Home › GK Questions Asked › Physics › Light - Refraction and Lenses

Light - Refraction and Lenses PYQs – Physics, SSC & Railway

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

📚 103 questionsSSC 3 · Railway 100
12
RRB Group D 2025-26 – 21 August 2026 Shift 1 · Numericals on refraction and lenses

Q1. A photographer uses a concave lens of focal length 12 cm. The object is placed 18 cm from the lens. Using the sign convention, calculate the image distance (v)?

  1. (a) -7.2 cm
  2. (b) +7.2 cm
  3. (c) +36 cm
  4. (d) -36 cm
Answer & solution

Answer: (a) -7.2 cm

Explanation:
For a concave (diverging) lens, focal length is negative: f = -12 cm. Object distance u = -18 cm (object on the same side as incoming light). Using the lens formula 1/v - 1/u = 1/f:
1/v = 1/f + 1/u = -1/12 - 1/18 = -(3+2)/36 = -5/36, so v = -36/5 = -7.2 cm.
Correct Answer: Option 1 – -7.2 cm
RRB Group D 2025-26 – 20 August 2026 Shift 1 · Concave lens - image formation and uses

Q2. For any object position, what can be said about the size of the image formed by a concave lens?

  1. (a) It is always magnified compared to the object
  2. (b) It is always diminished compared to the object
  3. (c) It is sometimes enlarged and sometimes diminished
  4. (d) It is always the same size as the object
Answer & solution

Answer: (b) It is always diminished compared to the object

Correct Answer:
Correct Answer: Option 2 – It is always diminished compared to the object
Explanation:
Aconcave (diverging) lensalways forms a virtual, erect, anddiminished(smaller) image of the object, regardless of the object's position — the image size decreases as the object moves farther away, but it never becomes magnified or equal to the object size.
Important Facts:
• Concave lens image is always formed between the optical centre and focus, on the same side as the object.
• This is unlike a convex lens, where image size and nature vary with object position.
RRB Group D 2025-26 – 19 August 2026 Shift 2 · Convex lens - image formation and uses

Q3. A real, inverted and diminished image of an object is formed by a convex lens. At which position is the object placed?

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

Answer: (a) Beyond 2F

Correct Answer:
Correct Answer: Option 1 – Beyond 2F
Explanation:
For a convex lens, areal, inverted, and diminishedimage is formed only when the object is placedbeyond 2F(farther than twice the focal length). The image then forms between F and 2F on the other side, smaller than the object.
Important Facts:
• Object at 2F → image at 2F, real, inverted, same size.
• Object between F and 2F → image beyond 2F, real, inverted, magnified.
• Object at F → image at infinity.
• Object between F and lens → image is virtual, erect, magnified.
RRB Group D 2025-26 – 19 August 2026 Shift 3 · Convex lens - image formation and uses

Q4. A convex lens is used to form an image of an object. When the object is placed at a distance greater than twice of the focal length (2f) of the lens, what is the nature of the image formed?

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

Answer: (d) Real, inverted and diminished

Correct Answer:
Option 4 – Real, inverted and diminished
Explanation:
For a convex lens, when the object is placedbeyond 2F(at a distance greater than twice the focal length), the image forms between F and 2F on the other side of the lens. This image isreal, inverted, and smaller (diminished)than the object.
Important Facts:
• Object at 2F → Image at 2F, real, inverted, same size.
• Object between F and 2F → Image beyond 2F, real, inverted, magnified.
• Object at F → Image at infinity.
• Object between F and lens → Image virtual, erect, magnified.
Correct Answer: Option 4 – Real, inverted and diminished
RRB Group D 2025-26 – 18 August 2026 Shift 1 · Numericals on refraction and lenses

Q5. An object is placed 15 cm from a concave lens of focal length 10 cm. What is the nature and position of the image formed by the lens?

  1. (a) The image is real, upright and 8 cm from the lens on the same side as the object
  2. (b) The image is virtual, upright and 6 cm from the lens on the same side as the object
  3. (c) The image is virtual, inverted and 12 cm from the lens on the opposite side to the object
  4. (d) The image is real, inverted and 20 cm from the lens on the opposite side to the object
Answer & solution

Answer: (b) The image is virtual, upright and 6 cm from the lens on the same side as the object

Image formed by a concave (diverging) lens
• Concave lens: focal length f = -10 cm (sign convention), object distance u = -15 cm
• Lens formula: 1/v - 1/u = 1/f
• 1/v = 1/f + 1/u = -1/10 + (-1/15) = -3/30 - 2/30 = -5/30 = -1/6
• v = -6 cm
A concave lens always forms a virtual, erect, and diminished image on the same side as the object. Here the image forms 6 cm from the lens on the same side as the object.
Correct Answer: The image is virtual, upright and 6 cm from the lens on the same side as the object
RRB Group D 2025-26 – 18 August 2026 Shift 2 · Convex lens - image formation and uses

Q6. A person wants to project a magnified real image of a small object onto a screen using a convex lens. Where should the object be placed relative to the lens?

  1. (a) At twice the focal length from the lens
  2. (b) Between the lens and its principal focus
  3. (c) At the principal focus of the lens
  4. (d) Between the focus and twice the focal length from the lens
Answer & solution

Answer: (d) Between the focus and twice the focal length from the lens

Convex Lens — Magnified Real Image
For a convex lens, a magnified real image on a screen is formed when the object is placedbetween F and 2F(the image then forms beyond 2F on the other side, magnified and real).
• Object at 2F → image at 2F, same size, real, inverted
• Object between F and 2F → image beyond 2F, magnified, real, inverted
• Object at F → image at infinity
• Object between lens and F → virtual, magnified image (not projectable on a screen)
Correct Answer: Between the focus and twice the focal length from the lens
RRB Group D 2025-26 – 18 August 2026 Shift 2 · Numericals on refraction and lenses

Q7. Two lenses, one converging with focal length 12 cm and one diverging with focal length −18 cm, are placed in contact. What is the combined focal length of this lens system?

  1. (a) 7.2 cm
  2. (b) 36 cm
  3. (c) 8 cm
  4. (d) 16 cm
Answer & solution

Answer: (b) 36 cm

Combined Focal Length of Lenses in Contact
For two thin lenses in contact, the combined power (and hence 1/f) adds algebraically:
• 1/f = 1/f₁ + 1/f₂ = 1/12 + 1/(−18)
• 1/f = 1/12 − 1/18 = (3 − 2)/36 = 1/36
• f = 36 cm
Correct Answer: 36 cm
RRB Group D 2025-26 – 18 August 2026 Shift 2 · Numericals on refraction and lenses

Q8. A convex lens forms a virtual image 15 cm on the same side as the object when the object is placed 10 cm from the lens. What is the focal length of the lens?

  1. (a) + 30 cm
  2. (b) −10 cm
  3. (c) −30 cm
  4. (d) + 10 cm
Answer & solution

Answer: (a) + 30 cm

Focal Length from Virtual Image Data
A virtual image on the same side as the object (formed by a convex lens acting as a magnifying glass) means both object and image distances are negative in the standard sign convention.
• u = −10 cm, v = −15 cm
• 1/f = 1/v − 1/u = 1/(−15) − 1/(−10) = −1/15 + 1/10 = 1/30
• f = +30 cm
A positive focal length confirms this is a converging (convex) lens, consistent with a magnified virtual image being formed.
Correct Answer: + 30 cm
RRB Group D 2025-26 – 17 August 2026 Shift 1 · Refractive index and optical density

Q9. A student observes a pencil half-immersed in water (refractive index 1.33) and notices the apparent bending is less than when the pencil is in a liquid with refractive index 1.5. What does this suggest about the relationship between refractive index and light bending?

  1. (a) Refractive index has no effect on bending
  2. (b) Greater refractive index reduces bending
  3. (c) Bending is same for all liquids
  4. (d) Greater refractive index increases bending
Answer & solution

Answer: (d) Greater refractive index increases bending

Refractive Index and Bending of Light
The amount of bending (refraction) of light at an interface increases with therefractive indexof the medium. A denser medium (higher refractive index) bends light rays more strongly toward the normal.
Since water (n = 1.33) showslessapparent bending than a liquid with n = 1.5, this confirms that agreater refractive index causes greater bendingof light — i.e., refractive index and bending are directly related.
Correct Answer: Greater refractive index increases bending
RRB Group D 2025-26 – 17 August 2026 Shift 3 · Lens formula and magnification

Q10. How is the object distance measured in the sign convention for spherical lenses?

  1. (a) It is measured from the focal point towards the position of the image.
  2. (b) It is measured from the optical center towards the position of the image.
  3. (c) It is measured from the focal point towards the position of the object.
  4. (d) It is measured from the optical center towards the position of the object.
Answer & solution

Answer: (d) It is measured from the optical center towards the position of the object.

Sign Convention for Spherical Lenses
In the New Cartesian sign convention used for spherical lenses, all distances are measured from theoptical centreof the lens.
• Object distance is measured from the optical centre to the position of the object.
• Image distance is measured from the optical centre to the position of the image.
• Distances measured against the direction of incident light are taken as negative, and in the direction of incident light as positive.
Correct Answer: It is measured from the optical center towards the position of the object.
RRB Group D 2025-26 – 17 August 2026 Shift 3 · Convex lens - image formation and uses

Q11. A student observes that the image formed by a lens is twice the size of the object and is real. Where is the object placed relative to the lens?

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

Answer: (b) Between F and 2F

Convex Lens Image Formation
For a convex lens, a real, enlarged (magnified) image is formed when the object is placed between the focus (F) and twice the focal length (2F).
• Object at 2F → image at 2F, same size, real, inverted.
• Object between F and 2F → image beyond 2F,real, enlarged, inverted.
• Object beyond 2F → image between F and 2F, real, diminished.
• Object at F → image at infinity.
Correct Answer: Between F and 2F
RRB Group D 2025-26 – 14 August 2026 Shift 1 · Convex lens - image formation and uses

Q12. When an object is placed at 2F₁ of a convex lens, where is the image formed?

  1. (a) At infinity
  2. (b) At F₂
  3. (c) At 2F₂
  4. (d) Between F₂ and 2F₂
Answer & solution

Answer: (c) At 2F₂

Answer: Option 3 – At 2F₂.
• For a convex lens with object at 2F₁, the image forms at2F₂(real, inverted, same size).
RRB Group D 2025-26 – 14 August 2026 Shift 1 · Optical instruments (microscope, telescope, camera)

Q13. The measure of an image's clarity is called _________, whereas _________ is the brightness difference between object parts.

  1. (a) contrast, resolution
  2. (b) illumination, magnification
  3. (c) magnification, illumination
  4. (d) resolution, contrast
Answer & solution

Answer: (d) resolution, contrast

Answer: Option 4 – resolution, contrast.
• Image clarity =resolution; brightness difference between parts =contrast.
RRB Group D 2025-26 – 14 August 2026 Shift 3 · Convex lens - image formation and uses

Q14. Which of the following statements is FALSE for a converging lens?

  1. (a) The light from the Sun, before reaching the converging lens, consists of parallel rays.
  2. (b) A converging lens can focus sunlight to form a sharp bright spot on paper.
  3. (c) The bright spot formed on the paper by the converging lens is a virtual image of the Sun.
  4. (d) The concentration of sunlight at a point due to the converging lens generates heat.
Answer & solution

Answer: (c) The bright spot formed on the paper by the converging lens is a virtual image of the Sun.

Answer: Option 3 – The bright spot is a virtual image of the Sun (FALSE statement).
• A converging lens forms arealimage of the Sun (a real bright spot), not virtual. So this statement is false.
RRB Group D 2025-26 – 13 August 2026 Shift 1 · Convex lens - image formation and uses

Q15. A student notices that the image formed by a lens is virtual, upright, and larger than the object. Which lens and object position is responsible for this observation?

  1. (a) Convex lens with object at twice the focal length
  2. (b) Concave lens with object at any position
  3. (c) Convex lens with object between optical centre of lens and focal point
  4. (d) Concave lens with object beyond twice the focal length
Answer & solution

Answer: (c) Convex lens with object between optical centre of lens and focal point

Answer: Option 3 – Convex lens with object between optical centre and focal point.
• A convex lens gives avirtual, erect, magnifiedimage when the object is within the focal length.
RRB Group D 2025-26 – 13 August 2026 Shift 3 · Refractive index and optical density

Q16. If the refractive index of medium 2 with respect to medium 1 is n₂₁, which formula correctly represents n₂₁?

