  3. (c) Between the pole and the focus of the mirror
  4. (d) At the centre of curvature of the mirror
Answer & solution

Answer: (c) Between the pole and the focus of the mirror

Answer: Option 3 — Between the pole and the focus of the mirror
A concave mirror gives avirtual, erect, magnifiedimage only when the object isbetween the pole and focus(shaving mirror) → Option 3.
RRB Group D 2025-26 – 3 August 2026 Shift 3 · Concave mirror - image formation and uses

Q27. A student observes that moving the object from beyond the center of curvature toward the focus of a concave mirror causes the image to change. What change is observed?

  1. (a) The image becomes larger and moves away from the mirror
  2. (b) The image remains the same size and position
  3. (c) The image becomes smaller and moves toward the mirror
  4. (d) The image becomes smaller and moves away from the mirror
Answer & solution

Answer: (a) The image becomes larger and moves away from the mirror

Answer: Option 1 — Image becomes larger and moves away from the mirror
For a concave mirror, as the object moves from beyond C toward F, thereal image moves awayfrom the mirror and growslarger→ Option 1.
RRB NTPC UG CBT-1 2025-26 – 19 June 2026 Shift 1 · Nature and properties of light

Q28. The transverse nature of light can be demonstrated using ______.

  1. (a) Prism
  2. (b) Polaroid
  3. (c) Lens
  4. (d) Mirror
Answer & solution

Answer: (b) Polaroid

Transverse Nature of Light
Light is an electromagnetic wave with electric and magnetic fields oscillating perpendicular to direction of propagation.
Polaroidallows only light waves vibrating in one plane to pass through, blocking others.
This selective transmission proves light waves are transverse (not longitudinal).
Prism shows dispersion, lens/mirror show reflection/refraction - none prove transverse nature.
Answer: Option 2 → Polaroid
RRB NTPC UG CBT-1 2025-26 – 7 May 2026 Shift 1 · Nature and properties of light

Q29. Which of the following optical properties of light enables sunglasses to reduce glare from sunlight reflected off surfaces such as water or roads?

  1. (a) Diffraction
  2. (b) Polarization
  3. (c) Interference
  4. (d) Refraction
Answer & solution

Answer: (b) Polarization

Topic:Optical Properties of Light - Polarization
Glare Formation:
Light reflected from flat surfaces (water, roads) becomes
partially polarized in horizontal plane.
Polarized Sunglasses:
Contain vertical polarizing filters that block horizontally
polarized light, thus reducing glare effectively.
Other Options:
Diffraction - bending around obstacles
Interference - wave superposition
Refraction - bending through media
Answer: Option 2 → Polarization
RRB NTPC GL CBT-1 2025-26 – 27 March 2026 Shift 1 · Nature and properties of light

Q30. What happens to light energy in interference and diffraction?

  1. (a) It decreases
  2. (b) It is redistributed
  3. (c) It reduces
  4. (d) It increases
Answer & solution

Answer: (b) It is redistributed

Interference and Diffraction
Law of Conservation of Energy applies to light.
In interference: constructive (bright) and destructive (dark) fringes form.
Energy from dark regions redistributes to bright regions.
In diffraction: light bends around obstacles, creating intensity patterns.
Total energy remains constant, only spatial distribution changes.
Energy is neither created nor destroyed, onlyredistributed.
Answer: Option 2 → It is redistributed
RRB NTPC GL CBT-1 2025-26 – 27 March 2026 Shift 2 · Spherical mirror terminology

Q31. Identify an example of a curved mirror.

  1. (a) Ladle
  2. (b) Prism
  3. (c) Wooden spatula
  4. (d) Shining spoon
Answer & solution

Answer: (d) Shining spoon

Curved Mirror:A reflective surface that is curved inward or outward.
Analysis of Options:
Ladle → Usually has a curved bowl but not reflective
Prism → Transparent glass object, not a mirror
Wooden spatula → Non-reflective flat surface
Shining spoon → Polished curved reflective surface
A shining spoon acts as both concave (inner side) and convex (outer side) mirror.
Answer: Option 4 → Shining spoon
RRB NTPC GL CBT-1 2025-26 – 24 March 2026 Shift 3 · Nature and properties of light

Q32. Polaroid glass is used in photographic cameras because:

  1. (a) it improves the resolution of camera
  2. (b) it improves image processing
  3. (c) it helps in auto focus of the lens
  4. (d) it reduces the light intensity to half on account of polarisation
Answer & solution

Answer: (d) it reduces the light intensity to half on account of polarisation

Topic: Polaroid Glass in Cameras
Key Facts:
1. Polaroid glass acts as a polarizing filter
2. It blocks one plane of light vibration
3. Unpolarized light has vibrations in all planes
Working Principle:
When unpolarized light passes through polaroid,
only one plane of vibration is transmitted
This reduces intensity to approximately half
Application in Cameras:
Controls excessive light entering the lens
Answer: Option 4 → it reduces the light intensity to half on account of polarisation
RRB NTPC GL CBT-1 2025-26 – 17 March 2026 Shift 3 · Concave mirror - image formation and uses

Q33. An object is placed between the focus and the pole of a concave mirror. The image formed is _______.

  1. (a) real and inverted
  2. (b) real and upright
  3. (c) virtual and inverted
  4. (d) virtual and upright
Answer & solution

Answer: (d) virtual and upright

Concave Mirror - Object Position: Between F and P
When object is placed between focus (F) and pole (P):
• Image formsbehind the mirror
• Image isvirtual(cannot be projected)
• Image iserect/upright(same orientation as object)
• Image ismagnified(larger than object)
This is the principle used inshaving mirrorsandmakeup mirrors.
Answer: Option 4 → virtual and upright
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Numericals on mirrors

Q34. A convex mirror of focal length 'X' cm forms an image at a distance of 8 cm when the object is placed at a distance of 40 cm in front of the mirror. What is the value of X?

  1. (a) + 20/3 cm
  2. (b) - 10 cm
  3. (c) + 10 cm
  4. (d) - 20/3 cm
Answer & solution

Answer: (c) + 10 cm

For a convex mirror, focal length (f) is positive.
Given
Object distance, u = -40 cm (object is in front, so negative convention)
Image distance, v = +8 cm (image is virtual and formed behind the mirror for convex mirror, so positive)
Mirror formula
(1)/(f) = (1)/(v + 1/u)
Substitute values
(1)/(f) = (1)/(8 + 1/(-40))
(1)/(f) = (1)/(8 - 1/40)
Find common denominator 40
(1)/(f) = (5)/(40 - 1/40) = (4)/(40) = (1)/(10)
So, focal length f = 10 cm
Since convex mirror has positive focal length, f = +10 cm
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Numericals on mirrors

Q35. A convex mirror produces a magnification of 1/3 when an object is placed at a distance of 30 cm from it. How much will be the image distance?

  1. (a) - 90 cm
  2. (b) - 10 cm
  3. (c) + 10 cm
  4. (d) + 90 cm
Answer & solution

Answer: (c) + 10 cm

Given
Magnification (m) = (1)/(3.)
Object distance (u) = -30 cm (since object distances are negative in mirror formula).
For a convex mirror, the magnification m = (v)/(u, where v is the image distance.)
Rearranging, v = m × u.
Substitute values
v = (1/3) × (-30) = -10 cm.
Sign of v is negative indicating the image is virtual and behind the mirror.
Therefore, the image distance is +10 cm (behind the mirror).
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Spherical mirror terminology

Q36. The centre of curvature (C) of a concave mirror is located ____________.

  1. (a) At infinity
  2. (b) Behind the mirror
  3. (c) At the pole (P)
  4. (d) In front of the mirror
Answer & solution

Answer: (d) In front of the mirror

The centre of curvature (C) of a concave mirror is the centre of the sphere from which the mirror is a part.
It always lies on the principal axis, in front of the mirror surface, as the reflective side faces inwards.
Hence, the centre of curvature is in front of the mirror.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Numericals on mirrors

Q37. The distance between the pole and the position of the object placed in front of a concave mirror is 120 cm. A real and inverted image is formed by this mirror on a screen at 40 cm from the pole of this mirror. What will be the focal length of this mirror?

  1. (a) + 60 cm
  2. (b) + 30 cm
  3. (c) - 60 cm
  4. (d) - 30 cm
Answer & solution

Answer: (d) - 30 cm

Given: Object distance (u) = -120 cm (since object is in front of concave mirror, distance is negative).
Image distance (v) = -40 cm (image is real and formed on the same side as object for concave mirror, so distance is negative).
Use mirror formula: 1/f = 1/v + 1/u.
Substitute values: 1/f = 1/(-40) + 1/(-120) = -1/40 − 1/120.
Find common denominator: 120.
1/f = (-3/120) − (1/120) = -4/120 = -1/30.
Therefore, focal length f = -30 cm.
Negative focal length confirms concave mirror.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Spherical mirror terminology

Q38. The principal focus of a concave mirror is:

  1. (a) The pole of the mirror
  2. (b) The point where rays parallel to the principal axis converge after reflection
  3. (c) A point on the mirror where all rays meet
  4. (d) The center of curvature of the mirror
Answer & solution

Answer: (b) The point where rays parallel to the principal axis converge after reflection

The principal focus of a concave mirror is the point where rays parallel to its principal axis meet after reflection.
This is the defining property of the principal focus in concave mirrors.
It is not simply the pole, centre of curvature, or any random point on the mirror.
Thus, the principal focus is the convergence point of these reflected rays.
Answer complete.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Concave mirror - image formation and uses

Q39. A concave mirror produces a virtual, erect and magnified image when the object is positioned _________.

  1. (a) Between the focus and the pole
  2. (b) Between the centre of curvature and the focus
  3. (c) Beyond the centre of curvature
  4. (d) At the principal focus
Answer & solution

Answer: (a) Between the focus and the pole

A concave mirror produces a virtual, erect and magnified image only when the object is placed between the focus and the pole of the mirror.
This region lies in front of the mirror but closer than the focal point.
In this position, reflected rays diverge and appear to come from behind the mirror, forming a virtual image.
Hence, the correct position is between the focus and the pole.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Spherical mirror terminology

Q40. When several rays parallel to the principal axis fall on a concave mirror, how do the reflected rays behave?

  1. (a) They spread out and appear to come from behind the mirror.
  2. (b) They are reflected back along the same path.
  3. (c) They travel parallel to the principal axis after reflection.
  4. (d) They all meet at a single point on the principal axis.
Answer & solution

Answer: (d) They all meet at a single point on the principal axis.

When rays parallel to the principal axis fall on a concave mirror,
they are reflected in such a way that they converge.
These reflected rays meet at a particular point on the principal axis called the focus.
Therefore, all reflected rays meet at a single point on the principal axis.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Numericals on mirrors

Q41. A convex mirror used as a rear-view mirror in a car has a radius of curvature of 4.00 m. A bus is located 6.00 m in front of the mirror. What will be the position and nature of the image formed?

  1. (a) Image is formed at 1.2 m behind the mirror
  2. (b) Image is formed at 1.5 m behind the mirror
  3. (c) Image is formed at 3.0 m in front of the mirror
  4. (d) The image is formed at 3.0 m behind the mirror
Answer & solution

Answer: (b) Image is formed at 1.5 m behind the mirror

For a convex mirror, radius of curvature (R) = +4.00 m (positive as convex mirrors have positive focal length).
Focal length (f)=(R)/(2)=(4.00)/(2) = +2.00 m.
Object distance (u) = -6.00 m (object in front of mirror is negative in mirror formula sign convention).
Mirror formula:(1)/(f)=(1)/(v)+(1)/(u)
Substituting:(1)/(2)=(1)/(v)+(1)/(-6)
1/2 = 1/v - 1/6
(1)/(v)=(1)/(2)+(1)/(6)=(3 + 1)/(6)=(4)/(6)=(2)/(3)
v =(3)/(2) = +1.5 m
Image distance is positive, means image is virtual and formed behind the mirror.
Hence, image is formed at 1.5 m behind the mirror and is virtual, erect, and diminished.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Concave mirror - image formation and uses

Q42. A dentist uses a mirror to obtain a large image of a patient’s tooth. Which mirror should the dentist use and why?

