Ray Optics — Previous Year Questions (Class 12 Physics)
Ray Optics covers lenses, mirrors, optical instruments, and combinations. CBSE and JEE test lens formula, magnification, eye defects, and complex systems.
TL;DR: Ray Optics covers lenses, mirrors, optical instruments, and combinations. CBSE and JEE test lens formula, magnification, eye defects, and complex syst…
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Ray Optics covers lenses, mirrors, optical instruments, and combinations. CBSE and JEE test lens formula, magnification, eye defects, and complex systems.
Ray Optics — Previous Year Questions with Solutions
Q (2023, 3 marks): A lens has a focal length of -15 cm. Is it a converging or diverging lens? If an object is placed 30 cm from the lens, find the image distance and magnification.
Answer: f = -15 cm (negative), so it is a Diverging Lens (concave).
u = -30 cm (object distance, real object, so negative in lens convention)
Using lens formula: 1/f = 1/u + 1/v
1/(-15) = 1/(-30) + 1/v
-1/15 = -1/30 + 1/v
1/v = -1/15 + 1/30 = (-2 + 1)/30 = -1/30
v = -30 cm
Image is virtual (v negative), 30 cm from the lens on the same side as the object.
Magnification m = -v/u = -(-30)/(-30) = -1
Image is upright, same size as object.
Final Answer: v = -30 cm (virtual), magnification = -1 (upright, same size)
Q (2022, 5 marks): A compound microscope has an objective lens of focal length 0.5 cm and an eyepiece of focal length 5 cm. The object is placed 0.6 cm from the objective. Find the magnification.
Answer: Objective: f_o = 0.5 cm, u_o = 0.6 cm (object distance)
Eyepiece: f_e = 5 cm
Using lens formula for objective: 1/f_o = 1/u_o + 1/v_o
1/0.5 = 1/0.6 + 1/v_o
2 = 10/6 + 1/v_o
1/v_o = 2 - 10/6 = (12 - 10)/6 = 2/6 = 1/3
v_o = 3 cm (real image from objective)
Magnification of objective: m_o = -v_o/u_o = -3/0.6 = -5
For eyepiece, the image from objective acts as object (at focal length if final image is at infinity):
For relaxed vision, image from objective is at focal point of eyepiece.
Magnification of eyepiece (at infinity): m_e = D/f_e where D = 25 cm (near point distance)
m_e = 25/5 = 5
Total magnification = |m_o| × |m_e| = 5 × 5 = 25
Final Answer: Total magnification = 25
Q (2021, 2 marks): A myopic person can see clearly up to 50 cm. What should be the focal length of the lens to correct this defect for the far point at infinity?
Answer: Myopia (short-sightedness): Far point is at 50 cm instead of infinity.
To correct, the lens should bring objects at infinity (v = ∞) to the far point at 50 cm (u = -50 cm, taking real object as negative).
Using lens formula: 1/f = 1/u + 1/v
1/f = 1/(-50) + 1/∞
1/f = -1/50
f = -50 cm
A diverging (concave) lens of focal length -50 cm is needed.
Power P = 1/f = 1/(-0.5) = -2 diopters
Final Answer: f = -50 cm (or power = -2 D)
Q (2023, 5 marks): Define and derive the expression for lateral magnification of a lens.
Answer: Lateral Magnification: The ratio of the image height (h') to the object height (h).
m = h'/h
Derivation using similar triangles:
Consider an object of height h at distance u and image of height h' at distance v.
By geometry and properties of similar triangles formed by the lens:
The triangles formed by the object and its rays through the optical center, and the image and its rays, are similar.
From similar triangles:
h'/h = v/u
m = h'/h = v/u (sign convention: object and image on opposite sides give opposite signs)
Alternatively, m = -v/u (with sign to indicate inversion)
Properties:
1. If m is positive, image is upright.
2. If m is negative, image is inverted.
3. If |m| > 1, image is magnified.
4. If |m| < 1, image is diminished.
5. If |m| = 1, image is same size.
Final Answer: m = h'/h = v/u; negative when inverted, positive when upright.
Q (2022, 3 marks): A telescope has an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm. Find the magnifying power and the length of the telescope in normal adjustment.
Answer: Telescope (refracting):
Objective: f_o = 100 cm
Eyepiece: f_e = 5 cm
Magnifying power (at infinity): M = f_o / f_e = 100 / 5 = 20
Length of telescope in normal adjustment (when final image is at infinity, i.e., relaxed viewing):
L = f_o + f_e = 100 + 5 = 105 cm
Alternative formula: M = -(f_o / f_e) if considering inversion (magnitude is 20).
Final Answer: Magnifying power = 20, Length = 105 cm
Q (2021, 3 marks): State the lens maker's formula and explain the terms.
Answer: Lens Maker's Formula: 1/f = (n - 1)[1/R1 - 1/R2]
Where:
f = focal length of the lens
n = refractive index of lens material (relative to surrounding medium, usually air where n_air ≈ 1)
R1 = radius of curvature of the first surface
R2 = radius of curvature of the second surface
Sign Convention:
- R is positive if center of curvature is on the side of the outgoing light.
- R is negative if center of curvature is on the side of the incoming light.
Example:
For a biconvex lens in air: both surfaces curve outward, so both R are positive.
1/f = (n - 1)[1/R1 + 1/R2], giving positive f (converging lens).
For a biconcave lens: both surfaces curve inward, so both R are negative.
1/f = (n - 1)[-1/R1 - 1/R2], giving negative f (diverging lens).
Final Answer: 1/f = (n - 1)[1/R1 - 1/R2]; allows calculation of focal length from geometry and material properties.
Frequently Asked Questions
What is the difference between linear and angular magnification?
Linear magnification (m) is for lenses and mirrors: ratio of image height to object height. Angular magnification is for optical instruments like telescopes and microscopes: ratio of angle subtended by image to angle subtended by object.
Why is a diverging lens used to correct myopia?
Myopia means the eye focuses light in front of the retina. A diverging lens spreads light rays, shifting the focal point backward onto the retina, allowing clear vision of distant objects.
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