  1. (a) n₂₁ = Speed of light in medium 2 minus speed of light in medium 1.
  2. (b) n₂₁ = Speed of light in medium 1 minus speed of light in medium 2.
  3. (c) n₂₁ = Speed of light in medium 2 divided by speed of light in medium 1.
  4. (d) n₂₁ = Speed of light in medium 1 divided by speed of light in medium 2.
Answer & solution

Answer: (d) n₂₁ = Speed of light in medium 1 divided by speed of light in medium 2.

Answer: Option 4 – Speed of light in medium 1 divided by speed of light in medium 2.
• n₂₁ = v₁/v₂ (speed in medium 1 ÷ speed in medium 2).
RRB Group D 2025-26 – 13 August 2026 Shift 3 · Numericals on refraction and lenses

Q17. A convex lens forms a real, inverted image of an object placed at a distance of 30 cm from the lens. The image is found to be formed at a distance of 60 cm on the other side of the lens. What is the focal length of the lens?

  1. (a) 18 cm
  2. (b) 20 cm
  3. (c) 12 cm
  4. (d) 15 cm
Answer & solution

Answer: (b) 20 cm

Answer: Option 2 – 20 cm.
• 1/f = 1/v − 1/u = 1/60 − 1/(−30) = 1/60 + 1/30 = 3/60 ⇒ f =20 cm.
RRB Group D 2025-26 – 13 August 2026 Shift 3 · Numericals on refraction and lenses

Q18. A 3 cm tall object is placed 9 cm in front of a concave lens with focal length 6 cm. What is the approximate height of the image formed?

  1. (a) 4 cm and upright
  2. (b) 3 cm and inverted
  3. (c) 1.2 cm and upright
  4. (d) 1 cm and inverted
Answer & solution

Answer: (c) 1.2 cm and upright

Answer: Option 3 – 1.2 cm and upright.
• Concave lens: u = −9, f = −6 ⇒ 1/v = 1/(−6) + 1/(−9)... using 1/v − 1/u = 1/f ⇒ v = −3.6 cm.
• Magnification m = v/u = (−3.6)/(−9) = 0.4 ⇒ image height = 0.4×3 =1.2 cm, upright(virtual).
RRB Group D 2025-26 – 12 August 2026 Shift 2 · Convex lens - image formation and uses

Q19. Which statement best describes the image formed by a convex lens when the object is placed between the focal point and the optical center of the lens?

  1. (a) The image is virtual, inverted and smaller than the object.
  2. (b) The image is real, upright and larger than the object.
  3. (c) The image is real, inverted and smaller than the object.
  4. (d) The image is virtual, upright and larger than the object.
Answer & solution

Answer: (d) The image is virtual, upright and larger than the object.

Answer: Option 4 – Virtual, upright and larger than the object.
• For a convex lens with the objectbetween F and the optical centre, the image isvirtual, erect and magnified(magnifying-glass action).
RRB Group D 2025-26 – 11 August 2026 Shift 1 · Laws of refraction and Snell's law

Q20. When a person standing underwater looks upward at a tree located on the bank, the tree appears to be at a position different from its original location. Which of the following principles is responsible for this apparent shift in the position?

  1. (a) Light rays from the tree bend due to refraction at the water-air interface.
  2. (b) The tree image is formed due to diffraction at the water surface.
  3. (c) Light rays from the tree are reflected at the water surface.
  4. (d) Water disperses the light from the tree into different colors.
Answer & solution

Answer: (a) Light rays from the tree bend due to refraction at the water-air interface.

Answer: Option 1 – Light rays from the tree bend due to refraction at the water-air interface.
• Light bends (refracts) as it passes from air into water, so the tree appears shifted from its real position.
RRB Group D 2025-26 – 11 August 2026 Shift 2 · Power of a lens and dioptre

Q21. A convex lens forms a real, inverted image three times larger than the object. If the object is placed 30 cm from the lens, what is the closest approximate value of the power of the lens?

  1. (a) +1 D
  2. (b) +3.2 D
  3. (c) +2 D
  4. (d) +4.44 D
Answer & solution

Answer: (d) +4.44 D

Answer: Option 4 – +4.44 D.
• Real inverted image, m = −3, u = −30 cm ⇒ v = 90 cm. Using 1/f = 1/v − 1/u = 1/90 + 1/30 = 4/90 ⇒ f = 22.5 cm.
• Power P = 100/f(cm) = 100/22.5 ≈+4.44 D.
RRB Group D 2025-26 – 11 August 2026 Shift 3 · Refractive index and optical density

Q22. When a ray of light passes from air into water, how does the refractive index of water with respect to refractive index of air relate to the speed of light in these two media?

  1. (a) The refractive index of water with respect to refractive index of air is the sum of the speeds in air and water
  2. (b) The refractive index of water with respect to refractive index of air is the ratio of the speed of light in air to that in water
  3. (c) The refractive index of water with respect to refractive index of air is the difference of the speeds in air and water
  4. (d) The refractive index of water with respect to refractive index of air is the ratio of the speed of light in water to that in air
Answer & solution

Answer: (b) The refractive index of water with respect to refractive index of air is the ratio of the speed of light in air to that in water

Answer: Option 2 – The ratio of the speed of light in air to that in water.
• Refractive index of water w.r.t. air = speed of light in air ÷ speed of light in water.
RRB Group D 2025-26 – 10 August 2026 Shift 1 · Convex lens - image formation and uses

Q23. If a ray of light passes through the optical center of a thin spherical lens, what is the path of the emergent ray?

  1. (a) It bends towards the principal axis
  2. (b) It is absorbed by the lens material
  3. (c) It reflects back along its original path
  4. (d) It continues in the same direction without deviation
Answer & solution

Answer: (d) It continues in the same direction without deviation

Answer: Option 4 — It continues in the same direction without deviation
A ray through theoptical centrepasses straight through, undeviated → Option 4.
RRB Group D 2025-26 – 10 August 2026 Shift 2 · Refractive index and optical density

Q24. A student measures the absolute refractive index of water as 1.33. What does this mean about the speed of light in water compared to air?

  1. (a) Light travels faster in water by a factor of 1.33 compared to air
  2. (b) Light can not travel through water at all.
  3. (c) Light speed is the same in water and air
  4. (d) Light travels slower in water by a factor of 1.33 compared to air
Answer & solution

Answer: (d) Light travels slower in water by a factor of 1.33 compared to air

Answer: Option 4 – Light travels slower in water by a factor of 1.33.
• Refractive index n = speed of light in air ÷ speed of light in medium = c/v.
• n = 1.33 means v = c/1.33, i.e. light travels1.33 times slowerin water than in air.
RRB Group D 2025-26 – 9 August 2026 Shift 1 · Numericals on refraction and lenses

Q25. A diamond has an absolute refractive index of 2.42. If the speed of light in vacuum is 3 x 10⁸ m/s, which value best estimates the speed of light in diamond?

  1. (a) 2 x 10⁸ m/s
  2. (b) 1.5 x 10⁸ m/s
  3. (c) 1.24 x 10⁸ m/s
  4. (d) 1 x 10⁸ m/s
Answer & solution

Answer: (c) 1.24 x 10⁸ m/s

Answer: Option 3 — 1.24 × 10⁸ m/s
v = c/n = (3 × 10⁸)/2.42 =1.24 × 10⁸ m/s→ Option 3.
RRB Group D 2025-26 – 9 August 2026 Shift 1 · Numericals on refraction and lenses

Q26. A convex lens forms a real image 30 cm from its optical centre. The object is placed 20 cm away from the lens. Using the sign convention, what is the correct sign and value for the focal length of the lens?

  1. (a) +60 cm
  2. (b) +12 cm
  3. (c) -60 cm
  4. (d) -12 cm
Answer & solution

Answer: (b) +12 cm

Answer: Option 2 — +12 cm
v = +30, u = −20. 1/f = 1/v − 1/u = 1/30 + 1/20 = 5/60 = 1/12 ⟹ f =+12 cm→ Option 2.
RRB Group D 2025-26 – 9 August 2026 Shift 2 · Lens formula and magnification

Q27. A lens forms an image three times larger than the object. If the image is erect, what is the magnification?

  1. (a) + 3
  2. (b) -3
  3. (c) + 0.33
  4. (d) -0.33
Answer & solution

Answer: (a) + 3

Answer: Option 1 — +3
An erect image 3× larger ⟹ magnification =+3→ Option 1.
RRB Group D 2025-26 – 6 August 2026 Shift 1 · Laws of refraction and Snell's law

Q28. Which of the following best explains why a pencil partially immersed in water appears bent at the surface?

  1. (a) Light reflects off the pencil and back to your eyes
  2. (b) The pencil absorbs light and emits it at a different angle
  3. (c) Light changes direction as it moves from air to water
  4. (d) Water makes the pencil look thicker
Answer & solution

Answer: (c) Light changes direction as it moves from air to water

Answer: Option 3 — Light changes direction as it moves from air to water
The pencil looks bent due torefractionat the air–water boundary → Option 3.
RRB Group D 2025-26 – 6 August 2026 Shift 2 · Convex lens - image formation and uses

Q29. During an experiment, a student places a candle beyond twice the focal length of a convex lens and records the image position and nature. Which of the following scenarios could result from this setup?

  1. (a) An inverted, real, and diminished image forms between focus and twice the focal length
  2. (b) A virtual, erect, and larger image appears on the same side as the object
  3. (c) An erect, real, and larger image forms beyond twice the focal length
  4. (d) A real, inverted, and same-sized image forms at twice the focal length
Answer & solution

Answer: (a) An inverted, real, and diminished image forms between focus and twice the focal length

Answer: Option 1 — Inverted, real, diminished image between F and 2F
Object beyond 2F of a convex lens ⟹real, inverted, diminishedimage between F and 2F → Option 1.
RRB Group D 2025-26 – 6 August 2026 Shift 3 · Convex lens - image formation and uses

Q30. A student places a convex lens in sunlight and notices a small bright spot forms on a paper held below the lens. What property of the lens allows this to happen?

  1. (a) The lens focuses parallel rays to a single point
  2. (b) The lens diverges the rays in all directions
  3. (c) The lens only works for colored light
  4. (d) The lens blocks all rays from passing through
Answer & solution

Answer: (a) The lens focuses parallel rays to a single point

Answer: Option 1 — The lens focuses parallel rays to a single point
A convex lensconverges parallel sun rays to its focal point, forming a bright spot → Option 1.
RRB Group D 2025-26 – 5 August 2026 Shift 2 · Convex lens - image formation and uses

Q31. A student moves a lighted candle from infinity towards the focus of a convex lens. How does the size of the image change?

  1. (a) The image remains the same size throughout
  2. (b) The image disappears
  3. (c) The image increases in size
  4. (d) The image decreases in size
Answer & solution

Answer: (c) The image increases in size

Answer: Option 3 — The image increases in size
As the object moves from infinity towards the focus of a convex lens, the real image moves away andgrows larger→ Option 3.
RRB Group D 2025-26 – 5 August 2026 Shift 3 · Convex lens - image formation and uses

Q32. A student places a candle beyond the center of curvature of a convex lens. What type of image is most likely to be formed on the other side of the lens?

  1. (a) A virtual and inverted image smaller than the object
  2. (b) A real and inverted image smaller than the object
  3. (c) A virtual and erect image larger than the object
  4. (d) A real and inverted image of the same size as the object
Answer & solution

Answer: (b) A real and inverted image smaller than the object

Answer: Option 2 — Real and inverted, smaller than the object
For a convex lens with the objectbeyond 2F (centre of curvature), the image isreal, inverted and diminished→ Option 2.
RRB Group D 2025-26 – 5 August 2026 Shift 3 · Laws of refraction and Snell's law

Q33. A student observes that the incident ray, the refracted ray, and the normal at the point of incidence all lie in one plane. Which law of refraction does this illustrate?

  1. (a) Law of dispersion
  2. (b) Second law of refraction
  3. (c) Law of superposition
  4. (d) First law of refraction
Answer & solution

Answer: (d) First law of refraction

Answer: Option 4 — First law of refraction
The incident ray, refracted ray and normal lying in one plane is thefirst law of refraction→ Option 4.
RRB Group D 2025-26 – 4 August 2026 Shift 1 · Numericals on refraction and lenses

Q34. A student places a candle 25 cm from a convex lens of focal length 15 cm and obtains a real image. What is the magnification of the image formed?

  1. (a) -2.5
  2. (b) 0.4
  3. (c) 1.67
  4. (d) - 1.5
Answer & solution

Answer: (d) - 1.5

Answer: Option 4 — −1.5
Lens: 1/v − 1/u = 1/f, u = −25, f = +15. 1/v = 1/15 − 1/25 = (5 − 3)/75 = 2/75 ⟹ v = 37.5 cm.
m = v/u = 37.5/(−25) =−1.5→ Option 4.
RRB Group D 2025-26 – 4 August 2026 Shift 3 · Numericals on refraction and lenses

Q35. A ray of light enters a rectangular glass slab from air at an angle of 40 degrees to the normal. If the refractive index of glass with respect to air is 1.5 and the velocity of light in air is 3 x 10⁸ m/s, what is the velocity of light inside the glass slab?