  1. (a) Concave mirror, because it gives a magnified image
  2. (b) Spherical mirror, because it gives a virtual image
  3. (c) Plane mirror, because it gives a real image
  4. (d) Convex mirror, because it gives a magnified image
Answer & solution

Answer: (a) Concave mirror, because it gives a magnified image

The dentist needs a magnified image to examine the tooth details clearly.
Concave mirrors are used because they can form magnified, real or virtual images depending on the object distance.
Specifically, when the tooth is placed between the focal point and the mirror, the concave mirror produces a magnified virtual image, which is ideal for examination.
Other mirrors do not produce a sufficiently magnified image for close detailed work.
Hence, a concave mirror is the correct choice.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Convex mirror - image formation and uses

Q43. How do rays parallel to the principal axis behave after reflection from a convex mirror?

  1. (a) They appear to diverge from a point on the principal axis.
  2. (b) They get reflected back along the same path.
  3. (c) They pass through the centre of curvature.
  4. (d) They actually converge at the focus.
Answer & solution

Answer: (a) They appear to diverge from a point on the principal axis.

Rays parallel to the principal axis of a convex mirror reflect in such a way that they appear to diverge.
After reflection, these rays seem to come from the focal point behind the mirror.
Since a convex mirror always forms virtual, erect, and diminished images, the rays do not actually meet but appear to diverge from a virtual focus.
Hence, among the options, the correct behaviour is that rays appear to diverge from a point on the principal axis.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Spherical mirror terminology

Q44. When using the sign convention for spherical mirrors, which direction along the principal axis is considered positive?

  1. (a) Downwards from the principal axis
  2. (b) Upwards from the principal axis
  3. (c) To the left of the origin (along - x-axis)
  4. (d) To the right of the origin (along + x-axis)
Answer & solution

Answer: (d) To the right of the origin (along + x-axis)

Solution:
In spherical mirrors,direction of incident light is taken as positive
Light travelsfrom left to right
Hence, distances measuredto the right are positive
Answer: To the right of the origin (along + x-axis)
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Convex mirror - image formation and uses

Q45. A security mirror in a store is required to provide a wide field of view and always form diminished, upright images regardless of the position of objects. Which type of mirror and principle should be used for its design?

  1. (a) Concave mirror using the mirror formula for real, inverted images
  2. (b) Concave mirror for magnification
  3. (c) Convex mirror using the mirror formula for virtual, diminished images
  4. (d) Plane mirror for undistorted images
Answer & solution

Answer: (c) Convex mirror using the mirror formula for virtual, diminished images

Option 3 is correct: Convex mirror using the mirror formula for virtual, diminished images.
Convex mirrors always form virtual, upright, and diminished images regardless of the object's position.
This property provides a wide field of view, making convex mirrors ideal for security purposes in stores.
Concave mirrors can form real, inverted images or magnified images but are not suitable here as the images need to be upright and diminished.
Plane mirrors produce images that are of the same size and not diminished, hence not suitable for wide field of view requirements.
Therefore, the design uses a convex mirror based on the mirror formula to get virtual, diminished images for better security coverage.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 1 · Spherical mirror terminology

Q46. If a point is located below the principal axis of a spherical mirror, what sign is assigned to its perpendicular distance according to the sign convention for spherical mirrors?

  1. (a) Depends on mirror type
  2. (b) Negative
  3. (c) Positive
  4. (d) Zero
Answer & solution

Answer: (b) Negative

According to the sign convention for spherical mirrors, distances measured below the principal axis are assigned a negative sign.
This applies for perpendicular distances from the principal axis.
Therefore, if a point lies below the principal axis, its perpendicular distance is negative.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 1 · Concave mirror - image formation and uses

Q47. When an object is placed at the center of curvature (C) of a concave mirror, what is the nature of the image formed?

  1. (a) Real and inverted
  2. (b) Highly diminished, point-sized
  3. (c) Virtual and erect
  4. (d) Enlarged and inverted
Answer & solution

Answer: (a) Real and inverted

When an object is placed at the centre of curvature (C) of a concave mirror,
the image forms at the centre of curvature itself.
The image is real because the light rays actually converge at this point.
It is inverted as the rays cross each other.
The size of the image is equal to the size of the object.
Hence, the correct nature of the image is: real and inverted.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 2 · Concave mirror - image formation and uses

Q48. A solar cooker uses a mirror to concentrate sunlight onto a vessel. Which type of mirror is used and what property makes it suitable?

  1. (a) A convex mirror because it provides a wide field of view
  2. (b) A plane mirror because it reflects light uniformly
  3. (c) A convex mirror because it spreads light over a large area
  4. (d) A concave mirror because it focuses parallel sunlight to a single point
Answer & solution

Answer: (d) A concave mirror because it focuses parallel sunlight to a single point

The solar cooker uses a mirror to concentrate sunlight onto the vessel.
A concave mirror is used because it can focus parallel rays of sunlight to a single point.
This focusing property increases the heat at the focal point, making it suitable for cooking.
Convex mirrors diverge light and are used for wide field of view, not concentration.
Plane mirrors reflect light uniformly but do not focus it.
Hence, the suitable mirror is a concave mirror due to its focusing ability.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 1 · Convex mirror - image formation and uses

Q49. If a convex mirror forms a point-sized image at its focus, where is the object located?

  1. (a) At the pole
  2. (b) Between P and F
  3. (c) At infinity
  4. (d) Between F and the pole
Answer & solution

Answer: (c) At infinity

A convex mirror always forms a virtual, erect, and diminished image of an object.
The image formed by a convex mirror is always located behind the mirror, and at the focus only when the object is at infinity.
When the object is at infinity, the reflected rays appear to come from the focus, forming a point-sized image there.
Therefore, the object must be located at infinity to form a point-sized image at the focus of a convex mirror.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 2 · Convex mirror - image formation and uses

Q50. When an object is placed at infinity in front of a convex mirror, what is the nature of the image formed?

  1. (a) Real and inverted
  2. (b) Real and erect
  3. (c) Virtual and inverted
  4. (d) Virtual and erect
Answer & solution

Answer: (d) Virtual and erect

The image formed by a convex mirror is always virtual and erect.
When the object is placed at infinity, the image forms at the focal point of the mirror.
This image is diminished in size but remains virtual and upright (erect).
Hence, the correct nature of the image is virtual and erect.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 3 · Spherical mirror terminology

Q51. Which type of spherical mirror has its reflecting surface curved inwards, facing towards the centre of the sphere?

  1. (a) Plane mirror
  2. (b) Parabolic mirror
  3. (c) Convex mirror
  4. (d) Concave mirror
Answer & solution

Answer: (d) Concave mirror

Correct option: 4
A spherical mirror where the reflecting surface curves inwards, towards the centre of the sphere, is called a concave mirror.
Concave mirrors converge light rays to a focal point.
This inward curvature distinguishes it from convex mirrors, whose reflecting surface bulges outwards.
Hence, the mirror described is a concave mirror.
Final answer: Option 4
SSC Stenographer 2025 – 6 December 2025 Shift 2 · Nature and properties of light

Q52. Which of the following phenomena occurs when a wave passes through a narrow slit?

  1. (a) Refraction
  2. (b) Reflection
  3. (c) Absorption
  4. (d) Diffraction
Answer & solution

Answer: (d) Diffraction

The correct answer is Diffraction . Diffraction is the phenomenon where waves bend or spread out when they pass through a narrow slit or around the edges of an obstacle. This effect is most prominent when the size of the slit is comparable to the wavelength of the wave. For example, light waves produce a diffraction pattern when passed through a narrow slit, showing alternate bright and dark fringes. Similarly, sound waves can bend around corners due to diffraction. This principle is important in optics, wave physics, and even in designing scientific instruments like diffraction gratings.
SSC CGL Tier-1 2025 – 24 September 2025 Shift 2 · Spherical mirror terminology

Q53. The minimum number of rays required to locate the image formed by a spherical mirror is ____.

  1. (a) One
  2. (b) Two
  3. (c) Three
  4. (d) Four
Answer & solution

Answer: (b) Two

To locate the image formed by a spherical mirror, the intersection point of at least two rays after reflection is required.
Hence the minimum number of rays needed is two.
SSC CGL Tier-1 2025 – 21 September 2025 Shift 1 · Concave mirror - image formation and uses

Q54. Which mirror type is used in solar furnaces to focus sunlight?

  1. (a) Convex
  2. (b) Plane
  3. (c) Concave
  4. (d) Biconvex
Answer & solution

Answer: (c) Concave

Concave mirrors converge parallel rays of sunlight to a single focal point, producing very high temperatures — ideal for solar furnaces.
Convex mirrors diverge light. Plane mirrors do not focus. Biconvex is a lens, not a mirror.
RRB NTPC UG CBT-1 2024-25 – 18 August 2025 Shift 3 · Wave optics - interference, diffraction and polarisation

Q55. What is the name of the principle which states that each point on a wavefront acts as a source of secondary wavelets that spread in all directions and form a new wavefront?

  1. (a) Fermat’s Principle
  2. (b) Superposition Principle
  3. (c) Doppler's Principle
  4. (d) Huygens' Principle
Answer & solution

Answer: (d) Huygens' Principle

SSC Selection Post 2025 (Phase XIII) – 29 July 2025 Shift 3 · Nature and properties of light

Q56. The phenomenon of interference in light waves is a direct consequence of:

  1. (a) Light being composed of particles (photons).
  2. (b) The diffraction of light around obstacles.
  3. (c) The superposition principle of waves.
  4. (d) The polarization of light waves.
Answer & solution

Answer: (c) The superposition principle of waves.

Correct Answer: Option 3 – The superposition principle of waves
Interference of light is a direct consequence of the superposition principle of waves, which states that when two or more waves overlap at a point, the resultant displacement is the algebraic sum of the individual displacements. When two coherent light waves (same frequency, constant phase difference) superpose: constructive interference occurs where crests meet crests (path difference = nλ ) → bright fringes; destructive interference occurs where crests meet troughs (path difference = (2n+1)λ/2) → dark fringes. Young's Double Slit Experiment (YDSE) classically demonstrates interference.
Option 1 (Particle nature/photons): Interference is a wave phenomenon — it was historically used by Thomas Young (1801) to establish the wave nature of light, contradicting the particle model.
Option 2 (Diffraction): Diffraction is a separate wave phenomenon (bending around obstacles); while both rely on superposition, interference specifically refers to superposition of two or more coherent wave sources, not diffraction.
Option 4 (Polarization): Polarization is a property of transverse waves (light is a transverse EM wave) showing directionality of oscillations — a separate phenomenon from interference.
RRB NTPC GL CBT-1 2024-25 – 23 June 2025 Shift 2 · Wave optics - interference, diffraction and polarisation

Q57. What is the nature of wavefront from a point source in an isotropic medium?

  1. (a) Plane
  2. (b) Cylindrical
  3. (c) Elliptical
  4. (d) Spherical
Answer & solution

Answer: (d) Spherical

RRB NTPC GL CBT-1 2024-25 – 20 June 2025 Shift 1 · Laws of reflection

Q58. According to the second law of reflection, all the incident ray, the normal to the mirror at the point of incidence and the reflected ray:

  1. (a) lie in the same plane
  2. (b) are mutually perpendicular to each other
  3. (c) lie in different planes
  4. (d) are parallel to each other
Answer & solution

Answer: (a) lie in the same plane

RRB Technician Grade III 2024-25 – 30 December 2024 Shift 1 · Spherical mirror terminology

Q59. An imaginary straight line passing through the pole and the centre of curvature of a spherical mirror is called as ______________.