  1. (a) 1.5 x 10⁸ m/s
  2. (b) 2.5 x 10⁸ m/s
  3. (c) 2 x 10⁸ m/s
  4. (d) 2.25 x 10⁸ m/s
Answer & solution

Answer: (c) 2 x 10⁸ m/s

Answer: Option 3 — 2 × 10⁸ m/s
v = c/n = (3 × 10⁸)/1.5 =2 × 10⁸ m/s→ Option 3.
RRB Group D 2025-26 – 3 August 2026 Shift 1 · Numericals on refraction and lenses

Q36. The speed of a light ray changes from 3 x 10⁸ m/s in air to 2 x 10⁸ m/s in a transparent solid. What is the refractive index of the solid with respect to air? (Take the refractive index of air as 1.)

  1. (a) 1.4
  2. (b) 1.5
  3. (c) 1.7
  4. (d) 1.2
Answer & solution

Answer: (b) 1.5

Answer: Option 2 — 1.5
Refractive indexn = speed in air / speed in medium = c/v.
n = (3 × 10⁸) / (2 × 10⁸) =1.5→ Option 2.
RRB Group D 2025-26 – 3 August 2026 Shift 2 · Convex lens - image formation and uses

Q37. A student places a candle at a distance greater than twice the focal length from a convex lens. She observes the image formed on a screen. Which of the following best describes the image's nature, position, and size?

  1. (a) The image is real, erect, formed beyond twice the focal length, and equal in size to the object.
  2. (b) The image is real, inverted, formed between the focal point and twice the focal length, and smaller than the object.
  3. (c) The image is virtual, inverted, formed at the focal point, and the same size as the object.
  4. (d) The image is virtual, erect, formed on the same side as the object, and larger than the object.
Answer & solution

Answer: (b) The image is real, inverted, formed between the focal point and twice the focal length, and smaller than the object.

Answer: Option 2 — Real, inverted, between F and 2F, smaller than object
For a convex lens, when the object is placedbeyond 2F, the image isreal, inverted, diminished, and formedbetween F and 2Fon the other side → Option 2. (This is the principle of a camera.)
RRB Group D 2025-26 – 3 August 2026 Shift 2 · Numericals on refraction and lenses

Q38. A student places an object 18 cm from a converging lens of focal length 12 cm. The image formed is real. What is the magnification produced by the lens?

  1. (a) + 2
  2. (b) + 1.5
  3. (c) - 2
  4. (d) - 1.5
Answer & solution

Answer: (c) - 2

Answer: Option 3 — −2
Lens formula 1/v − 1/u = 1/f, with u = −18, f = +12.
1/v = 1/12 + 1/(−18) = (3 − 2)/36 = 1/36 ⟹ v = +36 cm.
Magnification m = v/u = 36/(−18) =−2(real, inverted, magnified) → Option 3.
RRB Group D 2025-26 – 3 August 2026 Shift 3 · Laws of refraction and Snell's law

Q39. A student observes that a swimming pool appears shallower than its actual depth. Which concept best explains this effect?

  1. (a) Refraction of light at the water-air boundary
  2. (b) Scattering of light by water molecules
  3. (c) Reflection of light from the pool floor
  4. (d) Internal reflection of light within the water
Answer & solution

Answer: (a) Refraction of light at the water-air boundary

Answer: Option 1 — Refraction of light at the water–air boundary
Light from the pool bottom bends away from the normal when leaving water, making the pool lookshallower— this isrefraction→ Option 1.
RRB NTPC UG CBT-1 2025-26 – 20 June 2026 Shift 3 · Total internal reflection and optical fibre

Q40. Mirage is an optical illusion which can be explained based on which of the following phenomena of light?

  1. (a) Diffraction
  2. (b) Interference
  3. (c) Photoelectric Effect
  4. (d) Total Internal Reflection
Answer & solution

Answer: (d) Total Internal Reflection

Topic: Mirage - Optical Illusion
Mirage occurs due toTotal Internal Reflectionof light.
On hot days, air near ground is hotter and less dense than air above.
Light from sky travels from denser (cool) to rarer (hot) medium.
At critical angle, light undergoes total internal reflection.
This creates illusion of water on road/desert surface.
Answer: Option 4 → Total Internal Reflection
RRB NTPC UG CBT-1 2025-26 – 17 June 2026 Shift 1 · Laws of refraction and Snell's law

Q41. Which of the following properties is responsible for the bending of light when it moves from air into water?

  1. (a) Change in direction due to gravity
  2. (b) Change in speed of light
  3. (c) Change in color of light
  4. (d) Change in temperature
Answer & solution

Answer: (b) Change in speed of light

Topic:Refraction of Light
When light travels from one medium to another (air to water),
itsspeed changesdue to different optical densities.
Light slows down in denser medium (water).
This change in speed causes the light ray to bend at the interface.
This phenomenon is calledrefraction.
Snell's Law: n₁sin(θ₁) = n₂sin(θ₂) governs this bending.
Answer: Option 2 → Change in speed of light
RRB NTPC UG CBT-1 2025-26 – 14 June 2026 Shift 1 · Total internal reflection and optical fibre

Q42. Modern telecommunication systems use optical fibres to transmit data efficiently. Which optical phenomenon enables this high-speed data transmission?

  1. (a) Interference of light
  2. (b) Refraction of light
  3. (c) Total Internal Reflection
  4. (d) Dispersion of light
Answer & solution

Answer: (c) Total Internal Reflection

Topic: Optical Fibres in Telecommunication
Principle:Optical fibres work onTotal Internal Reflection (TIR)
How it works:
Light travels through core (denser medium) surrounded by cladding (rarer medium)
When light hits core-cladding boundary at angle > critical angle, it reflects back completely
This repeated TIR keeps light trapped inside fibre with minimal loss
Why TIR is ideal:
• No light escapes → Data transmitted over long distances
• High speed → Light travels at ~2×10⁸ m/s in fibre
• Low signal loss → Efficient data transmission
Answer: Option 3 → Total Internal Reflection
SSC GD 2026 – 27 May 2026 Shift 2 · Laws of refraction and Snell's law

Q43. A pencil partially submerged in water appears bent at the surface. Which phenomenon describes this?

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

Answer: (b) Refraction

Correct Answer: Refraction — Option (b)
The apparent bending of a pencil partially immersed in water is due to Refraction. About Refraction: Definition: The change in direction of light as it passes from one medium to another (due to change in speed of light); Cause: Light travels at different speeds in different media; Speed of light in vacuum: 3 × 10⁸ m/s; Speed of light in water: ~2.25 × 10⁸ m/s (slower); Speed of light in glass: ~2 × 10⁸ m/s (slower); Snell's Law: n₁ sin θ₁ = n₂ sin θ₂; When light passes from water to air (denser to rarer medium): It bends away from the normal; This makes the pencil in water appear to be at a different position than it actually is; Why the pencil appears bent: Light from the underwater part of the pencil bends (refracts) as it crosses the water-air interface; Our eyes trace the light rays back in straight lines; The brain perceives the pencil as being at the position where the extended straight lines meet — which appears bent or displaced; Other examples of Refraction: A swimming pool appears shallower than it is; Stars twinkle (due to refraction in atmosphere); A glass of water acts like a lens; Rainbow formation involves refraction; Lenses in spectacles/cameras work on refraction; About other options: Reflection: Light bouncing back from a surface (mirror effect); Diffraction: Bending of light around obstacles or through small openings; Interference: Superposition of two or more light waves causing bright and dark bands.
Option (a) Reflection — Incorrect: Reflection involves light bouncing back from a surface — it does NOT cause the apparent bending of a pencil in water; reflection is what happens in mirrors; the bent pencil in water is specifically due to refraction, which changes the direction of light as it moves from water to air.
Option (c) Diffraction — Incorrect: Diffraction is the bending of light waves around obstacles or through narrow openings — it is NOT responsible for the apparent bending of a pencil in water; diffraction produces interference patterns and is observed when light passes through very small openings; the pencil-in-water effect is due to refraction.
Option (d) Interference — Incorrect: Interference occurs when two or more waves superpose to produce regions of constructive or destructive interference — it is NOT responsible for the apparent bending of a pencil in water; interference produces patterns of bright and dark bands (like in thin films or double-slit experiments); the pencil-in-water effect is due to refraction.
RRB NTPC UG CBT-1 2025-26 – 8 May 2026 Shift 2 · Power of a lens and dioptre

Q44. A doctor prescribes a lens with a power of P=−2.5 D to a patient. This lens is _____.

  1. (a) A convex lens, used to correct Myopia.
  2. (b) A concave lens, used to correct Myopia.
  3. (c) A convex lens, used to correct Hypermetropia.
  4. (d) A concave lens, used to correct Hypermetropia.
Answer & solution

Answer: (b) A concave lens, used to correct Myopia.

Given: Power P = −2.5 D (negative)
Key Concept:
Negative power → Concave lens (diverging)
Positive power → Convex lens (converging)
Myopia (Near-sightedness):
Image forms before retina
Corrected by concave lens to diverge rays
Hypermetropia (Far-sightedness):
Image forms behind retina
Corrected by convex lens to converge rays
Answer: Option 2 → A concave lens, used to correct Myopia.
RRB NTPC GL CBT-1 2025-26 – 25 March 2026 Shift 1 · Laws of refraction and Snell's law

Q45. The phenomenon called ________ occurs when light passes from one transparent medium to another.

  1. (a) refraction
  2. (b) scattering
  3. (c) reflection
  4. (d) diffraction
Answer & solution

Answer: (a) refraction

Topic: Behavior of Light
Refractionis the bending of light when it passes from one transparent medium to another (e.g., air to water).
This occurs due to change in speed of light in different media.
Other phenomena:
• Reflection - bouncing back of light from a surface
• Scattering - light spreads in different directions
• Diffraction - bending around obstacles/edges
Answer: Option 1 → refraction
RRB NTPC GL CBT-1 2025-26 – 19 March 2026 Shift 1 · Convex lens - image formation and uses

Q46. __________ is primarily used to focus parallel rays of light.

  1. (a) Cylindrical lens
  2. (b) Concave lens
  3. (c) Bifocal lens
  4. (d) Convex lens
Answer & solution

Answer: (d) Convex lens

Topic: Optics - Lenses
Convex lens(converging lens) is thicker at center.
When parallel rays pass through convex lens, they converge at a single point calledfocal point.
This property is used in magnifying glasses, cameras, telescopes, and correcting hypermetropia.
Concave lensdiverges parallel rays (spreads them apart).
Answer: Option 4 → Convex lens
RRB NTPC GL CBT-1 2025-26 – 19 March 2026 Shift 3 · Optical instruments (microscope, telescope, camera)

Q47. Which type of lens is used in a simple microscope?

  1. (a) Plano-concave
  2. (b) Biconcave
  3. (c) Cylindrical
  4. (d) Biconvex
Answer & solution

Answer: (d) Biconvex

Simple Microscope (Magnifying Glass)
A simple microscope uses asingle convex lensto magnify objects.
The lens isbiconvex(convex on both sides) which converges light rays.
When object is placed between focus and lens, it forms a virtual, erect, and magnified image.
Concave lenses diverge light and cannot magnify.
Answer: Option 4 → Biconvex
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Numericals on refraction and lenses

Q48. A 2.0 cm tall object is placed 20 cm from a convex lens of focal length 15 cm. What is the magnification produced by the lens?

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

Answer: (a) -3

The magnification (M) produced by a convex lens is given by the formula: M = − v / u, where v is image distance and u is object distance.
Given: object height (h_o) = 2.0 cm, object distance (u) = -20 cm (since object distances are negative for lenses), focal length (f) = +15 cm (convex lens).
Using lens formula: 1/f = 1/v − 1/u
1/15 = 1/v − 1/(-20)
1/15 = 1/v + 1/20
1/v = 1/15 − 1/20
1/v = (4 − 3)/60 = 1/60
v = 60 cm
Magnification M = − v / u = -60 / (-20) = +3
Since the answer options include negative 3 and positive 3, the correct magnification according to signs is +3; but in lens sign convention magnification is negative if image is real and inverted.
Calculations show image distance positive and magnification positive indicating an upright image, which is unusual for convex lens with object outside focal length.
However, given the data and calculation, magnification = +3.
Check carefully; as per sign conventions, object distance is negative, focal length positive, image distance positive => real image formed on opposite side (inverted) => magnification negative.
So final magnification is -3.
Therefore, option 1 (-3) is correct.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Refractive index and optical density

Q49. If the speed of light in a medium is 2.25 × 10⁸ m/s, its refractive index is ________.