  1. (a) principal axis
  2. (b) reflected ray
  3. (c) emergent ray
  4. (d) incident ray
Answer & solution

Answer: (a) principal axis

Correct Answer: Option 1 — principal axis
The principal axis of a spherical mirror is the imaginary straight line joining the pole (P) — the geometrical centre of the reflecting surface — to the centre of curvature (C), the centre of the sphere of which the mirror is a part. Because it passes through C, the principal axis is always normal (perpendicular) to the mirror at the pole.
Every important point of the mirror lies on this line: the pole P, the principal focus F, and the centre of curvature C, with PC = R and PF = f, related by R = 2f. All object and image distances in the mirror formula 1/v + 1/u = 1/f are measured along the principal axis.
A ray travelling along the principal axis strikes the mirror normally and is reflected straight back on itself, while rays parallel to the principal axis converge at (or appear to diverge from) the focus.
❌ Why the other options are incorrect
✗ reflected ray — this is the ray that bounces off the mirror after striking it; it is a real light ray, not a fixed geometrical line of the mirror.
✗ emergent ray — this term belongs to refraction: it is the ray that comes out of a glass slab or a prism.
✗ incident ray — this is the ray falling on the mirror; its direction changes with the position of the source, so it cannot define the mirror's axis.
📌 SSC & Railway Exam Facts
• Key points of a spherical mirror: Pole (P), Centre of curvature (C), Principal focus (F), Radius of curvature (R), Aperture.
• R = 2f, so f = R/2.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
• Laws of reflection: ∠i = ∠r, and the incident ray, reflected ray and normal all lie in the same plane.
• A concave mirror is converging; a convex mirror is diverging.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Spherical mirror terminology

Q60. The centre of curvature of a concave mirror lies _____ the mirror, whereas it lies _______ the mirror for a convex mirror.

  1. (a) behind; behind
  2. (b) behind; in front of
  3. (c) in front of; in front of
  4. (d) in front of; behind
Answer & solution

Answer: (d) in front of; behind

Correct Answer: Option 4 — in front of; behind
The centre of curvature (C) is the centre of the imaginary sphere of which the mirror is a part. A concave mirror is silvered on its bulging outer face and reflects from the hollow inner face, so the centre of that sphere lies on the reflecting side — that is, in front of the mirror. C is therefore a real point for a concave mirror.
In a convex mirror the reflecting surface bulges outwards towards the object, so the centre of the sphere falls on the far side — behind the mirror. Both the centre of curvature C and the principal focus F of a convex mirror are virtual points lying behind it. The distance from the pole P to C is the radius of curvature R, and it is always twice the focal length: R = 2f.
❌ Why the other options are incorrect
✗ behind; behind — this would be true only if both mirrors were convex; a concave mirror's C is a real point in front of it.
✗ behind; in front of — this reverses the two mirrors completely.
✗ in front of; in front of — right for the concave mirror, but wrong for the convex mirror, whose C is virtual and behind the surface.
📌 SSC & Railway Exam Facts
• R = 2f, so f = R/2; a mirror of R = 20 cm has f = 10 cm.
• In the New Cartesian sign convention a concave mirror has f negative and a convex mirror f positive.
• A concave mirror is a converging mirror; a convex mirror is a diverging mirror.
• Convex mirrors are used as rear-view mirrors because they give an erect, diminished image with a wide field of view.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 2 · Spherical mirror terminology

Q61. Which mirror has negative focal length?

  1. (a) Plane mirror
  2. (b) Concave
  3. (c) Convex
  4. (d) Both convex and concave
Answer & solution

Answer: (b) Concave

Correct Answer: Option 2 — Concave
The sign of the focal length follows the New Cartesian sign convention. In it the pole P of the mirror is the origin, the incident light travels from left to right, and every distance measured against the incident light (i.e. to the left of P, in front of the mirror) is taken as negative.
A concave mirror converges parallel rays to a real focus F lying in front of the mirror, on the same side as the object. That distance is measured opposite to the incident light, so f is negative for a concave mirror.
A convex mirror spreads the rays out; its focus is a virtual point behind the mirror, measured along the direction of the incident light, so its f is positive.
❌ Why the other options are incorrect
✗ Plane mirror — it neither converges nor diverges light; its focus is at infinity, so f = ∞, which is neither positive nor negative in this sense.
✗ Convex — its focus is virtual and behind the mirror, giving a positive focal length; that is why R = +2f for it.
✗ Both convex and concave — cannot be true, since the two mirrors have focal lengths of opposite signs.
📌 SSC & Railway Exam Facts
• Concave mirror: f negative; Convex mirror: f positive; Plane mirror: f = infinity.
• For lenses the rule is the reverse in sign feel: convex (converging) lens f positive, concave (diverging) lens f negative.
• R = 2f for a spherical mirror of small aperture.
• Power of a mirror/lens P = 1/f (f in metres); unit dioptre (D).
• A concave mirror is used as a shaving mirror, dentist's mirror and in headlights and solar cookers.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Concave mirror - image formation and uses

Q62. Where should the object be placed around the concave mirror to ensure that the size of the image is the same as the size of the object?

  1. (a) Between focus F and pole P
  2. (b) At focus F
  3. (c) Between focus F and centre of curvature C
  4. (d) At C, centre of curvature
Answer & solution

Answer: (d) At C, centre of curvature

Correct Answer: Option 4 — At C, centre of curvature
For a concave mirror, when the object is placed exactly at the centre of curvature C, the reflected rays meet again at C itself. The image formed is real, inverted and exactly the same size as the object — magnification m = −1.
This follows from the mirror formula. With u = −R = −2f:
1/v + 1/u = 1/f gives 1/v = 1/f − 1/(−2f)… solving gives v = −2f, so the image is also at a distance 2f = R from the pole, and m = −v/u = −1.
This is the only position for a concave mirror that produces an image of the same size as the object, which is why it is used in the laboratory to find the radius of curvature of a concave mirror.
❌ Why the other options are incorrect
✗ Between focus F and pole P — the image is then virtual, erect and enlarged, formed behind the mirror (the shaving-mirror position).
✗ At focus F — the reflected rays become parallel, so the image is formed at infinity and is highly enlarged (used in searchlights and headlights).
✗ Between focus F and centre of curvature C — the image is real, inverted and enlarged, formed beyond C, not equal in size.
📌 SSC & Railway Exam Facts
• Concave mirror, object at C → image at C, real, inverted, same size (m = −1).
• Object at infinity → image at F, highly diminished; object beyond C → image between F and C, diminished.
• Object between P and F → image behind the mirror, virtual, erect, enlarged — the only erect image a concave mirror gives.
• Mirror formula 1/v + 1/u = 1/f; magnification m = −v/u = h'/h; R = 2f.
• A convex mirror always gives a virtual, erect, diminished image, whatever the object position.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Spherical mirror terminology

Q63. The centre of curvature of a concave mirror lies _____ the mirror, whereas it lies ________ the mirror for a convex mirror.

  1. (a) in front of, in front of
  2. (b) in front of, behind
  3. (c) behind, behind
  4. (d) behind, in front of
Answer & solution

Answer: (b) in front of, behind

Correct Answer: Option 2 — in front of, behind
The centre of curvature (C) is the centre of the imaginary sphere of which the mirror forms a part. Where it lies depends on which face of that sphere is silvered.
A concave mirror is silvered on its bulging outer surface and reflects from the hollow inner surface, so the centre of the sphere lies on the same side as the reflecting surface — that is, in front of the mirror. For a concave mirror C is therefore a real point that light can actually reach.
A convex mirror reflects from the bulging outer surface, so the centre of the sphere falls on the far side — behind the mirror. For a convex mirror both C and the principal focus F are virtual points behind the surface. In each case the distance PC is the radius of curvature R, with R = 2f.
❌ Why the other options are incorrect
✗ in front of, in front of — correct for the concave mirror but wrong for the convex one, whose C is virtual and behind it.
✗ behind, behind — true only if both mirrors were convex; a concave mirror's C is a real point in front of it.
✗ behind, in front of — this reverses the two mirrors completely.
📌 SSC & Railway Exam Facts
• R = 2f, so f = R/2; a mirror of R = 30 cm has f = 15 cm.
• In the New Cartesian sign convention, a concave mirror has f negative and a convex mirror f positive.
• A concave mirror is converging; a convex mirror is diverging.
• Convex mirrors are used as rear-view mirrors because they give an erect, diminished image with a wide field of view.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Concave mirror - image formation and uses

Q64. A dentist usually uses:

  1. (a) a concave mirror
  2. (b) a convex mirror
  3. (c) a plane mirror
  4. (d) a plano-convex mirror
Answer & solution

Answer: (a) a concave mirror

Correct Answer: Option 1 — a concave mirror
A dentist uses a concave mirror because it can form a virtual, erect and magnified (enlarged) image. When an object is placed between the pole (P) and the focus (F) of a concave mirror, the reflected rays diverge and appear to come from behind the mirror, producing an enlarged upright image.
The dentist holds the small concave mirror very close to the tooth — that is, the tooth lies within the focal length — so the tooth appears much bigger, and tiny cavities become easy to see. The same mirror also converges light from the dentist's lamp onto the tooth, so the area being examined is brightly illuminated. Both effects come from the fact that a concave mirror is a converging mirror.
For this reason a concave mirror is also used as a shaving mirror and a make-up mirror, and by an ENT specialist as a head mirror.
❌ Why the other options are incorrect
✗ a convex mirror — it always gives a virtual, erect but diminished image; a smaller tooth is useless for examination. It is used as a rear-view mirror.
✗ a plane mirror — it gives an image of the same size as the object, with no magnification at all.
✗ a plano-convex mirror — this is not a standard mirror used in dentistry; plano-convex refers to a lens shape, and a convex reflecting surface would still diminish the image.
📌 SSC & Railway Exam Facts
• Concave mirror: object between P and F → image is virtual, erect and magnified.
• Uses of a concave mirror — dentist's mirror, shaving mirror, torch and headlight reflectors, solar cookers, telescopes.
• Uses of a convex mirror — vehicle rear-view mirrors, shop security mirrors, street-light reflectors (wide field of view).
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h′/h; f = R/2.
• Concave mirror has negative f; convex mirror has positive f (New Cartesian Sign Convention).
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Convex mirror - image formation and uses

Q65. Convex mirrors are fitted on the sides of vehicles because they always give ___________.

  1. (a) an erect and an enlarged image
  2. (b) a narrower field of the view
  3. (c) a real image
  4. (d) an erect and diminished image
Answer & solution

Answer: (d) an erect and diminished image

Correct Answer: Option 4 — an erect and diminished image
A convex mirror is a diverging mirror — its reflecting surface bulges outward and its focus F and centre of curvature C both lie behind the mirror. Because of this, wherever the object is placed, the reflected rays always diverge and appear to meet behind the mirror. The image formed is therefore always virtual, erect and smaller than the object, and it lies between the pole P and the focus F.
Since the image is diminished, a very large area of the road can be squeezed into a small mirror. This gives a wide field of view, so the driver can see traffic over a much bigger stretch behind the vehicle than a plane mirror of the same size would show. The erect image also keeps the view natural — vehicles are not seen upside down.
The trade-off is that objects look farther away than they really are, which is why such mirrors carry the warning "objects in mirror are closer than they appear".
❌ Why the other options are incorrect
✗ an erect and an enlarged image — an enlarged image is given by a concave mirror when the object is between P and F; a convex mirror can never magnify.
✗ a narrower field of the view — exactly the opposite; the whole point of using a convex mirror is its wider field of view.
✗ a real image — a convex mirror never forms a real image for a real object; its image is always virtual and behind the mirror.
📌 SSC & Railway Exam Facts
• Convex mirror image: always virtual, erect, diminished, formed between P and F.
• Uses of a convex mirror — rear-view/side mirrors of vehicles, shop security mirrors, blind-turn road mirrors, street-light reflectors.
• For a convex mirror f and R are positive; for a concave mirror they are negative.
• f = R/2; mirror formula 1/v + 1/u = 1/f; magnification m = −v/u.
• For a convex mirror |m| < 1 always, and m is positive because the image is virtual and erect.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Numericals on mirrors

Q66. If an object is placed in front of a concave mirror at a distance of 20 cm and the image is formed on the same side at a distance of 40 cm, then the magnification of the mirror will be:

  1. (a) +0.4
  2. (b) −0.4
  3. (c) +2.0
  4. (d) −2.0
Answer & solution

Answer: (d) −2.0

Correct Answer: Option 4 — −2.0
📖 Given
• Object distance, u = −20 cm (the object is always in front of the mirror)
• Image distance, v = −40 cm (the image is on the same side, i.e. in front of the mirror)
• Mirror is concave
✍️ Formula
m = −v/u
🔢 Solution
• Apply the New Cartesian Sign Convention: distances measured against the incident light, i.e. in front of the mirror, are negative. So u = −20 cm and v = −40 cm.
• m = −v/u = −(−40)/(−20)
• m = −(40/20)
• m = −2.0
• The negative magnification means the image is real and inverted; the magnitude 2 means it is twice the size of the object.
In plain language, the object lies between F and C of the concave mirror, so a real, inverted, enlarged image is formed beyond C — exactly what m = −2.0 tells us.
❌ Why the other options are incorrect
✗ +0.4 — this comes from taking the ratio the wrong way round (20/40 = 0.5 mishandled) and also giving the image a virtual, erect nature it does not have.
✗ −0.4 — again u and v have been interchanged; that would describe a diminished image, but here the image is larger than the object.
✗ +2.0 — the magnitude is correct, but the positive sign would mean a virtual, erect image. An image formed on the same side as the object is real and inverted, so m must be negative.
📌 SSC & Railway Exam Facts
• m = −v/u = h′/h; m negative → real, inverted image; m positive → virtual, erect image.
• |m| > 1 means enlarged, |m| < 1 means diminished, |m| = 1 means the same size.
• Mirror formula 1/v + 1/u = 1/f; here 1/(−40) + 1/(−20) = 1/f gives f = −13.3 cm, confirming a concave mirror.
• A concave mirror gives an enlarged real image when the object lies between F and C.
• Plane mirror: m = +1 always — image is virtual, erect and of the same size, and laterally inverted.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Spherical mirror terminology

Q67. A ray of light passing or appear to pass through the principal focus of a concave/convex mirror after reflection ______________.

  1. (a) will emerge perpendicular to the principal axis, only in the case of concave mirror
  2. (b) will emerge perpendicular to the principal axis
  3. (c) will emerge perpendicular to the principal axis, only in the case of convex mirror
  4. (d) will emerge parallel to the principal axis
Answer & solution

Answer: (d) will emerge parallel to the principal axis

Correct Answer: Option 4 — will emerge parallel to the principal axis
This is one of the four standard ray-diagram rules for spherical mirrors. A ray that passes through the principal focus F (in a concave mirror) or appears to pass through F (in a convex mirror, where F lies behind the mirror) always comes back parallel to the principal axis after reflection.
The reason is the reversibility of light. The very first rule says that a ray travelling parallel to the principal axis is reflected through the focus. Reverse the direction of that same ray and you get exactly this rule — a ray directed at the focus returns parallel to the axis. Both are simply the laws of reflection (∠i = ∠r) applied at the mirror surface.
The rule holds for both concave and convex mirrors, which is why the answer carries no extra condition.
❌ Why the other options are incorrect
✗ will emerge perpendicular to the principal axis, only in the case of concave mirror — the emergent ray is parallel, never perpendicular, and no special condition applies.
✗ will emerge perpendicular to the principal axis — wrong direction; a perpendicular emergent ray would never form the images we actually observe.
✗ will emerge perpendicular to the principal axis, only in the case of convex mirror — wrong on both counts: wrong direction and a false restriction to convex mirrors.
📌 SSC & Railway Exam Facts
• Laws of reflection: ∠i = ∠r, and the incident ray, reflected ray and normal all lie in the same plane.
• For a spherical mirror f = R/2, where R is the radius of curvature.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h′/h.
• A ray passing through the centre of curvature C hits the mirror normally and retraces its own path.
• Concave mirrors are used in torches, headlights, shaving mirrors and solar cookers; convex mirrors as rear-view mirrors because they give a wide field of view.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Numericals on mirrors

Q68. The radius of curvature of any mirror is +20 cm. What is the focal length and nature of this mirror?

  1. (a) +40 cm, concave
  2. (b) +10 cm, concave
  3. (c) +10 cm, convex
  4. (d) −40 cm, convex
Answer & solution

Answer: (c) +10 cm, convex

Correct Answer: Option 3 — +10 cm, convex
📖 Given
• Radius of curvature, R = +20 cm
• To find: focal length f and the nature of the mirror
✍️ Formula
f = R/2
🔢 Solution
• f = R/2
• f = +20/2
• f = +10 cm
• Now the sign tells the nature. Under the New Cartesian Sign Convention, distances measured behind the mirror (in the direction of the incident light) are positive.
• A positive R and f mean the centre of curvature and the focus lie behind the mirror — this happens only for a convex mirror.
• So the mirror has f = +10 cm and is convex.
In short, the magnitude 10 cm comes from halving the radius, and the plus sign identifies it as a convex (diverging) mirror.
❌ Why the other options are incorrect
✗ +40 cm, concave — this is the error of writing f = 2R instead of f = R/2, and concave mirrors have a negative f.
✗ +10 cm, concave — the magnitude is right but the nature is wrong; for a concave mirror f would be −10 cm.
✗ −40 cm, convex — both wrong: the value comes from doubling instead of halving R, and the sign contradicts a convex mirror.
📌 SSC & Railway Exam Facts
• f = R/2 for every spherical mirror; for a plane mirror R = ∞, so f = ∞.
• Concave mirror: f and R are negative; convex mirror: f and R are positive.
• A convex mirror always forms a virtual, erect and diminished image, so it is used as a rear-view mirror.
• A concave mirror is used in torches, headlights, shaving mirrors, dentists' mirrors and solar cookers.
• Mirror formula 1/v + 1/u = 1/f; magnification m = −v/u.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 2 · Laws of reflection

Q69. The laws of reflection are applicable to ________.

  1. (a) neither plane surfaces nor spherical surfaces
  2. (b) both plane and spherical surfaces
  3. (c) only to plane surfaces and not to spherical surfaces
  4. (d) only to spherical surface and not to plane surfaces
Answer & solution

Answer: (b) both plane and spherical surfaces

Correct Answer: Option 2 — both plane and spherical surfaces
The laws of reflection are:
• First law — the angle of incidence is equal to the angle of reflection (∠i = ∠r).
• Second law — the incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
These laws are applied at the point of incidence, using the normal drawn at that point. For a plane mirror the normal is the same all over the surface. For a spherical mirror the normal at any point is the line joining that point to the centre of curvature, so the normal changes from point to point — but the two laws still hold exactly at every single point.
That is precisely why we can draw ray diagrams for concave and convex mirrors and derive the mirror formula 1/v + 1/u = 1/f. So the laws apply to all reflecting surfaces, plane as well as spherical (and even to rough surfaces, where they give diffuse reflection).
❌ Why the other options are incorrect
✗ neither plane surfaces nor spherical surfaces — if this were true, no mirror could form an image at all.
✗ only to plane surfaces and not to spherical surfaces — spherical mirror ray diagrams are built entirely on ∠i = ∠r, so the laws clearly apply there too.
✗ only to spherical surface and not to plane surfaces — the plane mirror is the simplest demonstration of the laws of reflection.
📌 SSC & Railway Exam Facts
• A plane mirror always forms a virtual, erect image of the same size, at the same distance behind the mirror; it is laterally inverted.
• Regular reflection occurs from a smooth polished surface; diffuse (irregular) reflection from a rough surface — the laws hold in both cases.
• f = R/2 for a spherical mirror; mirror formula 1/v + 1/u = 1/f.
• Magnification m = −v/u = h′/h; a negative m means a real, inverted image.
• A concave mirror gives an erect and magnified image only when the object is placed between the pole and the focus.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Concave mirror - image formation and uses

Q70. An object is placed in front of a concave mirror of focal length f, and it is found that an image of the same size as that of the object is formed. The object distance u is:

  1. (a) equal to f
  2. (b) equal to 2f
  3. (c) greater than 2f
  4. (d) less than f
Answer & solution

Answer: (b) equal to 2f

Correct Answer: Option 2 — equal to 2f
📖 Given
• Mirror is concave, focal length = f
• The image formed is of the same size as the object, i.e. |m| = 1
• To find: the object distance u
✍️ Formula
1/v + 1/u = 1/f and m = −v/u
🔢 Solution
• Same size means |m| = 1. For a concave mirror this image is real and inverted, so m = −1.
• From m = −v/u = −1, we get v = u.
• Substituting in the mirror formula: 1/u + 1/u = 1/f, i.e. 2/u = 1/f.
• Therefore u = 2f — the object is at a distance of twice the focal length.
• Since R = 2f, the object is exactly at the centre of curvature C, and the image is also formed at C.
In plain language: place the object at C of a concave mirror and you get a real, inverted image of the same size, also at C — the only position that gives unit magnification in a concave mirror.
❌ Why the other options are incorrect
✗ equal to f — with the object at the focus F, the reflected rays become parallel and the image is formed at infinity, highly enlarged.
✗ greater than 2f — the object beyond C gives a real, inverted but diminished image, formed between F and C.
✗ less than f — the object between P and F gives a virtual, erect and enlarged image behind the mirror, not a same-size one.
📌 SSC & Railway Exam Facts
• Concave mirror, object at C (u = 2f) → image at C, real, inverted, same size.
• Object beyond C → image between F and C, real, inverted, diminished.
• Object between F and C → image beyond C, real, inverted, enlarged.
• Object between P and F → image behind the mirror, virtual, erect, enlarged.
• f = R/2; mirror formula 1/v + 1/u = 1/f; magnification m = −v/u = h′/h.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Spherical mirror terminology

Q71. The distance from the focus point of a mirror to the pole is known as:

  1. (a) centre of curvature
  2. (b) aperture
  3. (c) focal length
  4. (d) radius of curvature
Answer & solution

Answer: (c) focal length

Correct Answer: Option 3 — focal length
The focal length (f) of a spherical mirror is defined as the distance between the principal focus F and the pole P of the mirror. The pole is the geometrical centre of the reflecting surface, and the principal focus is the point on the principal axis where rays parallel to the axis converge (concave mirror) or from which they appear to diverge (convex mirror) after reflection.
The focal length is related to the radius of curvature R by the simple relation f = R/2, i.e. the focus lies exactly midway between the pole and the centre of curvature.
Under the New Cartesian Sign Convention, f is negative for a concave mirror (the focus is in front of the mirror) and positive for a convex mirror (the focus is behind it). For a plane mirror R = ∞, so f = ∞.
❌ Why the other options are incorrect
✗ centre of curvature — this is a point, not a distance: the centre C of the imaginary sphere of which the mirror is a part.
✗ aperture — the effective diameter of the reflecting surface, i.e. how wide the mirror is; it decides how much light the mirror collects.
✗ radius of curvature — the distance from the pole P to the centre of curvature C, which is twice the focal length (R = 2f), not the focus-to-pole distance.
📌 SSC & Railway Exam Facts
• f = R/2 and R = 2f for every spherical mirror.
• Pole (P) — centre of the mirror surface; Centre of curvature (C) — centre of the sphere; Principal focus (F) — midpoint of PC.
• Concave mirror: f and R are negative; Convex mirror: f and R are positive.
• Mirror formula 1/v + 1/u = 1/f; magnification m = −v/u = h′/h.
• For a plane mirror, R = ∞ and f = ∞, and the image is always virtual, erect, same size and laterally inverted.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 1 · Laws of reflection

Q72. The laws of reflection are applicable to:

  1. (a) only curved reflecting surface
  2. (b) only plane reflecting surface
  3. (c) all types of reflecting surfaces excluding spherical surfaces
  4. (d) all types of reflecting surfaces including spherical surfaces
Answer & solution

Answer: (d) all types of reflecting surfaces including spherical surfaces

Correct Answer: Option 4 — all types of reflecting surfaces including spherical surfaces
The laws of reflection are:
• (1) The angle of incidence is equal to the angle of reflection, i.e. ∠i = ∠r.
• (2) The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.
These laws are universal — they hold for every reflecting surface, whether it is plane, concave, convex, cylindrical or irregular. The reason is that reflection is a point-by-point phenomenon: at each tiny point of the surface we draw the normal to the surface at that point, and the two laws are applied there.
For a spherical mirror the normal at any point is simply the line joining that point to the centre of curvature (C). Because the normal changes direction from point to point, a curved mirror converges or diverges the beam — yet at every single point ∠i = ∠r is still obeyed.
❌ Why the other options are incorrect
✗ only curved reflecting surface — the laws were first stated and are most easily demonstrated on a plane mirror, so they certainly apply there too.
✗ only plane reflecting surface — if this were true, ray diagrams for concave and convex mirrors, which are drawn using ∠i = ∠r, would be impossible.
✗ all types of reflecting surfaces excluding spherical surfaces — spherical mirrors are not an exception; the laws apply at each point using the normal through the centre of curvature.
📌 SSC & Railway Exam Facts
• Laws of reflection: ∠i = ∠r, and incident ray, reflected ray and normal are coplanar.
• They apply to regular (specular) as well as irregular (diffuse) reflection — a rough wall obeys them at every point.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u; R = 2f.
• A plane mirror forms a virtual, erect, same-sized image at an equal distance behind the mirror, with lateral inversion.
• Reflection does not change the wavelength, frequency or speed of light, since the medium is unchanged.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 2 · Concave mirror - image formation and uses

Q73. In a concave mirror, the object is placed between the pole and the focus point of the mirror. What is the nature of the image?