  1. (a) 1.33
  2. (b) 1.50
  3. (c) 1.75
  4. (d) 0.75
Answer & solution

Answer: (a) 1.33

Refractive index, n = speed of light in vacuum / speed of light in medium.
Speed of light in vacuum = 3 × 10^8 m/s.
Given speed of light in medium = 2.25 × 10^8 m/s.
Calculate n = (3 × 10^8) / (2.25 × 10^8) = 4/3 = 1.33.
Hence, the refractive index of the medium is 1.33.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Convex lens - image formation and uses

Q50. What happens to a ray of light that either passes through the principal focus (for a convex lens) or appears to meet at the principal focus (for a concave lens) before striking the lens?

  1. (a) It bends randomly after refraction.
  2. (b) It emerges parallel to the principal axis after refraction.
  3. (c) It always converges at the optical centre.
  4. (d) It forms an image at the focal plane of the lens.
Answer & solution

Answer: (b) It emerges parallel to the principal axis after refraction.

A ray of light passing through the principal focus of a convex lens refracts and emerges parallel to the principal axis.
Similarly, for a concave lens, a ray directed towards the principal focus appears to come from it and emerges parallel to the principal axis after refraction.
This is a fundamental property of lenses used in ray diagrams to locate images correctly.
Hence, the ray emerges parallel to the principal axis after refraction.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Concave lens - image formation and uses

Q51. A concave lens forms a virtual image when the object is placed __________.

  1. (a) Beyond 2F only
  2. (b) At 2F only
  3. (c) At infinity only
  4. (d) At any position in front of the lens.
Answer & solution

Answer: (d) At any position in front of the lens.

A concave lens always forms a virtual, erect and diminished image.
It produces this virtual image irrespective of the object's position in front of the lens.
Hence, whether the object is close or far, the image formed by a concave lens is always virtual.
Therefore, the correct option is that a concave lens forms a virtual image at any position in front of the lens.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Numericals on refraction and lenses

Q52. A concave lens has a focal length of 15 cm. If it forms an image 10 cm from the lens, how far away is the object placed from the lens?

  1. (a) +30 cm
  2. (b) +6 cm
  3. (c) -25 cm
  4. (d) -30 cm
Answer & solution

Answer: (d) -30 cm

Given: concave lens focal length, f = -15 cm (concave lens focal length is negative)
Image distance, v = -10 cm (image formed by concave lens is virtual and on same side as object, so negative)
Using lens formula: 1/f = 1/v − 1/u
Substitute values: 1/(-15) = 1/(-10) − 1/u
=> -1/15 = -1/10 − 1/u
=> -1/15 + 1/10 = -1/u
=> (-2 + 3)/30 = -1/u
=> 1/30 = -1/u
=> u = -30 cm
The negative sign means object is on the same side as incoming light (real object), so object distance = -30 cm
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Numericals on refraction and lenses

Q53. An object is placed at 100 cm on the left of a convex lens of focal length 40 cm. Where should a screen be placed to right of the lens so that the image formed is real, inverted and smaller in size as compared to the object size?

  1. (a) The screen should be placed anywhere between 20 cm and 40 cm from the lens.
  2. (b) The screen should be beyond 100 cm from the lens.
  3. (c) The screen should be placed at 40 cm from the lens.
  4. (d) The screen should be placed anywhere between 40 cm and 80 cm from the lens.
Answer & solution

Answer: (d) The screen should be placed anywhere between 40 cm and 80 cm from the lens.

Given: Object distance u = -100 cm (left side denotes negative in lens formula).
Focal length f = +40 cm (convex lens has positive focal length).
Using lens formula: 1/f = 1/v − 1/u
Substitute values: 1/40 = 1/v − 1/(-100)
That simplifies to 1/40 = 1/v + 1/100
1/v = 1/40 − 1/100 = (5/200) − (2/200) = 3/200
Therefore, v = 200/3 = approximately 66.67 cm (positive value means image on right side).
Magnification m = v/u = +66.67 / -100 = -0.6667 (negative sign means inverted image).
Since |m| < 1, image is smaller than the object.
Hence, the screen should be placed between 40 cm and 80 cm on the right side.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Lens formula and magnification

Q54. Which of the following is the correct relation between the object distance (u), image distance (v) and focal length (f) of a convex lens?

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

Answer: (d) (1/v) - (1/u) = 1/f

The relation between object distance (u), image distance (v), and focal length (f) of a convex lens is given by the lens formula.
Lens formula is (1/v) − (1/u) = 1/f, where all distances are measured from the optical centre of the lens.
This formula helps to relate the position of the object, image, and the focal length of the lens.
Hence, the correct relation is option 4: (1/v) − (1/u) = 1/f.
This is a basic formula used in optics problems in competitive exams.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Concave lens - image formation and uses

Q55. An object is placed at infinity in front of a concave lens. The image formed will be _____________ .

  1. (a) Virtual, erect, and highly diminished at the focus
  2. (b) Virtual, erect, and enlarged
  3. (c) Real, erect, and at the principal focus
  4. (d) Real, inverted, and at infinity
Answer & solution

Answer: (a) Virtual, erect, and highly diminished at the focus

A concave lens always forms a virtual image for objects placed at any distance.
When the object is at infinity, light rays are parallel and diverge after passing through the concave lens.
The image formed is virtual, erect, and highly diminished in size.
The image appears at the focus of the concave lens.
Hence, the correct description is virtual, erect, and highly diminished at the focus.
Final answer: Option 1
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Lens formula and magnification

Q56. If a lens forms a real and inverted image, the height of the image h' is generally taken as _____________.

  1. (a) Positive, as it is above the axis
  2. (b) Negative, as it is below the axis
  3. (c) Negative, only if the object is virtual
  4. (d) Positive, but less than the object height
Answer & solution

Answer: (b) Negative, as it is below the axis

For a real and inverted image formed by a lens, the image is below the principal axis.
Thus, the height of the image, h', is considered negative due to the inverted position.
Therefore, the convention is: real inverted image height = negative value.
Final answer: Option 2
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Convex lens - image formation and uses

Q57. If an object is placed between F₁ and 2F₁ of a convex lens, the image produced is ________. (F₁ is the position of the principal focus.)

  1. (a) Enlarged, real, inverted
  2. (b) Diminished, real, inverted
  3. (c) Enlarged, virtual, erect
  4. (d) Same size, real, inverted
Answer & solution

Answer: (a) Enlarged, real, inverted

When an object is placed between the principal focus (F₁) and twice the focal length (2F₁) of a convex lens:
The image formed is real.
It is inverted.
The image is enlarged compared to the object.
This is a fundamental property of convex lenses in this object position range.
Final answer: Option 1
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Power of a lens and dioptre

Q58. The extent of bending of light rays achieved by a lens is expressed in terms of its ___________.

  1. (a) Power
  2. (b) Aperture
  3. (c) Magnification
  4. (d) Image Distance
Answer & solution

Answer: (a) Power

The extent of bending of light rays by a lens is called its power.
Power indicates the lens's ability to converge or diverge light.
It is measured in diopters (D).
Thus, the correct term is 'Power'.
Final answer: Option 1
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Numericals on refraction and lenses

Q59. Two lens of power (+ 4 D) and (- 6 D) are placed in contact with each other. What will be their combined focal length in cm?

  1. (a) + 0.5 cm
  2. (b) + 0.1 cm
  3. (c) - 50 cm
  4. (d) - 10 cm
Answer & solution

Answer: (c) - 50 cm

Power P of lens = +4 D and -6 D respectively.
Combined power P_total = P1 + P2 = +4 + (-6) = -2 D.
Focal length f (in metres) = 1 / Power (in dioptres).
So, f = 1 / -2 = -0.5 m = -50 cm.
Check carefully: power = -2 D means focal length is -50 cm.
Options are in length units (cm), so answer is -50 cm.
Option 3 says -50 cm, Option 4 says -10 cm.
Re-examining the closest combined focal length corresponds to power -2 D, which is -50 cm.
Therefore, correct combined focal length is -50 cm.
Final answer: Option 3
RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Convex lens - image formation and uses

Q60. A student uses a convex lens to project an image of a tree onto a wall. If the image is found to be smaller than the tree, what can be inferred about the tree's position relative to the lens?

  1. (a) The tree is between the lens and its focal point.
  2. (b) The tree is at the focal point of the lens.
  3. (c) The tree is at twice the focal length from the lens.
  4. (d) The tree is placed beyond twice the focal length (beyond 2F) of the lens.
Answer & solution

Answer: (d) The tree is placed beyond twice the focal length (beyond 2F) of the lens.

Solution:
Aconvex lensforms areal, inverted, and diminished (smaller)image when the object is placedbeyond 2F (twice the focal length).
Since the image issmaller than the tree, it must be adiminished image.
Such images are formed only when the object isbeyond 2F.
Hence, the correct answer is:The tree is placed beyond twice the focal length (beyond 2F) of the lens.
RRB JE CBT-1 2025-26 – 25 February 2026 Shift 3 · Lens formula and magnification

Q61. What is the sign of the focal length for a convex lens as per the sign convention for spherical lenses?

  1. (a) Positive
  2. (b) Negative
  3. (c) Zero
  4. (d) Variable
Answer & solution

Answer: (a) Positive

For spherical lenses, the sign convention states:
The focal length (f) is positive for a convex lens.
This is because a convex lens converges light rays, hence it has a positive focal length.
Concave lenses have negative focal lengths as per the sign convention.
Therefore, the focal length of a convex lens is positive.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 3 · Concave lens - image formation and uses

Q62. A designer wants to create a device that spreads light from a small bulb to illuminate a wide area uniformly. Which lens should be used and why?

  1. (a) A concave lens, because it diverges light rays outward
  2. (b) A convex lens, because it converges light rays
  3. (c) A plane glass, because it transmits light without bending
  4. (d) A prism, because it bends light at an angle
Answer & solution

Answer: (a) A concave lens, because it diverges light rays outward

Correct option: 1
A concave lens diverges light rays outward.
This divergence helps spread light from a small bulb over a wide area.
Hence, concave lens is suitable to illuminate a wide area uniformly.
Convex lenses converge rays, focusing light on a small spot, which is not ideal here.
Plane glass simply passes light without spreading it.
Prisms bend light but do not uniformly spread it.
Therefore, the best choice is a concave lens for spreading light uniformly.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Laws of refraction and Snell's law

Q63. When a light ray enters a rectangular glass slab obliquely from air, what happens to its direction at the first interface?

  1. (a) It gets reflected back
  2. (b) It bends away from the normal
  3. (c) It travels straight
  4. (d) It bends towards the normal
Answer & solution

Answer: (d) It bends towards the normal

When light passes from a less dense medium (air) to a denser medium (glass), it slows down.
Due to this change in speed, it bends towards the normal at the interface.
Hence, the ray bends towards the normal when entering the glass slab obliquely from air.
Option 4 is correct.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 3 · Convex lens - image formation and uses

Q64. A bi-convex lens forms a real, inverted image twice as large as the object. Where is the object located relative to the lens?

  1. (a) At twice the focal length from the lens
  2. (b) Beyond twice the focal length from the lens
  3. (c) Between one focal length and twice the focal length from the lens
  4. (d) At the focal length from the lens
Answer & solution

Answer: (c) Between one focal length and twice the focal length from the lens

Solution:
Abi-convex lens (convex lens)forms areal and inverted imagewhen the object is placedbeyond the focal length.
If theimage is twice the size of the object, themagnification |m| = 2.
For a convex lens, this happens when theobject lies between the focal length (F) and twice the focal length (2F).
In this position, theimage forms beyond 2F, real, inverted, and magnified.
Answer: Between one focal length and twice the focal length from the lens
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 1 · Convex lens - image formation and uses

Q65. A physicist uses a convex lens to focus sunlight onto a piece of paper and notices the smallest, brightest spot forms at a certain distance from the lens. Which property of the lens does this experiment help to determine?

  1. (a) The experiment helps to determine the focal length of the lens.
  2. (b) The experiment helps to determine the principal axis of the lens.
  3. (c) The experiment helps to determine the optical center of the lens.
  4. (d) The experiment helps to determine the diameter of the lens.
Answer & solution

Answer: (a) The experiment helps to determine the focal length of the lens.

The smallest and brightest spot formed by focusing sunlight is the focal point of the convex lens.
This spot appears at the focal length distance from the lens.
Hence, the experiment helps to determine the focal length of the lens.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 1 · Refractive index and optical density

Q66. If a material has an absolute refractive index of 1.7, what does this imply about the speed of light in that material compared to vacuum?