  1. (a) Real and inverted
  2. (b) Virtual and erect
  3. (c) Virtual and inverted
  4. (d) Real and erect
Answer & solution

Answer: (b) Virtual and erect

Correct Answer: Option 2 — Virtual and erect
When an object is placed between the pole (P) and the principal focus (F) of a concave mirror, the reflected rays diverge after reflection. They never actually meet, so they must be produced backwards; they appear to meet behind the mirror.
The image so formed is therefore virtual, erect and enlarged (magnified), and it lies behind the mirror. This is the only position of the object for which a concave mirror gives an erect image.
This is exactly why a concave mirror is used as a shaving mirror and as a dentist's mirror — the face or the tooth is kept closer than the focus so that a big, upright image is seen.
❌ Why the other options are incorrect
✗ Real and inverted — a concave mirror gives a real, inverted image only when the object is at or beyond F, not between P and F.
✗ Virtual and inverted — impossible: a virtual image formed by a mirror is always erect.
✗ Real and erect — impossible for a single mirror: a real image formed by a mirror is always inverted.
📌 SSC & Railway Exam Facts
• Concave mirror, object between P and F → virtual, erect, enlarged image behind the mirror.
• Object at C → image at C, real, inverted, same size; object at F → image at infinity.
• A convex mirror always forms a virtual, erect, diminished image, whatever the object position.
• Mirror formula 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
• For a spherical mirror, R = 2f (radius of curvature is twice the focal length).
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 2 · Concave mirror - image formation and uses

Q74. Consider a ray passing through the principal focus and not parallel to the principal axis of a concave mirror, after reflection, it will:

  1. (a) emerge parallel to the principal axis
  2. (b) emerge obliquely to the principal axis
  3. (c) pass through the principal focus
  4. (d) pass through the centre of curvature
Answer & solution

Answer: (a) emerge parallel to the principal axis

Correct Answer: Option 1 — emerge parallel to the principal axis
This is one of the standard rules of ray diagrams for spherical mirrors. A ray that passes through the principal focus (F) of a concave mirror strikes the mirror and, after reflection, travels parallel to the principal axis.
The reason is the principle of reversibility of light — light retraces its own path. A ray coming parallel to the principal axis is reflected through F; therefore a ray going through F must be reflected parallel to the axis.
This property is used in torches, headlights and searchlights, where the bulb is placed at the focus of a concave reflector so that a strong parallel beam comes out.
❌ Why the other options are incorrect
✗ emerge obliquely to the principal axis — the reflected ray is exactly parallel, not slanting; the rule is definite, not approximate.
✗ pass through the principal focus — this happens to a ray that arrives parallel to the principal axis, which is the reverse case.
✗ pass through the centre of curvature — a ray retraces its path through C only when it strikes the mirror normally, i.e. it was already directed at C.
📌 SSC & Railway Exam Facts
• Ray parallel to axis → reflected through F; ray through F → reflected parallel to axis.
• Ray through C → retraces its own path (angle of incidence = 0°).
• Laws of reflection: ∠i = ∠r, and the incident ray, reflected ray and normal lie in the same plane.
• A concave mirror is a converging mirror; a convex mirror is a diverging mirror.
• f = R/2 for a spherical mirror; a concave mirror's f is taken negative in the sign convention.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Mirror formula and magnification

Q75. A negative sign in magnification of spherical mirrors indicates that the __________.

  1. (a) image is enlarged
  2. (b) image is diminished
  3. (c) image is real
  4. (d) image is virtual
Answer & solution

Answer: (c) image is real

Correct Answer: Option 3 — image is real
For a spherical mirror the magnification is
m = h'/h = −v/u
where h' is the image height and h the object height.
The sign of m tells us about the nature and orientation of the image, while the numerical value of m tells us about the size.
• m is negative → the image is real and inverted (it is formed in front of the mirror and can be caught on a screen).
• m is positive → the image is virtual and erect (it is formed behind the mirror and cannot be caught on a screen).
Since a real image formed by a mirror is always inverted, h' is measured below the axis and comes out negative, which is exactly why m is negative. A concave mirror gives a negative m whenever the object lies beyond the focus.
❌ Why the other options are incorrect
✗ image is enlarged — enlargement is decided by the magnitude of m, not its sign: |m| > 1 means enlarged, whatever the sign.
✗ image is diminished — again a matter of magnitude: |m| < 1 means diminished; a diminished image may be real (negative m) or virtual (positive m).
✗ image is virtual — a virtual image always has a positive magnification, since it is erect; this is the opposite of what a negative sign means.
📌 SSC & Railway Exam Facts
• Magnification m = h'/h = −v/u for mirrors; for lenses m = h'/h = v/u.
• Negative m → real, inverted; positive m → virtual, erect.
• |m| > 1 → enlarged, |m| = 1 → same size, |m| < 1 → diminished.
• A convex mirror always gives 0 < m < 1 — a virtual, erect, diminished image.
• Mirror formula: 1/v + 1/u = 1/f; R = 2f; magnification has no unit.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Mirror formula and magnification

Q76. Which of the following correctly represents the relationship between u(object distance), v (image distance) and f (focal length) of a spherical mirror?

  1. (a) 1/v − 1/u = 1/2f
  2. (b) 1/v + 1/u = 1/f
  3. (c) 1/v − 1/u = 1/f
  4. (d) 1/v + 1/u = 1/2f
Answer & solution

Answer: (b) 1/v + 1/u = 1/f

Correct Answer: Option 2 — 1/v + 1/u = 1/f
The relation connecting the object distance u, the image distance v and the focal length f of a spherical mirror is called the mirror formula:
1/v + 1/u = 1/f
It holds for both concave and convex mirrors, provided the New Cartesian Sign Convention is used: all distances are measured from the pole, distances measured in the direction of the incident light are positive and those opposite to it are negative. So for a concave mirror f is negative, and for a convex mirror f is positive.
Remember also that f = R/2, i.e. the focal length is half the radius of curvature, and the magnification is m = −v/u.
A common trap is to mix this up with the lens formula, 1/v − 1/u = 1/f, which has a minus sign — mirrors use plus, lenses use minus.
❌ Why the other options are incorrect
✗ 1/v − 1/u = 1/2f — wrong on two counts: it uses the lens-type minus sign and an extra factor of 2 that does not belong in the formula.
✗ 1/v − 1/u = 1/f — this is the lens formula, not the mirror formula; using it for a mirror gives a wrong image position.
✗ 1/v + 1/u = 1/2f — the right-hand side must be 1/f, not 1/2f; the factor 2 belongs to the separate relation R = 2f.
📌 SSC & Railway Exam Facts
• Mirror formula: 1/v + 1/u = 1/f; Lens formula: 1/v − 1/u = 1/f.
• f = R/2 for a spherical mirror; R is the radius of curvature.
• Magnification: mirror m = −v/u, lens m = v/u.
• Power of a lens P = 1/f (metre), SI unit dioptre (D).
• A concave mirror is converging (f negative); a convex mirror is diverging (f positive) in the New Cartesian convention.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 3 · Spherical mirror terminology

Q77. A spherical mirror, whose reflecting surface faces towards the centre of the sphere, is called a _________.

  1. (a) concave mirror
  2. (b) convex mirror
  3. (c) plano-convex mirror
  4. (d) plane mirror
Answer & solution

Answer: (a) concave mirror

Correct Answer: Option 1 — concave mirror
A spherical mirror is a piece cut from a hollow sphere with one surface silvered. If the reflecting (polished) surface is the inner, hollow side — the side facing the centre of the sphere — the mirror is called a concave mirror. If the outer, bulging side reflects, it is a convex mirror.
A concave mirror is a converging mirror: a beam of light parallel to the principal axis is brought together at the principal focus F after reflection. Its focal length f = R/2, and in the New Cartesian sign convention f is negative.
Because of this converging property a concave mirror is used as a shaving mirror, a dentist's mirror, in torches, headlights and searchlights (bulb at the focus gives a parallel beam) and in solar cookers and reflecting telescopes.
❌ Why the other options are incorrect
✗ convex mirror — here the reflecting surface is the outer bulged side, away from the centre of the sphere; it is a diverging mirror and always gives a virtual, erect, diminished image.
✗ plano-convex mirror — 'plano-convex' describes a lens with one flat and one bulging face, not a spherical mirror of the kind described.
✗ plane mirror — a plane mirror is flat, so it has no centre of curvature and no focus; its focal length is taken as infinity.
📌 SSC & Railway Exam Facts
• Concave = converging (reflecting surface inward); Convex = diverging (reflecting surface outward).
• f = R/2; mirror formula 1/v + 1/u = 1/f; magnification m = −v/u.
• A convex mirror is used as a rear-view mirror because it gives a wide field of view and an erect image.
• A concave mirror is used in torches, solar cookers, shaving mirrors and reflecting telescopes.
• Only a concave mirror can give a real image; a convex mirror and a plane mirror always give virtual images.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 2 · Mirror formula and magnification

Q78. If the magnification of any mirror is +2.3, which type of mirror is it?

  1. (a) Planoconvex mirror
  2. (b) Plane mirror
  3. (c) Convex mirror
  4. (d) Concave mirror
Answer & solution

Answer: (d) Concave mirror

Correct Answer: Option 4 — Concave mirror
📖 Given
• Magnification, m = +2.3
• The positive sign means the image is virtual and erect
• The magnitude 2.3 > 1 means the image is enlarged
✍️ Formula
m = h′/h = −v/u
🔢 Solution
• m is positive ⇒ image is virtual and erect
• |m| = 2.3, which is greater than 1 ⇒ image is bigger than the object
• Only a concave mirror can give a virtual, erect and magnified image — it does so when the object is placed between the pole P and the focus F
• Hence the mirror is a concave mirror
This is exactly the arrangement used in a shaving mirror and a dentist's mirror, where a magnified upright image is needed.
❌ Why the other options are incorrect
✗ Planoconvex mirror — a plano-convex shape is a lens term; it is not one of the standard spherical mirrors studied here.
✗ Plane mirror — a plane mirror always gives m = +1, an image exactly the same size as the object, never 2.3.
✗ Convex mirror — it always forms a virtual, erect but diminished image, so 0 < m < 1; it can never give m = 2.3.
📌 SSC & Railway Exam Facts
• m = −v/u = h′/h; positive m → virtual, erect, negative m → real, inverted.
• Mirror formula: 1/v + 1/u = 1/f, and R = 2f.
• Concave mirror uses: torch, headlight, solar cooker, shaving mirror, dentist's mirror.
• Convex mirror is used as a rear-view mirror for its wide field of view.
• A concave mirror gives a real, inverted image for every object position beyond the focus F.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 2 · Convex mirror - image formation and uses

Q79. A convex mirror created a highly diminished, point-sized virtual image, then the position of the image is:

  1. (a) in front of the mirror at the centre of curvature C
  2. (b) behind the mirror at the focus F
  3. (c) behind the mirror at the centre of curvature C
  4. (d) in front of the mirror at the focus F
Answer & solution

Answer: (b) behind the mirror at the focus F

Correct Answer: Option 2 — behind the mirror at the focus F
A convex mirror always forms a virtual, erect and diminished image, and that image always lies behind the mirror, between the pole P and the focus F.
The image becomes highly diminished and point-sized only in one limiting case — when the object is placed at infinity. Rays coming from an object at infinity reach the convex mirror as a parallel beam; after reflection they diverge and appear to come from the principal focus F situated behind the mirror.
So the point-sized virtual image forms exactly at the focus F, behind the mirror. As the object is brought nearer from infinity, the image moves from F towards the pole P and grows a little in size, but it always stays virtual, erect and smaller than the object.
❌ Why the other options are incorrect
✗ in front of the mirror at the centre of curvature C — an image formed in front of a mirror is real; a convex mirror can never form a real image of a real object.
✗ behind the mirror at the centre of curvature C — the image of a convex mirror never goes beyond F; it stays in the region between P and F.
✗ in front of the mirror at the focus F — for a convex mirror the focus itself lies behind the mirror, not in front, and the image is virtual.
📌 SSC & Railway Exam Facts
• Convex mirror: image is always virtual, erect and diminished, formed between P and F.
• Object at infinity → image at F, point-sized; object anywhere else → image between P and F.
• For a convex mirror, f and R are taken positive by the sign convention, and f = R/2.
• Convex mirrors are used as rear-view mirrors and in shop security mirrors for a wide field of view.
• Concave mirror forms a real inverted image except when the object lies between P and F.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Concave mirror - image formation and uses

Q80. What is the position of the image formed by a concave mirror if the object is placed just beyond the centre of curvature of the mirror?

  1. (a) At the centre of curvature
  2. (b) At infinity
  3. (c) At the focus
  4. (d) Between the centre of curvature and the focus point
Answer & solution

Answer: (d) Between the centre of curvature and the focus point

Correct Answer: Option 4 — Between the centre of curvature and the focus point
For a concave mirror, the position of the image depends entirely on where the object is placed with respect to the pole P, the focus F and the centre of curvature C.
When the object is placed beyond C, the reflected rays meet at a point lying between F and C. The image so formed is real, inverted and diminished (smaller than the object).
This can be checked with the mirror formula 1/v + 1/u = 1/f. Taking f = −10 cm (so C is at −20 cm) and an object at u = −30 cm, we get v = −15 cm — a point lying between F (−10 cm) and C (−20 cm), exactly as stated.
❌ Why the other options are incorrect
✗ At the centre of curvature — the image forms at C only when the object itself is placed at C; then the image is real, inverted and of the same size.
✗ At infinity — the image goes to infinity only when the object is placed at the focus F; the reflected rays then emerge parallel.
✗ At the focus — the image forms at F only when the object is at infinity; the image is then highly diminished and point-sized.
📌 SSC & Railway Exam Facts
• Concave mirror ray table: object at ∞ → image at F; beyond C → between F and C; at C → at C; between C and F → beyond C; at F → at ∞; between P and F → behind the mirror, virtual and enlarged.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u; R = 2f.
• A concave mirror forms a virtual, erect, enlarged image only when the object lies between P and F.
• Concave mirrors are used in torches, headlights, solar cookers, shaving mirrors and dentists' mirrors.
• By the sign convention, distances measured against the incident light are negative.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Convex mirror - image formation and uses

Q81. In the rear-view mirror of cars, people use which of the following?

  1. (a) Plano-convex lens
  2. (b) Biconvex lens
  3. (c) Concave mirror
  4. (d) Convex mirror
Answer & solution

Answer: (d) Convex mirror

Correct Answer: Option 4 — Convex mirror
A convex mirror is used as the rear-view (wing) mirror in vehicles. Its reflecting surface bulges outwards, so it always forms an image that is virtual, erect and diminished, formed between the pole P and the focus F behind the mirror.
Two properties make it ideal for this job:
• It gives an erect image, so the driver sees the traffic behind the right way up.
• It has a much wider field of view than a plane mirror of the same size, because the diminished image lets a large region behind fit into a small mirror.
The only drawback is that objects look smaller and therefore farther away than they really are — which is why such mirrors carry the warning "objects in the mirror are closer than they appear".
❌ Why the other options are incorrect
✗ Plano-convex lens — a lens refracts light and is used in optical instruments; a rear-view mirror must reflect light, not transmit it.
✗ Biconvex lens — again a converging lens, used in magnifying glasses, cameras and to correct hypermetropia, not as a mirror.
✗ Concave mirror — for distant objects it gives a real, inverted image and has a narrow field of view, so it would show the traffic upside down.
📌 SSC & Railway Exam Facts
• Convex mirror: image is always virtual, erect and diminished; used in rear-view mirrors and shop security mirrors.
• Concave mirror: used in torches, headlights, solar cookers, shaving mirrors and dentists' mirrors.
• Plane mirror gives a virtual, erect, same-sized, laterally inverted image.
• Mirror formula: 1/v + 1/u = 1/f; m = −v/u; f = R/2.
• For a convex mirror, f is positive; for a concave mirror, f is negative.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Spherical mirror terminology

Q82. The diameter of the reflecting surface of a spherical mirror is same as its:

  1. (a) aperture
  2. (b) inverse of the power
  3. (c) radius of curvature
  4. (d) focal length
Answer & solution

Answer: (a) aperture

Correct Answer: Option 1 — aperture
The aperture of a spherical mirror is the effective diameter of its reflecting surface — that is, the width of the portion of the mirror from which reflection actually takes place.
The aperture decides how much light the mirror can collect. A mirror with a large aperture gathers more light and forms a brighter image, which is why the mirrors of reflecting telescopes are made very wide.
In all the standard mirror formula derivations the mirror is assumed to have a small aperture, so that the relation f = R/2 holds and spherical aberration stays negligible.
❌ Why the other options are incorrect
✗ inverse of the power — the inverse of power is the focal length, f = 1/P; it has nothing to do with the diameter.
✗ radius of curvature — R is the radius of the hollow sphere of which the mirror is a part, measured from the centre of curvature to the pole, not across the rim.
✗ focal length — f is the distance from the pole to the principal focus, and for a spherical mirror f = R/2.
📌 SSC & Railway Exam Facts
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h′/h.
• For a spherical mirror, R = 2f.
• Concave mirror — used in torches, headlights, shaving mirrors, solar cookers and dentists' mirrors.
• Convex mirror — used as a rear-view mirror because it gives an erect, diminished image and a wide field of view.
• The pole (P) is the centre of the reflecting surface and lies on the mirror itself.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Concave mirror - image formation and uses

Q83. A concave mirror has a radius of curvature of 4.00 m. If any object is located at 2.00 m from this mirror, find the position of the image.

  1. (a) Zero
  2. (b) Infinity
  3. (c) Between pole and focus point
  4. (d) At centre of curvature
Answer & solution

Answer: (b) Infinity

Correct Answer: Option 2 — Infinity
📖 Given
• Radius of curvature, R = 4.00 m
• Object distance from the mirror, u = 2.00 m
• Mirror is concave
✍️ Formula
f = R/2 and 1/v + 1/u = 1/f
🔢 Solution
• f = R/2 = 4.00/2 = 2.00 m
• So the object distance u = 2.00 m is exactly equal to the focal length f — the object is placed at the principal focus F.
• Using the mirror formula: 1/v = 1/f − 1/u = 1/2.00 − 1/2.00
• 1/v = 0
• v = infinity
When an object is kept at the focus of a concave mirror, all the reflected rays become parallel to one another, so they meet only at infinity. The image is real, inverted and highly enlarged.
❌ Why the other options are incorrect
✗ Zero — v = 0 would mean the image is formed on the mirror itself, which happens for no ordinary object position.
✗ Between pole and focus point — an image in the P–F region belongs to a convex mirror, which always forms a virtual, diminished image between its pole and focus. A concave mirror never forms a real image there.
✗ At centre of curvature — the image forms at C only when the object is also placed at C (u = R), here that would need u = 4.00 m.
📌 SSC & Railway Exam Facts
• For a spherical mirror, R = 2f.
• Object at F → image at infinity; object at infinity → image at F.
• This is the working principle of a torch, headlight and searchlight, where the bulb is placed at the focus.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u.
• A concave mirror is used in solar cookers, shaving mirrors and dentists' mirrors.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 3 · Mirror formula and magnification

Q84. A positive and negative sign in the magnification indicates that the images are:

  1. (a) real and virtual, respectively
  2. (b) virtual and real, respectively
  3. (c) both real in nature
  4. (d) both virtual in nature
Answer & solution

Answer: (b) virtual and real, respectively

Correct Answer: Option 2 — virtual and real, respectively
The magnification of a mirror is m = h′/h = −v/u, where h′ is the image height and h the object height. The sign of m carries the information about the nature of the image:
• Positive magnification (m > 0) — the image is erect, and an erect image formed by a mirror or lens is always virtual.
• Negative magnification (m < 0) — the image is inverted, and an inverted image is always real.
So a positive sign means a virtual image and a negative sign means a real image — in that order. A plane mirror always gives m = +1 (virtual, erect, same size), while a concave mirror with the object beyond F gives a negative m (real, inverted).
❌ Why the other options are incorrect
✗ real and virtual, respectively — this reverses the rule; positive goes with virtual, not with real.
✗ both real in nature — then every image would be inverted, which is false for a plane mirror and a convex mirror.
✗ both virtual in nature — then no image could ever be caught on a screen, but real images certainly can be.
📌 SSC & Railway Exam Facts
• m = h′/h = −v/u for mirrors; m = h′/h = v/u for lenses.
• |m| > 1 → enlarged, |m| < 1 → diminished, |m| = 1 → same size.
• A convex mirror always forms a virtual, erect and diminished image, so m is positive and less than 1.
• A real image can be taken on a screen; a virtual image cannot.
• Mirror formula: 1/v + 1/u = 1/f, with sign conventions measured from the pole.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Concave mirror - image formation and uses

Q85. The image formed by a concave mirror is real, inverted and with the same size as the object. The position of the object is kept:

  1. (a) at the focus
  2. (b) far away from the centre of curvature
  3. (c) at the centre of curvature
  4. (d) between the focus and the centre of curvature
Answer & solution

Answer: (c) at the centre of curvature

Correct Answer: Option 3 — at the centre of curvature
For a concave mirror, when the object is placed exactly at the centre of curvature (C), the image is also formed at C and it is real, inverted and exactly the same size as the object. The magnification is then m = −1, where the minus sign shows that the image is inverted and the magnitude 1 shows that the sizes are equal.
This can be checked with the mirror formula. Taking u = −R and f = −R/2, we get 1/v = 1/f − 1/u = −2/R + 1/R = −1/R, so v = −R and m = −v/u = −1. It is the only object position for a concave mirror that gives an image of the same size.
❌ Why the other options are incorrect
✗ at the focus — an object at F gives an image at infinity, highly enlarged; this is the arrangement used in a torch or a searchlight, where a parallel beam is wanted.
✗ far away from the centre of curvature — that is beyond C; the image then forms between F and C and is real, inverted but diminished, not the same size.
✗ between the focus and the centre of curvature — the image then forms beyond C and is real, inverted and enlarged; this is the arrangement used in a projector.
📌 SSC & Railway Exam Facts
• Object at C → image at C, real, inverted, same size, m = −1.
• Object at infinity → image at F, point-sized; object at F → image at infinity.
• Object between P and F → image virtual, erect and enlarged, formed behind the mirror.
• Mirror formula 1/v + 1/u = 1/f; f = R/2; magnification m = −v/u = h'/h.
• Concave mirrors are used in torches, headlights, solar cookers, shaving mirrors and by dentists.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Concave mirror - image formation and uses

Q86. When you look at the front side of a table spoon, an inverted image of yours is formed. The front side of table spoon works as a:

  1. (a) plane mirror
  2. (b) concave lens
  3. (c) convex mirror
  4. (d) concave mirror
Answer & solution

Answer: (d) concave mirror

Correct Answer: Option 4 — concave mirror
The front (inner) surface of a spoon curves inwards, exactly like a concave mirror. A concave mirror is a converging mirror: when you hold it at a distance greater than its focal length, it forms a real and inverted image of your face. That is why your reflection in the bowl of a spoon appears upside down.
If you hold the spoon very close to your face — nearer than the focus — the image becomes virtual, erect and enlarged, which is the same principle used in a shaving mirror. The back (outer) surface of the same spoon bulges outwards and behaves as a convex mirror, always giving an erect, diminished image.
❌ Why the other options are incorrect
✗ plane mirror — a plane mirror always forms a virtual, erect image of the same size; it can never give an inverted image.
✗ convex mirror — a convex mirror always forms a virtual, erect and diminished image; that is the back of the spoon, not the front.
✗ concave lens — a spoon is opaque and works by reflection, not by refraction, so it cannot act as a lens; besides, a concave lens gives an erect image.
📌 SSC & Railway Exam Facts
• Inner surface of a spoon = concave mirror; outer surface of a spoon = convex mirror.
• A concave mirror gives a real, inverted image when the object is beyond F, and a virtual, erect, enlarged image when it is between P and F.
• A convex mirror gives a virtual, erect and diminished image for every object position.
• A plane mirror gives a virtual, erect, same-size image with lateral inversion, formed as far behind the mirror as the object is in front.
• f = R/2; mirror formula 1/v + 1/u = 1/f.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Spherical mirror terminology

Q87. Which of the following statements about the principal focus of a spherical mirror is/are correct?
(a) It always lies midway between the pole and the centre of curvature.
(b) Its position depends on the size of the mirror.
(c) It always lies at the centre of curvature.