  1. (a) Light travels at the same speed in both.
  2. (b) Light travels 1.7 times faster in the material than in vacuum.
  3. (c) The speed difference cannot be determined from this value.
  4. (d) Light travels 1.7 times slower in the material than in vacuum.
Answer & solution

Answer: (d) Light travels 1.7 times slower in the material than in vacuum.

The absolute refractive index (n) is the ratio of speed of light in vacuum (c) to speed of light in the material (v).
n = c / v
Given n = 1.7, so light travels in the material slower than in vacuum.
Speed in material = c / 1.7, which means light is 1.7 times slower in the material than in vacuum.
Therefore, option 4 is correct.
Final answer: Option 4
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 3 · Laws of refraction and Snell's law

Q67. What does Snell's law of refraction state about the ratio of the sine of the angle of incidence to the sine of the angle of refraction for a given colour and pair of media?

  1. (a) It depends on the type of mirror.
  2. (b) It varies with the normal.
  3. (c) It changes with the angle of incidence.
  4. (d) It remains constant.
Answer & solution

Answer: (d) It remains constant.

Snell's law states that for a given colour and pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
This constant is called the refractive index of the second medium with respect to the first medium.
Therefore, the ratio does not depend on the angle of incidence or normal but remains constant for a specific colour and media pair.
Hence, the correct option is that it remains constant.
Final answer: Option 4
SSC CGL Tier-1 2025 – 13 September 2025 Shift 2 · Refractive index and optical density

Q68. When light enters from air into a medium, its speed changes. Which of the following properties of light remains constant during this transition?

  1. (a) Wavelength
  2. (b) Frequency
  3. (c) Speed
  4. (d) Amplitude
Answer & solution

Answer: (b) Frequency

• Physics - Light refraction:
• When light travels from one medium to another, speed changes (that is what causes refraction).
• Wavelength also changes when light enters a new medium.
• Speed changes (not constant).
• Amplitude changes (intensity may vary).
• FREQUENCY remains constant — it is determined by the source of light, not the medium.
• Relationship: v = fλ; when v changes and f stays same, λ must change.
• Answer: Frequency
RRB NTPC UG CBT-1 2024-25 – 9 September 2025 Shift 2 · Total internal reflection and optical fibre

Q69. The critical angle for diamond is approximately _______.

  1. (a) 45°
  2. (b) 24.4°
  3. (c) 60°
  4. (d) 31.8°
Answer & solution

Answer: (b) 24.4°

RRB NTPC UG CBT-1 2024-25 – 4 September 2025 Shift 2 · Total internal reflection and optical fibre

Q70. In an optical fibre, the core is made of a _________.

  1. (a) material with higher refractive index
  2. (b) material with lower refractive index
  3. (c) material with higher density
  4. (d) material with lower density
Answer & solution

Answer: (a) material with higher refractive index

RRB NTPC UG CBT-1 2024-25 – 2 September 2025 Shift 2 · Total internal reflection and optical fibre

Q71. Optical fibres work on the principle of ____________.

  1. (a) reflection only
  2. (b) refraction only
  3. (c) total internal reflection
  4. (d) total internal refraction
Answer & solution

Answer: (c) total internal reflection

RRB NTPC UG CBT-1 2024-25 – 22 August 2025 Shift 1 · Total internal reflection and optical fibre

Q72. What is the condition for total internal reflection to occur?

  1. (a) Light must travel from a denser to a rarer medium at an angle greater than the critical angle.
  2. (b) Light must travel from a rarer medium to a denser medium at an angle of 90°.
  3. (c) Light must strike the boundary at 90° to the normal separating any two mediums of different refractive indices.
  4. (d) The refractive indices of both media must be equal.
Answer & solution

Answer: (a) Light must travel from a denser to a rarer medium at an angle greater than the critical angle.

RRB NTPC UG CBT-1 2024-25 – 18 August 2025 Shift 2 · Refractive index and optical density

Q73. Which of the following can have higher optical density but lower mass density?

  1. (a) Water
  2. (b) Turpentine
  3. (c) Germanium
  4. (d) Glass
Answer & solution

Answer: (b) Turpentine

SSC Selection Post 2025 (Phase XIII) – 28 July 2025 Shift 3 · Total internal reflection and optical fibre

Q74. The phenomenon of total internal reflection can occur when light travels from:

  1. (a) A rarer medium to a denser medium at any angle of incidence.
  2. (b) A denser medium to a rarer medium at any angle of incidence.
  3. (c) A rarer medium to a denser medium at an angle of incidence greater than the critical angle.
  4. (d) A denser medium to a rarer medium at an angle of incidence greater than the critical angle.
Answer & solution

Answer: (d) A denser medium to a rarer medium at an angle of incidence greater than the critical angle.

Correct Answer: Option 4 – A denser medium to a rarer medium at an angle of incidence greater than the critical angle
Total Internal Reflection (TIR) requires two conditions: (1) Light must travel from a denser (optically denser, higher refractive index) medium to a rarer (optically rarer, lower refractive index) medium. (2) The angle of incidence must be greater than the critical angle (θc). When both conditions are met, no refraction occurs and all light is reflected back into the denser medium. Applications: optical fibres, mirages, diamonds' brilliance, endoscopes.
Option 1 (Rarer to denser, any angle): TIR cannot occur when going from rarer to denser — refraction always occurs as light bends toward the normal.
Option 2 (Denser to rarer, any angle): At angles less than the critical angle, refraction (not TIR) occurs; TIR requires angle GREATER than critical angle.
Option 3 (Rarer to denser, angle > critical angle): TIR never occurs going from rarer to denser medium regardless of angle.
RRB Technician Grade III 2024-25 – 30 December 2024 Shift 3 · Refractive index and optical density

Q75. The cause of being the refractive index(μ) of transparent medium > 1, is:

  1. (a) Speed of light in vacuum > speed of light in transparent medium
  2. (b) Speed of light in vacuum = speed of light in transparent medium
  3. (c) Wavelength of light gets change as it moves from denser to rarer medium
  4. (d) Speed of light in vacuum < speed of light in transparent medium
Answer & solution

Answer: (a) Speed of light in vacuum > speed of light in transparent medium

Correct Answer: Option 1 — Speed of light in vacuum > speed of light in transparent medium
The absolute refractive index of a medium is defined as
μ = c/v
where c is the speed of light in vacuum (3 × 10⁸ m/s) and v is its speed in the medium.
A fraction is greater than 1 only when the numerator is larger than the denominator. So μ > 1 exactly when c > v — that is, when light travels slower inside the medium than in vacuum. Light always slows down on entering any material medium, because it is repeatedly absorbed and re-emitted by the atoms, so every transparent medium has μ greater than 1.
For example, for water v = 2.25 × 10⁸ m/s, giving μ = 3/2.25 = 1.33; for ordinary glass μ ≈ 1.5, and for diamond μ = 2.42, the largest among common substances.
❌ Why the other options are incorrect
✗ Speed of light in vacuum = speed of light in transparent medium — then μ = c/v = 1 exactly, which is the case for vacuum itself, not for a transparent medium.
✗ Wavelength of light gets change as it moves from denser to rarer medium — the wavelength does change on refraction, but that is a consequence of the change of speed (frequency stays constant), not the defining cause of μ > 1.
✗ Speed of light in vacuum < speed of light in transparent medium — this would give μ < 1, meaning light travels faster than c in matter, which is impossible.
📌 SSC & Railway Exam Facts
• μ = c/v = sin i / sin r (Snell's law); μ has no unit — it is a pure ratio.
• c = 3 × 10⁸ m/s in vacuum, the maximum possible speed in nature.
• μ (water) = 1.33, μ (glass) ≈ 1.5, μ (diamond) = 2.42 — highest among common media.
• On refraction the frequency stays the same; the speed and wavelength change.
• Higher μ means optically denser; light bends towards the normal entering a denser medium and away entering a rarer one.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Power of a lens and dioptre

Q76. If the focal length of a lens is −25 cm, the power of the lens will be:

  1. (a) +0.04 D
  2. (b) −0.04 D
  3. (c) −4 D
  4. (d) +4 D
Answer & solution

Answer: (c) −4 D

Correct Answer: Option 3 — −4 D
📖 Given
• Focal length of the lens, f = −25 cm
• To find: power P of the lens
✍️ Formula
P = 1/f, where f must be in metres and P is in dioptre (D)
🔢 Solution
• First convert the focal length into metres: f = −25 cm = −25/100 m = −0.25 m.
• P = 1/f = 1/(−0.25)
• P = −4 D
• The negative sign shows that the lens is a concave (diverging) lens.
In plain language, a lens of focal length 25 cm bends light fairly strongly, and the minus sign tells us it spreads the rays out instead of bringing them to a point.
❌ Why the other options are incorrect
✗ +0.04 D — this comes from dividing 1 by 25 without converting cm to metres, and the sign has also been dropped.
✗ −0.04 D — the sign is right, but the centimetre value was used directly instead of −0.25 m; 1/(−25) = −0.04.
✗ +4 D — the magnitude is correct but the sign is wrong; +4 D would mean a convex (converging) lens with f = +0.25 m.
📌 SSC & Railway Exam Facts
• P = 1/f (in metres); SI unit of power is the dioptre (D), and 1 D = 1 m⁻¹.
• Convex (converging) lens → f positive, P positive; concave (diverging) lens → f negative, P negative.
• For lenses in contact, powers simply add: P = P₁ + P₂ + P₃ …
• Lens formula 1/v − 1/u = 1/f; magnification m = v/u.
• Myopia (short-sightedness) is corrected by a concave lens; hypermetropia (long-sightedness) by a convex lens.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 1 · Power of a lens and dioptre

Q77. If the power of a lens is +1 Dioptre, then the focal length of the lens will be:

  1. (a) 0.75 m
  2. (b) 0.5 m
  3. (c) 0.25 m
  4. (d) 1 m
Answer & solution

Answer: (d) 1 m

Correct Answer: Option 4 — 1 m
📖 Given
• Power of the lens, P = +1 D
• To find: focal length f in metres
✍️ Formula
P = 1/f, so f = 1/P (f in metres, P in dioptre)
🔢 Solution
• f = 1/P
• f = 1/(+1)
• f = +1 m = 100 cm
• The positive sign of both P and f shows that the lens is a convex (converging) lens.
In plain language, a 1-dioptre lens is defined as the lens whose focal length is exactly 1 metre — this is the very definition of the dioptre, so the calculation is simply reading the definition backwards.
❌ Why the other options are incorrect
✗ 0.75 m — this would correspond to P = 1/0.75 ≈ 1.33 D, not 1 D.
✗ 0.5 m — this is the focal length of a +2 D lens, since 1/0.5 = 2 D.
✗ 0.25 m — this is the focal length of a +4 D lens, since 1/0.25 = 4 D; it comes from confusing the value with a much stronger lens.
📌 SSC & Railway Exam Facts
• 1 dioptre is the power of a lens of focal length 1 metre; 1 D = 1 m⁻¹.
• P = 1/f (metres) — never put f in centimetres in this formula.
• Convex lens → f and P positive; concave lens → f and P negative.
• Powers of thin lenses in contact add up: P = P₁ + P₂.
• An optician's prescription such as "−2.5 D" means a concave lens of focal length 40 cm, used to correct myopia.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Concave lens - image formation and uses

Q78. An object is placed between F and 2F of a concave lens. Which of the following options represents the characteristics of the image formed?

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

Answer: (d) Virtual and diminished

Correct Answer: Option 4 — Virtual and diminished
A concave lens is a diverging lens — it is thinner at the middle and thicker at the edges, so it spreads out the rays falling on it instead of bringing them together. Because the refracted rays always diverge, they can never actually meet on a screen. They only appear to come from a point on the same side as the object.
For this reason a concave lens forms, for every position of a real object, an image that is:
• virtual (it cannot be taken on a screen),
• erect (the same way up as the object), and
• diminished (smaller than the object),
• located between the optical centre O and the focus F, on the same side as the object.
So placing the object between F and 2F makes no difference to the nature of the image — unlike a convex lens, a concave lens has no position that gives a real or an enlarged image. The image simply lies somewhere within the focal length, smaller and upright.
❌ Why the other options are incorrect
✗ Virtual and enlarged — this is what a convex lens gives when the object is placed within its focal length (the magnifying-glass case); a concave lens never enlarges.
✗ Real and enlarged — this is a convex lens with the object between F and 2F; the option is a trap for students who read "concave" as "convex".
✗ Real and diminished — a convex lens with the object beyond 2F; a concave lens can never form a real image of a real object.
📌 SSC & Railway Exam Facts
• Concave (diverging) lens — image is always virtual, erect and diminished, formed between O and F.
• Convex lens, object between F and 2F → image beyond 2F, real, inverted, enlarged.
• f is negative for a concave lens, so its power P = 1/f is negative too.
• Lens formula 1/v − 1/u = 1/f; magnification m = v/u; for a concave lens m is positive and less than 1.
• A concave lens corrects myopia (short-sightedness); a convex lens corrects hypermetropia (long-sightedness).
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Refractive index and optical density

Q79. The speed of light is ____________.

  1. (a) greater in glass than in water
  2. (b) greater in water than in air
  3. (c) greater in water than in glass
  4. (d) same in air, glass and water
Answer & solution

Answer: (c) greater in water than in glass

Correct Answer: Option 3 — greater in water than in glass
The speed of light in a medium is decided by the refractive index of that medium, through the relation n = c/v, where c = 3 × 10⁸ m/s is the speed of light in vacuum and v is its speed in the medium. Rearranging, v = c/n — so the higher the refractive index, the slower the light.
Standard values are: air ≈ 1.0003 (≈ 1), water = 1.33, glass = 1.5. Since n(water) = 1.33 is less than n(glass) = 1.5, light travels faster in water than in glass.
Working it out: in water v = 3 × 10⁸/1.33 ≈ 2.25 × 10⁸ m/s, while in glass v = 3 × 10⁸/1.5 = 2 × 10⁸ m/s. So the order of speed is air > water > glass, exactly the reverse of the order of refractive index.
This is also why glass is called optically denser than water, and why a ray bends more on entering glass than on entering water.
❌ Why the other options are incorrect
✗ greater in glass than in water — the exact reverse of the truth; glass has the higher refractive index, so light is slower in glass.
✗ greater in water than in air — light is fastest in air (n ≈ 1) and slows down in water (n = 1.33); water can never beat air.
✗ same in air, glass and water — if the speed were the same everywhere, refraction would not occur at all and lenses and prisms would not work.
📌 SSC & Railway Exam Facts
• n = c/v; c = 3 × 10⁸ m/s in vacuum — the maximum possible speed in nature.
• Refractive index: air ≈ 1, water = 1.33, glass = 1.5, diamond = 2.42 (the highest among common media).
• Higher refractive index → optically denser → lower speed of light and greater bending.
• Snell's law: n₁ sin i = n₂ sin r, i.e. n = sin i / sin r for light going from air into a medium.
• Diamond sparkles because of total internal reflection, its critical angle being only about 24°.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 1 · Refractive index and optical density

Q80. Ramu has been given a tub of water with a refractive index of 1.33, a mug of oil with a refractive index of 1.46, a bottle of glycerine with a refractive index of 1.5 and a glass-slab with a refractive index of 1.65. Light will travel fastest in:

  1. (a) glass-slab
  2. (b) water
  3. (c) oil
  4. (d) glycerine
Answer & solution

Answer: (b) water

Correct Answer: Option 2 — water
📖 Given
• Refractive index of water, n = 1.33
• Refractive index of oil, n = 1.46
• Refractive index of glycerine, n = 1.5
• Refractive index of glass-slab, n = 1.65
✍️ Formula
n = c / v, so v = c / n
(c = speed of light in vacuum = 3 × 10⁸ m/s)
🔢 Solution
• Since c is the same for all four media, v is inversely proportional to n.
• The smallest refractive index gives the largest speed.
• Comparing: 1.33 < 1.46 < 1.5 < 1.65, so the smallest n is 1.33 (water).
• v in water = 3 × 10⁸ / 1.33 ≈ 2.26 × 10⁸ m/s — the highest of the four.
So light travels fastest in water, because water is optically the rarest of the four given media.
❌ Why the other options are incorrect
✗ glass-slab — n = 1.65 is the largest value, so light is slowest here: v ≈ 1.82 × 10⁸ m/s. Choosing it comes from wrongly assuming higher n means higher speed.
✗ oil — n = 1.46 gives v ≈ 2.05 × 10⁸ m/s, slower than in water.
✗ glycerine — n = 1.5 gives v ≈ 2.0 × 10⁸ m/s, again slower than in water.
📌 SSC & Railway Exam Facts
• Refractive index n = c/v; it has no unit and is always greater than 1 for a material medium.
• Higher n → optically denser → light slower → ray bends more.
• Standard values: air = 1.0003, water = 1.33, glass ≈ 1.5, diamond = 2.42.
• Diamond has the highest refractive index among common materials and the smallest critical angle (24.4°).
• Snell's Law: n₁ sin i = n₂ sin r.
RRB Technician Grade III 2024-25 – 28 December 2024 Shift 2 · Laws of refraction and Snell's law

Q81. According to one of the laws of refraction, the ______ ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

  1. (a) Incident
  2. (b) Emergent
  3. (c) Parallel
  4. (d) Reflected
Answer & solution

Answer: (a) Incident

Correct Answer: Option 1 — Incident
Refraction is the bending of light when it passes from one transparent medium into another, caused by the change in the speed of light in the two media.
The First Law of Refraction states that the incident ray, the refracted ray and the normal drawn to the interface at the point of incidence all lie in the same plane.
The Second Law of Refraction, known as Snell's Law, states that for a given pair of media the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant:
sin i / sin r = n₂₁ (the refractive index of medium 2 with respect to medium 1).
❌ Why the other options are incorrect
✗ Emergent — the emergent ray is the ray that finally leaves a glass slab or prism; it is not named in the first law of refraction.
✗ Parallel — 'parallel ray' is not a term of the laws of refraction; in a glass slab the emergent ray only happens to be parallel to the incident ray.
✗ Reflected — the reflected ray belongs to the laws of reflection, where ∠i = ∠r at a mirror surface.
📌 SSC & Railway Exam Facts
• Snell's Law: n₁ sin i = n₂ sin r, or sin i / sin r = constant.
• Refractive index n = c/v, where c = 3 × 10⁸ m/s is the speed of light in vacuum.
• Denser to rarer → ray bends away from the normal; rarer to denser → bends towards the normal.
• Refraction explains a stick looking bent in water, the twinkling of stars and the apparent flattening of the setting Sun.
• Refractive index of water = 1.33, glass ≈ 1.5, diamond = 2.42 (the highest of common materials).
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 1 · Power of a lens and dioptre

Q82. A student has an eye-power of −0.5D. What is the focal length and type of lens of spectacles that she should use for correcting this defect?

  1. (a) 2.5 m, convex lens
  2. (b) −2.0 m, concave lens
  3. (c) −2.5 m, concave lens
  4. (d) 2.0 m, convex lens
Answer & solution

Answer: (b) −2.0 m, concave lens

Correct Answer: Option 2 — −2.0 m, concave lens
📖 Given
• Power of the corrective lens, P = −0.5 D
• Focal length is required in metres
✍️ Formula
P = 1/f, so f = 1/P (with f in metres and P in dioptres)
🔢 Solution
• f = 1/P
• f = 1/(−0.5)
• f = −2.0 m
• A negative focal length always belongs to a concave (diverging) lens
• So she needs a concave lens of focal length −2.0 m
A negative power means the defect is myopia (near-sightedness), in which distant objects look blurred; a concave lens diverges the incoming rays so that the image is pushed back onto the retina.
❌ Why the other options are incorrect
✗ 2.5 m, convex lens — a convex lens has positive power; it is used for hypermetropia, not for a −0.5 D defect.
✗ −2.5 m, concave lens — the lens type is right, but −2.5 m would correspond to a power of −0.4 D, not −0.5 D.
✗ 2.0 m, convex lens — the magnitude is right but the sign and the lens type are wrong; +2.0 m corresponds to +0.5 D.
📌 SSC & Railway Exam Facts
• Power P = 1/f (f in metres); SI unit dioptre (D), and 1 D = 1 m⁻¹.
• Convex lens → positive power; concave lens → negative power.
• Myopia → concave lens; Hypermetropia → convex lens; Presbyopia → bifocal lens.
• Lens formula: 1/v − 1/u = 1/f; magnification m = v/u.
• Power of lenses in contact adds up: P = P₁ + P₂.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 1 · Lens formula and magnification

Q83. Magnification produced by lens is related to the object-distance u, and the image-distance v. This relationship is given by:

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

Answer: (b) v/u

Correct Answer: Option 2 — v/u
The magnification of a lens tells us how many times larger (or smaller) the image is compared with the object. It is defined as
m = height of image / height of object = h′/h
and for a lens it is related to the distances by
m = v/u
where v is the image distance and u the object distance, both measured from the optical centre of the lens with the usual sign convention.
If |m| > 1 the image is enlarged, if |m| < 1 it is diminished, and if m = 1 it is the same size as the object. A positive m means a virtual and erect image, a negative m means a real and inverted image.
❌ Why the other options are incorrect
✗ u/v — this is the ratio turned upside down; it is the reciprocal of the magnification.
✗ u − v — subtracting two distances gives a length in metres, not a pure ratio; magnification has no unit.
✗ v − u — same error as above: a difference of distances can never measure how many times the image is enlarged.
📌 SSC & Railway Exam Facts
• For a lens: m = v/u; for a mirror: m = −v/u.
• Lens formula: 1/v − 1/u = 1/f; mirror formula: 1/v + 1/u = 1/f.
• Magnification is a pure number with no unit.
• Power of a lens P = 1/f (metres), unit dioptre (D).
• A convex lens can form both real and virtual images; a concave lens always forms a virtual, erect, diminished image.
RRB Technician Grade III 2024-25 – 27 December 2024 Shift 3 · Power of a lens and dioptre

Q84. A combination of two lenses of power +2.0 D and −0.25 D is equivalent to a single lens of power:

  1. (a) +2.25 D
  2. (b) +1.25 D
  3. (c) +1.75 D
  4. (d) −3.5 D
Answer & solution

Answer: (c) +1.75 D

Correct Answer: Option 3 — +1.75 D
📖 Given
• Power of the first lens, P₁ = +2.0 D (a converging / convex lens)
• Power of the second lens, P₂ = −0.25 D (a diverging / concave lens)
• The two lenses are placed in contact
✍️ Formula
P = P₁ + P₂ — powers of thin lenses in contact simply add, with their signs
🔢 Solution
• P = (+2.0) + (−0.25)
• P = 2.0 − 0.25
• P = +1.75 D
The positive sign shows that the combination still behaves as a converging (convex) lens. Its focal length is f = 1/P = 1/1.75 ≈ 0.57 m, that is about 57 cm.
❌ Why the other options are incorrect
✗ +2.25 D — this is 2.0 + 0.25; the minus sign of the second lens has been ignored while adding.
✗ +1.25 D — this would follow from subtracting 0.75 D, a value that is simply not given in the question.
✗ −3.5 D — this has both the wrong magnitude and the wrong sign; a +2.0 D lens dominates a −0.25 D lens, so the result must be positive.
📌 SSC & Railway Exam Facts
• Power P = 1/f (f in metres); SI unit dioptre (D), and 1 D = 1 m⁻¹.
• For lenses in contact: P = P₁ + P₂ + P₃ … and 1/f = 1/f₁ + 1/f₂ + ….
• Convex lens → positive power; concave lens → negative power.
• Lens formula: 1/v − 1/u = 1/f; magnification m = v/u.
• Myopia is corrected by a concave lens and hypermetropia by a convex lens.
RRB Technician Grade III 2024-25 – 26 December 2024 Shift 2 · Laws of refraction and Snell's law

Q85. For a given colour and given pair of media, and light is going from rarer to denser media, the ratio of sine of the angle of incidence to the sine of angle of refraction is:

  1. (a) always 1
  2. (b) always zero
  3. (c) a constant greater than 1
  4. (d) a constant less than 1
Answer & solution

Answer: (c) a constant greater than 1

Correct Answer: Option 3 — a constant greater than 1
According to Snell's law of refraction, for a given colour of light and a given pair of media,
sin i / sin r = n₂₁ = constant
This constant is the refractive index of the second medium with respect to the first.
When light travels from a rarer medium to a denser medium, it bends towards the normal, so the angle of refraction is smaller than the angle of incidence (r < i). Since sin i > sin r, the ratio sin i / sin r is greater than 1. For example, for air → glass the value is about 1.5, and for air → water about 1.33.
❌ Why the other options are incorrect
✗ always 1 — the ratio is 1 only when the two media are optically identical, so that light goes straight without bending.
✗ always zero — sin r can never make the ratio zero; a zero ratio would mean no refracted ray at all.
✗ a constant less than 1 — that happens the other way round, when light goes from a denser to a rarer medium and bends away from the normal.
📌 SSC & Railway Exam Facts
• Snell's law: n = sin i / sin r; the incident ray, refracted ray and normal all lie in one plane.
• Refractive index n = c/v, where c = 3 × 10⁸ m/s is the speed of light in vacuum.
• Diamond has the highest common refractive index, about 2.42.
• Refraction explains a pencil looking bent in water and the apparent raising of a pond's bottom.
• Light going from denser to rarer beyond the critical angle shows total internal reflection, used in optical fibres.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 1 · Numericals on refraction and lenses