  1. (a) (a) and (b)
  2. (b) Only (c)
  3. (c) (b) and (c)
  4. (d) Only (a)
Answer & solution

Answer: (d) Only (a)

Correct Answer: Option 4 — Only (a)
For a spherical mirror of small aperture the focal length is half the radius of curvature: f = R/2. Since the pole P is at one end and the centre of curvature C at a distance R, the principal focus F lies exactly midway between P and C. Hence statement (a) is correct.
Statement (b) is wrong: the focal length depends only on the radius of curvature (the curvature of the surface), not on how large or small a piece of the sphere the mirror is. Two mirrors cut from the same sphere have the same focal length whatever their sizes. Statement (c) is also wrong, because F lies at R/2, i.e. half way to C, and never at C itself.
❌ Why the other options are incorrect
✗ (a) and (b) — (a) is right, but (b) is wrong; the size (aperture) of a mirror does not shift its principal focus.
✗ Only (c) — (c) is false; the focus lies at R/2, and the point at distance R is the centre of curvature, not the focus.
✗ (b) and (c) — both of these statements are wrong, so this combination cannot be correct.
📌 SSC & Railway Exam Facts
• f = R/2, so R = 2f; the principal focus lies midway between the pole and the centre of curvature.
• Pole (P) — centre of the reflecting surface; aperture — effective diameter of the mirror.
• For a concave mirror F is in front of the mirror (real focus); for a convex mirror F is behind it (virtual focus).
• Mirror formula 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
• Laws of reflection: angle of incidence = angle of reflection, and the incident ray, reflected ray and normal all lie in the same plane.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Spherical mirror terminology

Q88. Suppose a number of rays parallel to the principal axis are falling on a concave mirror, then all the reflected rays will intersect:

  1. (a) in between the pole and infinity
  2. (b) at infinity
  3. (c) at the pole
  4. (d) at the principal focus
Answer & solution

Answer: (d) at the principal focus

Correct Answer: Option 4 — at the principal focus
A concave mirror is a converging mirror. When a beam of rays parallel to the principal axis strikes it, every ray obeys the laws of reflection and, after reflection, they all pass through one single point on the principal axis. That point is called the principal focus (F) of the mirror.
The distance of this point from the pole (P) of the mirror is the focal length (f), and for a spherical mirror it is half the radius of curvature: f = R/2. Because the reflected rays actually meet there, the focus of a concave mirror is a real focus — a screen placed at F shows a bright point. This is why concave mirrors are used in solar cookers, torches, headlights and shaving mirrors.
❌ Why the other options are incorrect
✗ in between the pole and infinity — too vague; the rays meet at one exact point, the focus, which lies midway between the pole and the centre of curvature.
✗ at infinity — the rays are coming from infinity; a converging mirror brings them to a finite point, it does not send them back to infinity.
✗ at the pole — the pole is the centre of the mirror surface itself; a ray hitting the pole is reflected making equal angles with the principal axis.
📌 SSC & Railway Exam Facts
• f = R/2 for a spherical mirror; mirror formula 1/v + 1/u = 1/f.
• Concave mirror — real focus, converging; convex mirror — virtual focus, diverging.
• Magnification of a mirror, m = −v/u = h′/h.
• Convex mirrors are used as rear-view mirrors because they give an erect, diminished image and a wide field of view.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Spherical mirror terminology

Q89. Which of the following is true regarding the principal axis of a spherical mirror?

  1. (a) It is parallel to the surface of the mirror.
  2. (b) It is normal to the mirror at its pole.
  3. (c) It is located at the edge of the mirror.
  4. (d) It intersects the surface of the mirror at an angle of 45 degrees.
Answer & solution

Answer: (b) It is normal to the mirror at its pole.

Correct Answer: Option 2 — It is normal to the mirror at its pole.
The principal axis of a spherical mirror is the straight line joining the pole (P) and the centre of curvature (C) of the mirror. Since the centre of curvature is the centre of the sphere from which the mirror was cut, this line is a radius of that sphere — and a radius always meets the spherical surface at right angles. Hence the principal axis is normal (perpendicular) to the mirror at its pole.
This is why a ray travelling along the principal axis hits the pole head-on, with angle of incidence 0°, and is reflected straight back along the same path. The whole sign convention of mirrors is measured along this axis, with the pole as the origin.
❌ Why the other options are incorrect
✗ It is parallel to the surface of the mirror. — the opposite is true; it is perpendicular to the mirror surface at the pole.
✗ It is located at the edge of the mirror. — the principal axis passes through the middle of the mirror, i.e. the pole, not through its rim.
✗ It intersects the surface of the mirror at an angle of 45 degrees. — the angle is 90°, not 45°, because a radius is always normal to a sphere.
📌 SSC & Railway Exam Facts
• Pole (P) — the geometric centre of the reflecting surface; Centre of curvature (C) — centre of the sphere.
• Aperture — the diameter of the reflecting surface; f = R/2.
• A ray through C retraces its path; a ray through F becomes parallel to the principal axis.
• Laws of reflection: ∠i = ∠r, and the incident ray, reflected ray and normal lie in the same plane.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Spherical mirror terminology

Q90. The aperture of a spherical mirror is:

  1. (a) the diameter of the reflecting surface of the mirror
  2. (b) the distance between the centre of curvature and the focus
  3. (c) the distance between the object and the mirror
  4. (d) the distance between the image and the mirror
Answer & solution

Answer: (a) the diameter of the reflecting surface of the mirror

Correct Answer: Option 1 — the diameter of the reflecting surface of the mirror
A spherical mirror is a small piece cut out of a hollow sphere and silvered on one side. The aperture of such a mirror is the effective diameter of its reflecting surface — that is, the width of the light-collecting face measured across the rim. A mirror with a large aperture gathers more light and gives a brighter image, which is why the mirrors of reflecting telescopes are made as wide as possible.
In all the standard mirror formulas we assume a small aperture compared with the radius of curvature. This keeps the reflected rays close to the principal axis (paraxial rays) and lets us use f = R/2 and 1/v + 1/u = 1/f without the blurring called spherical aberration.
❌ Why the other options are incorrect
✗ the distance between the centre of curvature and the focus — this distance is equal to the focal length f itself (since PF = FC = R/2); it is not the aperture.
✗ the distance between the object and the mirror — that is the object distance u, measured from the pole, which changes every time the object is moved.
✗ the distance between the image and the mirror — that is the image distance v; it is found from the mirror formula and is not a fixed property of the mirror.
📌 SSC & Railway Exam Facts
• Pole (P) — centre of the reflecting surface; Centre of curvature (C) — centre of the sphere; Radius of curvature R = PC.
• f = R/2; mirror formula 1/v + 1/u = 1/f; magnification m = −v/u.
• A large aperture gives a brighter but more aberrated image.
• Concave mirror — converging, used in solar cookers, headlights, shaving mirrors.
• Convex mirror — diverging, used as a rear-view mirror for its wide field of view.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 2 · Numericals on mirrors

Q91. The focal length of a spherical mirror is 12 cm, then the radius of curvature of this mirror will be:

  1. (a) 3cm
  2. (b) 6cm
  3. (c) 2cm
  4. (d) 24cm
Answer & solution

Answer: (d) 24cm

Correct Answer: Option 4 — 24cm
📖 Given
• Focal length of the spherical mirror, f = 12 cm
• To find: radius of curvature, R
✍️ Formula
R = 2f (equivalently f = R/2)
🔢 Solution
• R = 2 × f
• R = 2 × 12
• R = 24 cm
So the mirror is a part of a sphere of radius 24 cm, and its focus lies exactly halfway between the pole and the centre of curvature.
❌ Why the other options are incorrect
✗ 6cm — this is f/2; the focal length has been halved instead of doubled.
✗ 3cm — this is f/4, obtained by dividing by 2 twice.
✗ 2cm — this is just the number 2 from the formula; it has no connection with 12 cm.
📌 SSC & Railway Exam Facts
• For a spherical mirror, f = R/2, i.e. the focal length is half the radius of curvature.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
• A concave mirror has a negative focal length and a convex mirror a positive one in the sign convention.
• Concave mirrors are used in torches, headlights, shaving mirrors and solar cookers.
• Convex mirrors are used as rear-view mirrors because they give an erect, diminished image and a wide field of view.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 2 · Spherical mirror terminology

Q92. Which of the following points always lie(s) on the principal axis of a spherical mirror?
i)Principal Focus
ii)Point of incidence
iii)Pole

  1. (a) Both (ii) and (iii)
  2. (b) Only (iii)
  3. (c) Both (i) and (iii)
  4. (d) Only (i)
Answer & solution

Answer: (c) Both (i) and (iii)

Correct Answer: Option 3 — Both (i) and (iii)
The principal axis of a spherical mirror is the straight line joining the pole (P) of the mirror and its centre of curvature (C). By this very definition the pole always lies on the principal axis, so statement (iii) is true.
The principal focus (F) is the point on the principal axis where rays coming parallel to the principal axis meet (concave) or appear to come from (convex) after reflection. It lies midway between P and C, so (i) is also true. The point of incidence, however, is wherever a particular ray happens to strike the mirror surface — it can be anywhere on the reflecting surface and normally is not on the principal axis, so (ii) is false.
❌ Why the other options are incorrect
✗ Both (ii) and (iii) — it wrongly includes the point of incidence, which changes with every incident ray.
✗ Only (iii) — the pole does lie on the axis, but this option leaves out the principal focus, which also always lies on it.
✗ Only (i) — the principal focus lies on the axis, but the pole, being the centre of the mirror surface, always lies on it too.
📌 SSC & Railway Exam Facts
• Pole (P) — the centre of the reflecting surface; Centre of curvature (C) — the centre of the sphere of which the mirror is a part.
• Aperture is the effective diameter of the reflecting surface; f = R/2.
• Laws of reflection: angle of incidence = angle of reflection, and the incident ray, reflected ray and normal lie in the same plane.
• For a spherical mirror the normal at any point passes through the centre of curvature.
• A concave mirror placed with the object between P and F gives a virtual, erect and enlarged image.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Convex mirror - image formation and uses

Q93. A spherical mirror diverges a beam of light, which is incident on it. Which of the following statement(s) about the mirror is/are true?
i)The mirror used is convex.
ii)The mirror has positive focal length.
iii) The image formed by this mirror can be seen on a screen.

  1. (a) Only (i)
  2. (b) Only(iii)
  3. (c) Both (ii) and (iii)
  4. (d) Both (i) and (ii)
Answer & solution

Answer: (d) Both (i) and (ii)

Correct Answer: Option 4 — Both (i) and (ii)
A mirror that spreads out (diverges) a beam of light falling on it is a convex mirror. Its reflecting surface bulges outwards, so rays parallel to the principal axis, after reflection, appear to come from a focus behind the mirror. Hence statement (i) is true — a convex mirror is also called a diverging mirror.
In the New Cartesian sign convention the focus of a convex mirror lies behind the mirror, on the same side as the outgoing light, so its focal length is positive — statement (ii) is true. A convex mirror always forms a virtual, erect and diminished image, and a virtual image can never be caught on a screen, so statement (iii) is false.
❌ Why the other options are incorrect
✗ Only (i) — (i) is indeed true, but this option leaves out (ii); a convex mirror also has a positive focal length.
✗ Only(iii) — (iii) is the one statement that is false, because the image formed by a convex mirror is virtual.
✗ Both (ii) and (iii) — it wrongly includes the false statement (iii) and drops (i), the statement that the mirror is convex.
📌 SSC & Railway Exam Facts
• Convex mirror = diverging mirror; concave mirror = converging mirror.
• Sign convention: f is positive for a convex mirror and negative for a concave mirror.
• Convex mirrors are used as rear-view mirrors — erect, diminished image and a wide field of view.
• Real images can be taken on a screen; virtual images cannot.
• Mirror formula 1/v + 1/u = 1/f, magnification m = −v/u, and f = R/2.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Concave mirror - image formation and uses

Q94. Rays of the Sun converge at a point of 30 cm in front of a concave mirror. Where should the object be placed so that the size of the image is equal to the size of the object?

  1. (a) Between 30 and 60 cm in front of the mirror
  2. (b) More than 30 cm in front of the mirror
  3. (c) 30 cm in front of the mirror
  4. (d) 60 cm in front of the mirror
Answer & solution

Answer: (d) 60 cm in front of the mirror

Correct Answer: Option 4 — 60 cm in front of the mirror
📖 Given
• Rays from the Sun are effectively parallel to the principal axis and meet 30 cm in front of the mirror
• The point where parallel rays meet is the principal focus, so f = 30 cm
• Required: the object position for which image size = object size
✍️ Formula
R = 2f; for a concave mirror an object placed at the centre of curvature (C) gives an image of the same size (magnification m = −1).
🔢 Solution
• f = 30 cm
• R = 2 × f = 2 × 30
• R = 60 cm
• So the object must be placed at C, i.e. 60 cm in front of the mirror
• Check: u = −60 cm, f = −30 cm → 1/v = 1/f − 1/u = −1/30 + 1/60 = −1/60, so v = −60 cm and m = −v/u = −1
The image is formed at C itself — real, inverted and exactly the same size as the object.
❌ Why the other options are incorrect
✗ 30 cm in front of the mirror — this is the focus F; an object placed at F gives an image at infinity, highly enlarged, not of equal size.
✗ Between 30 and 60 cm in front of the mirror — that is between F and C; the image then forms beyond C and is enlarged, not equal.
✗ More than 30 cm in front of the mirror — too vague; beyond C (more than 60 cm) the image is diminished, so only the exact distance of 60 cm works.
📌 SSC & Railway Exam Facts
• Sun's rays are parallel, so they converge exactly at the principal focus of a concave mirror; f = R/2.
• Object at C → image at C, real, inverted, same size (m = −1).
• Object beyond C → image between F and C, diminished; object between P and F → virtual, erect, enlarged.
• Mirror formula 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
• Concave mirrors are used in solar cookers, torches, headlights and shaving mirrors.
RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Spherical mirror terminology

Q95. Which of the following statements is/are NOT true about the principal focus of a spherical mirror?
i) It always lies on the principal axis.
ii) An incident ray passing through the focus after reflection becomes parallel to the principal axis.
iii) Principal focus is a point where all the incident rays after reflection from the mirror meet.

  1. (a) Only (iii)
  2. (b) Only (ii) and (iii)
  3. (c) Only (i)
  4. (d) Only (i) and (ii)
Answer & solution

Answer: (a) Only (iii)

Correct Answer: Option 1 — Only (iii)
The principal focus (F) of a spherical mirror is the point on the principal axis at which rays parallel to the principal axis converge after reflection (concave mirror) or from which they appear to diverge (convex mirror). The crucial words are "rays parallel to the principal axis".
Statement (i) is true — the principal focus always lies on the principal axis, midway between the pole P and the centre of curvature C, so f = R/2.
Statement (ii) is true — this is one of the standard ray-diagram rules: a ray that passes through the focus before striking the mirror comes back parallel to the principal axis (the reverse of the defining property).
Statement (iii) is NOT true — only rays parallel to the principal axis meet at the focus, not all incident rays. Rays coming from any other direction meet somewhere else, and parallel rays inclined to the axis meet on the focal plane, not at F. Hence only (iii) is untrue.
❌ Why the other options are incorrect
✗ Only (ii) and (iii) — statement (ii) is a correct ray-diagram rule, so it cannot be included.
✗ Only (i) — statement (i) is true; the focus is by definition a point on the principal axis.
✗ Only (i) and (ii) — both of these statements are true, so neither of them qualifies as "NOT true".
📌 SSC & Railway Exam Facts
• Focal length f = R/2, where R is the radius of curvature.
• A concave mirror has a real focus; a convex mirror has a virtual focus.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u = h'/h.
• Concave mirrors are used as shaving mirrors, dentist's mirrors, headlight reflectors and solar concentrators.
• Convex mirrors are used as rear-view mirrors because they give an erect, diminished image and a wider field of view.
RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Concave mirror - image formation and uses

Q96. If the position of an object is in between the principal focus and pole of a concave mirror, then the image formed will be:

  1. (a) real and diminished
  2. (b) virtual and diminished
  3. (c) real and enlarged
  4. (d) virtual, erect and enlarged
Answer & solution

Answer: (d) virtual, erect and enlarged

Correct Answer: Option 4 — virtual, erect and enlarged
For a concave mirror, when the object is placed between the pole (P) and the principal focus (F), the reflected rays diverge after reflection and never actually meet in front of the mirror. They only appear to come from a point behind the mirror, so the image formed is:
virtual, erect (upright) and enlarged (magnified), and it lies behind the mirror.
This is exactly why a concave mirror is used as a shaving mirror and as a dentist's mirror — the face or the tooth is kept closer to the mirror than its focus, giving a large, upright image.
This is the only position of an object in front of a concave mirror that gives a virtual image; for every other position (at F, between F and C, at C, beyond C) the image is real and inverted.
❌ Why the other options are incorrect
✗ real and diminished — a real, diminished image is formed when the object is placed beyond the centre of curvature C.
✗ virtual and diminished — a virtual and diminished image is the signature of a convex mirror, not a concave one.
✗ real and enlarged — a real, enlarged image is formed when the object lies between F and C, not between P and F.
📌 SSC & Railway Exam Facts
• Concave mirror image positions: object at infinity → at F, real, highly diminished; beyond C → between F and C, real, diminished; at C → at C, real, same size; between C and F → beyond C, real, enlarged; at F → at infinity; between P and F → behind mirror, virtual, erect, enlarged.
• Mirror formula: 1/v + 1/u = 1/f; magnification m = −v/u.
• m is positive for a virtual, erect image and negative for a real, inverted image; |m| > 1 means enlarged.
• A convex mirror always gives a virtual, erect and diminished image, whatever the object's position.
• f = R/2, so a mirror with R = 20 cm has a focal length of 10 cm.
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 2 · Concave mirror - image formation and uses

Q97. A concave mirror can form a real, inverted and magnified image when the object is placed ____________.

  1. (a) beyond the centre of curvature
  2. (b) between the focus and the centre of curvature
  3. (c) at the pole
  4. (d) between the pole and the focus
Answer & solution

Answer: (b) between the focus and the centre of curvature

Answer: Option 2 — between the focus and the centre of curvature
With the object between F and C, the image forms beyond C.
It is real, inverted and magnified — the only position that gives all three.
Revision note — Images formed by a concave mirror
• Object at infinity → image at F, real, inverted, highly diminished (point-sized).
• Object beyond C → image between F and C, real, inverted, diminished.
• Object at C → image at C, real, inverted, same size.
• Object between F and C → image beyond C, real, inverted, enlarged.
• Object at F → image at infinity, real, inverted, highly enlarged.
• Object between P and F → image behind the mirror, virtual, erect, enlarged (shaving / dentist's mirror).
Do not mix these up
• Between F and C vs between P and F — both give a magnified image.
• Only F to C gives a real, inverted one; P to F gives a virtual, erect one.
Also asked from this topic: the image when the object is between the pole and focus (virtual and upright); where to place the face in a shaving mirror (between the pole and the focus).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 3 · Numericals on mirrors

Q98. An object is placed at 30 cm from a convex mirror with a focal length of 15 cm. Calculate the image distance (v) using the mirror formula.

  1. (a) 10 cm
  2. (b) −10 cm
  3. (c) 15 cm
  4. (d) −15 cm
Answer & solution

Answer: (a) 10 cm

Mirror formula.
1/v + 1/u = 1/f
Signs (New Cartesian convention).
Object is in front of the mirror ⇒ u = −30 cm
Focus of a convex mirror is behind it ⇒ f = +15 cm
Find v.
1/v = 1/f − 1/u
1/v = 1/15 − 1/(−30)
1/v = 1/15 + 1/30
1/v = 2/30 + 1/30 = 3/30 = 1/10
v = +10 cm
The + sign means the image is behind the mirror — virtual and erect.
Answer: 10 cm
Revision note — Mirror formula and sign convention
• Mirror formula: 1/v + 1/u = 1/f; for a spherical mirror f = R/2.
• Object distance u is always negative (object in front of the mirror).
• Concave mirror: f negative; convex mirror: f positive.
• Positive v ⇒ image behind the mirror (virtual); negative v ⇒ image in front (real).
• Magnification m = −v/u; here m = −10/(−30) = +1/3 — erect and diminished.
Also asked from this topic: a convex mirror of R = 2 m with a car 3 m away (image 0.75 m behind the mirror); where a parallel beam appears to diverge from a convex mirror of R = 50 cm (25 cm behind the mirror).
Option 2 (−10 cm) comes from taking f as −15 cm — a convex mirror's focal length is positive.
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 1 · Mirror formula and magnification

Q99. If an image is formed by a spherical mirror with a magnification of +1, which of the following statements is true?

  1. (a) The image is virtual and equal in size to the object.
  2. (b) The image is real and equal in size to the object.
  3. (c) The image is real and twice the size of the object.
  4. (d) The image is virtual and half the size of the object.
Answer & solution

Answer: (a) The image is virtual and equal in size to the object.

Answer: Option 1 — The image is virtual and equal in size to the object.
Positive magnification means the image is erect, and an erect image from a single mirror is always virtual.
Magnitude 1 means the image is the same size as the object.
Revision note — Magnification by a spherical mirror
• m = height of image ÷ height of object = −v/u.
• m positive ⇒ virtual and erect; m negative ⇒ real and inverted.
• |m| > 1 enlarged, |m| = 1 same size, |m| < 1 diminished.
• A convex mirror always gives m positive and less than 1.
• A concave mirror gives m = −1 with the object at C, and m positive and greater than 1 with the object between P and F.
• A plane mirror always gives m = +1.
Do not mix these up
• m = +1 vs m = −1 — +1 is virtual and erect (plane mirror); −1 is real and inverted (concave mirror, object at C).
Also asked from this topic: the sign and size of magnification for an inverted, smaller image (negative, less than one); the magnification of a concave mirror of f = 10 cm with the object at 15 cm (−2).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 1 · Numericals on mirrors

Q100. An object is placed 12 cm in front of a convex mirror with a focal length of 8 cm. What is the image distance (v)?

  1. (a) +20 cm
  2. (b) −20 cm
  3. (c) +4.8 cm
  4. (d) −4.8 cm
Answer & solution

Answer: (c) +4.8 cm

Mirror formula.
1/v + 1/u = 1/f
So 1/v = 1/f − 1/u
Signs (New Cartesian convention).
Object in front of the mirror ⇒ u = −12 cm
Convex mirror ⇒ focus behind ⇒ f = +8 cm
Substitute.
1/v = 1/8 − 1/(−12)
= 1/8 + 1/12
= 3/24 + 2/24
= 5/24
v = 24/5 = +4.8 cm
Answer: +4.8 cm
Revision note — Convex mirror
• A convex mirror always gives a virtual, erect, diminished image behind the mirror, between P and F.
• f and R of a convex mirror are positive; R = 2f.
• Magnification here: m = −v/u = −4.8/(−12) = +0.4 (erect, smaller).
• Used as rear-view mirrors in vehicles — gives a wider field of view.
Also asked from this topic: image of a car 3 m from a convex mirror of radius 2 m (0.75 m behind the mirror); focus of a convex mirror of radius 50 cm (25 cm behind the mirror).
A positive v for a convex mirror just means the image is behind it — −4.8 cm comes from forgetting that f is positive.
12

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