Q86. An object of height 30 cm is placed in front of a concave lens and an image of height 10 cm is formed. The magnification of the lens is:

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

Answer: (b) 1/3

Correct Answer: Option 2 — 1/3
📖 Given
• Height of the object, h = 30 cm
• Height of the image, h' = 10 cm
• The lens is concave (diverging), so the image is virtual, erect and diminished
✍️ Formula
m = h'/h = v/u
🔢 Solution
• m = h'/h
• m = 10/30
• m = 1/3
A magnification of 1/3 (a positive fraction less than 1) tells us that the image is erect and only one-third the size of the object — exactly what a concave lens always produces.
❌ Why the other options are incorrect
✗ 3 — this is h/h' = 30/10, the ratio taken upside down; magnification is always image height divided by object height.
✗ 20 — this is 30 − 10, a subtraction of the two heights; magnification is a ratio, never a difference.
✗ −20 — this is 10 − 30, the same subtraction mistake with the sign reversed; the value is neither a ratio nor correct in sign, since a concave lens gives a positive magnification.
📌 SSC & Railway Exam Facts
• Magnification of a lens: m = h'/h = v/u; it has no unit.
• m positive ⇒ virtual and erect image; m negative ⇒ real and inverted image.
• |m| > 1 means enlarged, |m| < 1 means diminished, |m| = 1 means the same size.
• A concave (diverging) lens always gives a virtual, erect, diminished image, whatever the object position.
• Lens formula 1/v − 1/u = 1/f; power P = 1/f (in metres), measured in dioptre (D) — negative for a concave lens.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Power of a lens and dioptre

Q87. Which of the following is correct about the relationship between power and focal length of a lens?

  1. (a) Power is inversely proportional to focal length.
  2. (b) Power is inversely proportional to square of the focal length.
  3. (c) Power is directly proportional to focal length.
  4. (d) Power is directly proportional to square of the focal length.
Answer & solution

Answer: (a) Power is inversely proportional to focal length.

Correct Answer: Option 1 — Power is inversely proportional to focal length.
The power (P) of a lens measures its ability to converge or diverge a beam of light. It is defined as the reciprocal of the focal length taken in metres: P = 1/f. So a lens with a short focal length bends light sharply and therefore has a large power, while a lens with a long focal length bends light only gently and has a small power.
The SI unit of power of a lens is the dioptre (D), where 1 D = 1 m⁻¹ — the power of a lens whose focal length is 1 m. The power of a convex (converging) lens is positive and that of a concave (diverging) lens is negative.
❌ Why the other options are incorrect
✗ Power is inversely proportional to square of the focal length. — wrong; power depends on 1/f, not 1/f². The inverse-square rule belongs to the intensity of light, not to lens power.
✗ Power is directly proportional to focal length. — this reverses the relation; a larger focal length actually gives a weaker lens.
✗ Power is directly proportional to square of the focal length. — no such relation exists anywhere in ray optics.
📌 SSC & Railway Exam Facts
• P = 1/f with f in metres; SI unit dioptre (D).
• A convex lens of f = 50 cm = 0.5 m has power +2 D.
• For thin lenses in contact, P = P₁ + P₂.
• Myopia is corrected by a concave lens (negative power); hypermetropia by a convex lens (positive power).
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Power of a lens and dioptre

Q88. Which of the following statements is NOT correct?

  1. (a) The power of convex lens is positive.
  2. (b) Larger the focal length, smaller is the power.
  3. (c) Larger the focal length, larger the power.
  4. (d) The power of concave lens is negative.
Answer & solution

Answer: (c) Larger the focal length, larger the power.

Correct Answer: Option 3 — Larger the focal length, larger the power.
This statement is false, and since the question asks which one is NOT correct, it is the answer. The power of a lens is P = 1/f (f in metres), so power and focal length are inversely related. A larger focal length always means a smaller power, never a larger one.
By the New Cartesian sign convention, a convex lens has a positive focal length and hence a positive power, while a concave lens has a negative focal length and hence a negative power. The unit of power is the dioptre (D), equal to m⁻¹.
❌ Why the other options are incorrect
✗ The power of convex lens is positive. — this is true; a convex lens converges light and carries a + sign, so it cannot be the answer.
✗ Larger the focal length, smaller is the power. — true, and it is just the statement of P = 1/f in words.
✗ The power of concave lens is negative. — true; a concave lens diverges light and is given a − sign.
📌 SSC & Railway Exam Facts
• P = 1/f (metre); unit dioptre (D), given by Monoyer.
• f = +25 cm convex lens ⇒ P = +4 D; f = −50 cm concave lens ⇒ P = −2 D.
• Lens formula: 1/v − 1/u = 1/f; magnification m = v/u = h′/h.
• A thick, highly curved lens has a short focal length and therefore a high power.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 2 · Numericals on refraction and lenses

Q89. An object of height 30 cm is placed in front of a concave lens and an image of height 10 cm is formed. The magnification of the lens is:

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

Answer: (d) 1/3

Correct Answer: Option 4 — 1/3
📖 Given
• Height of object, h = 30 cm
• Height of image, h′ = 10 cm
• The lens is a concave (diverging) lens
✍️ Formula
Magnification, m = h′/h = v/u
🔢 Solution
• m = h′/h = 10/30
• m = 1/3
A concave lens always forms a virtual, erect and diminished image, so the magnification comes out positive and less than 1 — here the image is exactly one-third the size of the object.
❌ Why the other options are incorrect
✗ −20 — obtained by subtracting the heights (10 − 30 = −20); magnification is a ratio, never a difference.
✗ 20 — the same subtraction done the other way (30 − 10 = 20); again a difference, not a ratio.
✗ 3 — the ratio taken upside down, h/h′ = 30/10 = 3. That would mean the image is 3 times bigger, which a concave lens can never do.
📌 SSC & Railway Exam Facts
• For a lens, m = h′/h = v/u; for a mirror, m = h′/h = −v/u.
• |m| > 1 ⇒ enlarged image; |m| < 1 ⇒ diminished image; m = 1 ⇒ same size.
• A concave lens always gives a virtual, erect, diminished image between the optical centre and F.
• Lens formula 1/v − 1/u = 1/f; power P = 1/f (m) in dioptre.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Refractive index and optical density

Q90. The refractive index is related to:

  1. (a) the speed of propagation of light in two different media
  2. (b) the frequency of light in two different media
  3. (c) the magnetic moment of two different media
  4. (d) the dielectric polarizability of two media
Answer & solution

Answer: (a) the speed of propagation of light in two different media

Correct Answer: Option 1 — the speed of propagation of light in two different media
When light passes from one transparent medium into another it changes its speed, and this change of speed is what bends the ray. The refractive index is simply the number that measures this change. For light going from medium 1 into medium 2,
n₂₁ = speed of light in medium 1 / speed of light in medium 2 = v₁/v₂
When the first medium is vacuum (or air), we get the absolute refractive index, n = c/v, where c = 3 × 10⁸ m/s. A larger refractive index means light travels more slowly in that medium, so the medium is optically denser and bends the ray towards the normal.
By Snell's law, sin i / sin r = n₂₁, which is exactly the ratio of the two speeds. Refractive index is a ratio of two speeds, so it has no unit.
❌ Why the other options are incorrect
✗ the frequency of light in two different media — the frequency never changes during refraction; only the speed and wavelength change, so frequency cannot define the refractive index.
✗ the magnetic moment of two different media — a magnetic property of matter; it has nothing to do with how fast light travels through a transparent medium.
✗ the dielectric polarizability of two media — an electrical property; it influences the optical behaviour of a material indirectly, but refractive index is defined through speed, not polarizability.
📌 SSC & Railway Exam Facts
• n = c/v; refractive index is a pure number (no unit) and is always ≥ 1 for a real medium.
• n(water) = 1.33, n(glass) = 1.5, n(diamond) = 2.42 — the highest among common materials.
• Snell's law: n₁ sin i = n₂ sin r; it was given by Willebrord Snellius.
• Light travels fastest in vacuum, c = 3 × 10⁸ m/s; a denser medium slows it down.
RRB Technician Grade III 2024-25 – 24 December 2024 Shift 3 · Laws of refraction and Snell's law

Q91. According to the laws of refraction, which of the following is/are true?
(A) The incident ray, the refracted ray and the normal to the interface of two transparent media do not lie in the same plane.
(B) For a given colour of light and a given pair of media, the ratio of the sine of angle of incidence to sine of angle of refraction is a constant.

  1. (a) Both ‘A’ and ‘B’ are false.
  2. (b) ‘A’ is false and ‘B’ is true.
  3. (c) Both ‘A’ and ‘B’ are true.
  4. (d) ‘A’ is true and ‘B’ is false.
Answer & solution

Answer: (b) ‘A’ is false and ‘B’ is true.

Correct Answer: Option 2 — ‘A’ is false and ‘B’ is true.
There are exactly two laws of refraction, and the two statements test them one by one.
Statement (A) is false. The first law of refraction says the incident ray, the refracted ray and the normal to the interface at the point of incidence all lie in the SAME plane. Statement (A) says they do not, so it reverses the law.
Statement (B) is true. This is the second law of refraction, Snell's law: for a given colour (wavelength) of light and a given pair of media,
sin i / sin r = constant = n₂₁
This constant is the refractive index of the second medium with respect to the first. The conditions "given colour" and "given pair of media" matter, because the refractive index changes with wavelength — that is what causes dispersion.
❌ Why the other options are incorrect
✗ Both ‘A’ and ‘B’ are false. — wrong, because (B) is a correct statement of Snell's law.
✗ Both ‘A’ and ‘B’ are true. — wrong, because (A) contradicts the first law of refraction; the three do lie in one plane.
✗ ‘A’ is true and ‘B’ is false. — this reverses both judgements and is the exact opposite of the correct answer.
📌 SSC & Railway Exam Facts
• First law — incident ray, refracted ray and normal are coplanar.
• Second law (Snell's law) — sin i / sin r = n₂₁ = v₁/v₂.
• A ray entering a denser medium bends towards the normal; entering a rarer medium it bends away from the normal.
• A ray falling normally (i = 0°) goes straight through, undeviated.
• Dispersion happens because n is different for different colours — violet bends most, red least.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 1 · Refractive index and optical density

Q92. What is the unit of refractive index?

  1. (a) m³
  2. (b) m/s²
  3. (c) Unitless
  4. (d) m/s
Answer & solution

Answer: (c) Unitless

Correct Answer: Option 3 — Unitless
The refractive index of a medium tells us how much light slows down and hence bends when it enters that medium. It is defined as
n = speed of light in vacuum / speed of light in the medium = c/v
Both the numerator and the denominator are speeds measured in m/s, so the units cancel out completely. The refractive index is therefore a pure number — dimensionless and unitless.
The same conclusion follows from Snell's law, n = sin i / sin r: the sine of an angle is itself a ratio of two lengths, so it has no unit either. A medium with a higher refractive index is optically denser and bends light more towards the normal.
❌ Why the other options are incorrect
✗ m³ — the cubic metre is the SI unit of volume, which has nothing to do with a ratio of speeds.
✗ m/s² — this is the unit of acceleration, not of an optical ratio.
✗ m/s — this is the unit of speed itself; but refractive index is the ratio of two speeds, so the unit cancels.
📌 SSC & Railway Exam Facts
• n = c/v and Snell's law: n = sin i/sin r — both give a dimensionless number.
• Speed of light in vacuum, c = 3 × 10⁸ m/s — the maximum possible speed in nature.
• Refractive indices: air ≈ 1.0003, water = 1.33, glass ≈ 1.5, diamond = 2.42 (the highest among common media).
• A higher n means a smaller speed of light in that medium and greater bending towards the normal.
• Other unitless quantities: relative density (specific gravity), strain, magnification, coefficient of friction, angle in radian.
RRB Technician Grade III 2024-25 – 23 December 2024 Shift 3 · Laws of refraction and Snell's law

Q93. Which of the following statements is NOT correct?

  1. (a) The angle of incidence and refraction are equal.
  2. (b) The speed of light in the given medium is different for different colours,
  3. (c) The incident ray, refracted ray and normal to the surface at incidence lie in the same plane.
  4. (d) The speed of light in different media is different.
Answer & solution

Answer: (a) The angle of incidence and refraction are equal.

Correct Answer: Option 1 — The angle of incidence and refraction are equal.
This statement is wrong. When light passes from one transparent medium into another it bends — and bending means precisely that the angle of refraction (r) is different from the angle of incidence (i). The two are related by Snell's law: sin i / sin r = n, where n is the refractive index of the second medium with respect to the first.
The two angles become equal only in the trivial cases of normal incidence (i = r = 0°) or when both media are optically identical. "Angle of incidence = angle of reflection" is a law of reflection, and the question is about refraction, so this statement cannot be correct.
❌ Why the other options are incorrect
✗ The speed of light in the given medium is different for different colours — this is true; violet travels slowest and red fastest in glass, which is why white light splits into seven colours (dispersion) in a prism.
✗ The incident ray, refracted ray and normal to the surface at incidence lie in the same plane — true; this is the first law of refraction.
✗ The speed of light in different media is different — true; light travels at 3 × 10⁸ m/s in vacuum but more slowly in water and in glass.
📌 SSC & Railway Exam Facts
• Snell's law: n = sin i / sin r = c/v; refractive index has no unit.
• Refractive index of water = 1.33, glass = 1.5, diamond = 2.42 (highest among common materials).
• Light bends towards the normal entering a denser medium and away from the normal entering a rarer medium.
• Total internal reflection occurs from denser to rarer medium beyond the critical angle — glass ≈ 42°, water ≈ 48.6°, diamond ≈ 24°.
• Twinkling of stars, the apparent bending of a stick in water and the early sunrise are all effects of refraction.
RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Lens formula and magnification

Q94. According to the sign convention, the focal length of a convex lens is _______ and that of a concave lens is _______.

  1. (a) negative; positive
  2. (b) positive; positive
  3. (c) negative; negative
  4. (d) positive; negative
Answer & solution

Answer: (d) positive; negative

Correct Answer: Option 4 — positive; negative
In the New Cartesian Sign Convention for spherical lenses, all distances are measured from the optical centre of the lens, distances measured in the direction of the incident light are taken positive and those measured against it are taken negative.
A convex lens is a converging lens — a beam of light parallel to the principal axis meets at a real focus on the far side of the lens, i.e. on the side where the light is travelling. Hence its focal length f is positive.
A concave lens is a diverging lens — the parallel beam spreads out and only appears to come from a virtual focus on the same side as the incoming light. Hence its focal length f is negative.
The same signs carry over to power, P = 1/f (in metres): a convex lens has positive power and a concave lens negative power.
❌ Why the other options are incorrect
✗ negative; positive — this reverses the two lenses; it would make the converging lens diverging and vice versa.
✗ positive; positive — a concave lens can never have a positive focal length because its focus is virtual.
✗ negative; negative — a convex lens forms a real focus, so its focal length cannot be negative.
📌 SSC & Railway Exam Facts
• Lens formula: 1/v − 1/u = 1/f; mirror formula: 1/v + 1/u = 1/f.
• Power of a lens, P = 1/f (m); SI unit dioptre (D), and 1 D = 1 m⁻¹.
• Convex lens → f positive, used to cure hypermetropia (long sight); concave lens → f negative, used to cure myopia (short sight).
• For mirrors, a concave mirror has negative f and a convex mirror positive f in the same convention.
• Magnification of a lens, m = v/u = h'/h.
RRB Technician Grade III 2024-25 – 20 December 2024 Shift 3 · Numericals on refraction and lenses

Q95. What is the power of this combination of lens placed together when the power of the converging lens is 4.50 D and diverging lens is 3.00 D?

  1. (a) −7.5D
  2. (b) −1.5D
  3. (c) +1.5D
  4. (d) −13.5D
Answer & solution

Answer: (c) +1.5D

Correct Answer: Option 3 — +1.5D
📖 Given
• Power of the converging (convex) lens, P₁ = +4.50 D (positive, because a convex lens converges light)
• Power of the diverging (concave) lens, P₂ = −3.00 D (negative, because a concave lens diverges light)
• The two lenses are placed in contact with each other
✍️ Formula
For thin lenses in contact, P = P₁ + P₂ (powers simply add algebraically)
🔢 Solution
• P = (+4.50) + (−3.00)
• P = 4.50 − 3.00
• P = +1.5 D
The positive sign tells us the combination as a whole still behaves like a converging lens, but a weaker one; its focal length is f = 1/P = 1/1.5 = 0.67 m ≈ 66.7 cm.
❌ Why the other options are incorrect
✗ −7.5D — obtained by adding the two magnitudes and then wrongly making the answer negative (4.5 + 3.0 = 7.5).
✗ −1.5D — the right magnitude but the wrong sign; it would mean the combination diverges, which is wrong because the converging lens is the stronger of the two.
✗ −13.5D — obtained by multiplying 4.5 by 3, which is not how powers combine.
📌 SSC & Railway Exam Facts
• Lenses in contact: P = P₁ + P₂ + P₃ … and 1/f = 1/f₁ + 1/f₂ + …
• Power P = 1/f (in metres); SI unit dioptre (D), so f = 1/P.
• Convex lens → power positive; concave lens → power negative.
• A lens of focal length 50 cm = 0.5 m has a power of +2 D.
• An optician's prescription of −2.5 D means a concave lens of focal length 40 cm, used for myopia.
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 2 · Laws of refraction and Snell's law

Q96. Which law of refraction states that light bends due to a change in speed when passing from one medium to another?

  1. (a) Huygens' principle
  2. (b) Law of reflection
  3. (c) Snell's law
  4. (d) Fermat’s principle
Answer & solution

Answer: (c) Snell's law

Answer: Option 3 — Snell's law
Snell's law links the bending of light to the ratio of its speeds in the two media: sin i / sin r = v₁ / v₂ = constant (the refractive index).
Revision note — Refraction of light
• Refraction: bending of light when it passes from one medium to another, because its speed changes.
• First law: incident ray, refracted ray and the normal lie in one plane.
• Second law (Snell's law): sin i / sin r = constant = refractive index of medium 2 w.r.t. medium 1.
• Refractive index n = c / v (speed in vacuum ÷ speed in the medium); n has no unit.
• Rarer → denser: ray bends towards the normal; denser → rarer: away from the normal.
• Examples: a pencil looks bent in water; a pool looks shallower; stars twinkle.
Do not mix these up
• Snell's law vs Huygens' principle — Snell's law is the refraction law; Huygens' principle is a wave construction used to derive it.
Also asked from this topic: the cause of light bending from air into water (change in speed); the law that incident ray, refracted ray and normal lie in one plane (first law of refraction).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 3 · Convex lens - image formation and uses

Q97. A lens has ________ principal foci.

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

Answer: (c) 2

Answer: Option 3 — 2
Light can pass through a lens from either side, so a lens has two principal foci, F₁ and F₂, one on each side of the optical centre.
Revision note — Terms of a spherical lens
• Optical centre (O) — the central point of the lens; a ray through it goes undeviated.
• Principal focus — convex lens: parallel rays converge to it (real); concave lens: they appear to diverge from it (virtual).
• Focal length (f) — distance from the optical centre to a principal focus.
• Lens formula: 1/v − 1/u = 1/f; magnification m = v/u.
• Power P = 1/f (f in metres); unit dioptre (D); convex lens +, concave lens −.
• Convex lens: object between F and O ⇒ image virtual, erect, magnified (magnifying glass).
Do not mix these up
• Lens vs mirror — a lens has two principal foci; a spherical mirror has only one, because light never passes through a mirror.
Also asked from this topic: a ray through the optical centre (undeviated); lens that focuses parallel rays (convex); image of an object between O and F of a convex lens (virtual, erect, magnified).
RRB JE CBT-1 2024-25 – 18 December 2024 Shift 3 · Concave lens - image formation and uses

Q98. What type of image does a concave lens form when an object is placed at infinity?

  1. (a) Real, inverted and same size
  2. (b) Virtual, erect and larger
  3. (c) Real, erect and point sized
  4. (d) Virtual, erect and point sized
Answer & solution

Answer: (d) Virtual, erect and point sized

Answer: Option 4 — Virtual, erect and point sized
Parallel rays from an object at infinity diverge after passing through a concave lens.
They appear to come from the focus F_(1).
So the image is virtual, erect and highly diminished (point-sized), on the same side as the object.
Revision note — Image formation by a concave lens
• Object at infinity → image at F_(1); virtual, erect, highly diminished (point-sized).
• Object anywhere between infinity and the lens → image between F_(1) and the optical centre; virtual, erect, diminished.
• A concave lens always gives a virtual, erect, diminished image — it can never give a real image of a real object.
• Sign convention: focal length and power of a concave lens are negative.
• Uses: correcting myopia, peepholes in doors, the upper part of bifocal lenses.
Do not mix these up
• Concave lens vs convex lens (object at infinity) — convex gives a real, inverted, point-sized image at F_(2) on the other side; concave gives a virtual, erect one at F_(1) on the same side.
Also asked from this topic: image size by a concave lens for any object (always diminished); a lens of power −2.5 D (concave); the lens used to correct myopia (concave).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 1 · Convex lens - image formation and uses

Q99. Which of the following describes the relationship between the aperture of a lens and its radius of curvature in the context of thin lenses?

  1. (a) The aperture is much smaller than the radius of curvature.
  2. (b) The aperture is independent of the radius of curvature.
  3. (c) The aperture equals the radius of curvature.
  4. (d) The aperture is much larger than the radius of curvature.
Answer & solution

Answer: (a) The aperture is much smaller than the radius of curvature.

Answer: Option 1 — The aperture is much smaller than the radius of curvature.
NCERT studies only lenses whose aperture is much less than the radius of curvature.
Such lenses are called thin lenses with small apertures; the lens formula and ray rules hold for them.
Revision note — Spherical lens terms
• Aperture: the effective diameter of the circular outline of a lens.
• Centre of curvature (C): centre of the sphere of which a lens surface is a part; a lens has two (C_(1), C_(2)).
• Principal axis: the line through both centres of curvature.
• Optical centre (O): the central point of the lens; a ray through it passes undeviated.
• Power P = 1/f (f in metres); SI unit dioptre (D).
• Convex lens converges light (f positive); concave lens diverges light (f negative).
Do not mix these up
• Aperture vs focal length — aperture is a width of the lens; focal length is a distance along the axis.
Also asked from this topic: the path of a ray through the optical centre of a thin lens (it goes straight on without deviation).
RRB JE CBT-1 2024-25 – 17 December 2024 Shift 2 · Refractive index and optical density

Q100. Which of the following statements is true regarding the refractive index?

  1. (a) Refractive index of glass is less than 1.
  2. (b) Refractive index of a given medium is independent of the wavelength of the light.
  3. (c) Refractive index depends only on the speed of light in vacuum.
  4. (d) Refractive index of water is always greater than 1.
Answer & solution

Answer: (d) Refractive index of water is always greater than 1.

Answer: Option 4 — Refractive index of water is always greater than 1.
Refractive index n = speed of light in vacuum ÷ speed of light in the medium.
Light is slower in water than in vacuum, so n of water (1.33) is more than 1.
Revision note — Refractive index
• n = c ÷ v; it is a ratio of two speeds, so it has no unit.
• Values: vacuum 1, air 1.0003, water 1.33, crown glass 1.52, diamond 2.42 (highest among common media).
• Every material medium has n > 1, because light is fastest in vacuum.
• n depends on the wavelength — violet light has a larger n than red, which is why a prism causes dispersion.
• Higher n = optically denser medium; light going into it bends towards the normal.
• Relative index n₂₁ = v₁ ÷ v₂ (speed in medium 1 ÷ speed in medium 2).
Do not mix these up
• Optical density vs mass density — kerosene is optically denser than water but lighter than water.
Also asked from this topic: the refractive index of a solid where light slows from 3 × 10⁸ m/s to 2 × 10⁸ m/s (1.5); what a refractive index of 1.33 for water means (light travels 1.33 times slower in water than in air).
12

More Physics chapters

Work, Energy and Power196 questionsElectricity187 questionsMagnetism and Magnetic Effects of Current129 questionsMotion123 questionsSound and Waves123 questionsForce and Laws of Motion112 questionsLight - Reflection and Mirrors106 questionsGravitation90 questionsHuman Eye and the Colourful World85 questionsUnits, Measurement & Dimensions56 questionsHeat and Thermodynamics47 questionsSources of Energy38 questionsPhysics Miscellaneous33 questionsPressure, Fluids and Properties of Matter30 questionsModern Physics and Electronics24 questions
📥 PDF of this paper (all subjects, English & Hindi): All previous year papers PDF · ▶ Attempt previous year papers as free online tests · 🏆 Free live mock tests · 📰 Daily current affairs quiz