CBSE Class 12 Physics Optics Objective Questions

CBSE Class 12 Physics Optics Objective Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

OBJECTIVE
CLASS 12 PHYSICS : OPTICS

1. A beam of light is coverging towards a point on the screen. If a plane parallel plate of glass of refractive index μ and thickness t is introduced in the path of the beam, the covergent point is shifted
(a) t \( \left[ 1 - \frac{1}{\mu} \right] \) away
(b) t \( \left[ 1 + \frac{1}{\mu} \right] \) nearer
(c) t \( \left[ 1 + \frac{1}{\mu} \right] \) away
(d) t \( \left[ 1 - \frac{1}{\mu} \right] \) nearer

Answer: (a) t \( \left[ 1 - \frac{1}{\mu} \right] \) away

1. A glass plate of thickness t and refractive index μ increases the optical path of the beam; the convergent point moves forward, away from the plate.
2. The normal shift produced by a plane parallel plate is t(1 - 1/μ).
3. Hence, the point shifts by t(1 - 1/μ) away and the correct option is (a).

Teacher's Note:
a) The shift is in the direction of propagation of the light.
b) The formula for the shift is the same as that for the apparent shift of an object seen through a slab.

2. A ray of light from a denser medium strikes a rarer medium at ∠i. The reflected and refracted rays make an angle of 90º with each other. The angle of reflection and refraction are r and r' respectively. The critical angle is
(a) sin^-1 (tan r)
(b) sin^-1 (tan i)
(c) sin^-1 (tan r')
(d) tan^-1(sin i)

Answer: (b) sin^-1 (tan i)

1. The reflected and refracted rays are at 90°, so r + r' = 90°, and since r = i we get r' = 90° - i.
2. By Snell's law, μ_d sin i = μ_r sin r' = μ_r cos i, so μ_r/μ_d = tan i.
3. The critical angle is given by sin C = μ_r/μ_d = tan i.
4. C = sin⁻¹(tan i). Hence, the correct option is (b).

Teacher's Note:
a) By the law of reflection r = i, so option (a) is numerically the same as option (b); the option written directly in terms of the angle of incidence is chosen.
b) The ratio of refractive indices is the sine of the critical angle.

3. A lens behaves as a converging lens in air and diverging lens in water. The refractive index of material is
(a) equal to 1
(b) between 1 and 1.33
(c) 1.33
(d) > 1.33

Answer: (b) between 1 and 1.33

1. A lens is converging in air only if its refractive index is greater than that of air, so μ is greater than 1.
2. The lens becomes diverging in water when the refractive index of its material is less than that of water, so μ is less than 1.33.
3. Hence, 1 is less than μ which is less than 1.33, and the correct option is (b).

Teacher's Note:
a) A lens behaves as converging when the refractive index of the lens is greater than that of the surrounding medium.
b) If μ were equal to 1.33, the lens would disappear in water.

4. A point source of light is placed at the bottom of a vessel containing a liquid of refractive index 5/3. A person is viewing the source from above the surface. There is an opaque disc of radius 1 cm floating on the surface. The centre of the disc lies vertically above the source. The liquid from the vessel is gradually drained out through a tap. The maximum height of the liquid for which the source cannot be seen at all from above is
(a) \( \frac{3}{2} \) cm
(b) \( \frac{4}{3} \) cm
(c) \( \frac{2}{3} \) cm
(d) \( \frac{3}{4} \) cm

Answer: (b) \( \frac{4}{3} \) cm

1. The source cannot be seen if all the rays that leave the surface outside the disc are totally reflected. The critical angle satisfies sin C = 1/μ = 3/5.
2. So tan C = 3/4.
3. The ray at the critical angle just reaches the edge of the disc: radius = h tan C, so 1 = h × 3/4.
4. h = 4/3 cm.
5. Hence, the correct option is (b).

Teacher's Note:
a) Rays hitting the surface at more than the critical angle cannot come out.
b) The maximum height is when the disc just covers the circle from which light could escape.

5. A glass cube of each side 4 cm and μ equal to 1.5 has a point object at its centre. The circular area which should be covered on each face of the cube to prevent the spot from being observed from any direction for viewing is
(a) 2.51 cm²
(b) 5.02 cm²
(c) 10.04 cm²
(d) 20.08 cm²

Answer: (c) 10.04 cm²

1. sin C = 1/1.5 = 2/3, so tan C = 2/√5 = 0.894.
2. The distance from the centre to each face is 2 cm, so the radius of the circle through which light can escape is r = 2 tan C = 1.79 cm.
3. Area = πr² = π × 3.2 = 10.05 cm².
4. Hence, the correct option is (c).

Teacher's Note:
a) The area to be covered is the circle of radius h tan C on each face.
b) Use h = 2 cm, which is half the side of the cube.

6. An object is placed 20 cm in front of a block of glass 10 cm thick having its farther end silvered. The image is formed 33.2 cm behind the silvered surface. The refractive index of glass is
(a) 1.41
(b) 1.46
(c) 1.16
(d) 1.61

Answer: (c) 1.16

1. The silvered face acts as a plane mirror. For an observer inside the glass, the object appears at a distance 20μ in front of the front face (the apparent depth is multiplied by μ).
2. The distance of this apparent object from the silvered face = 20μ + 10.
3. A plane mirror forms an image as far behind it as the object is in front, so 20μ + 10 = 33.2.
4. μ = 23.2/20 = 1.16.
5. Hence, the correct option is (c).

Teacher's Note:
a) The refraction at the front face has to be treated separately from the reflection.
b) Check: 20 × 1.16 + 10 = 33.2.

7. An air bubble inside a glass slab of refractive index 1.5 appears to be 6 cm deep when viewed from one side and 4 cm deep when viewed from the other side. The thickness of slab is
(a) 10 cm
(b) 6.67 cm
(c) 9 cm
(d) 15 cm

Answer: (d) 15 cm

1. For viewing from air, the apparent depth = real depth/μ.
2. From one side: real depth d₁ = 6 × 1.5 = 9 cm. From the other side: d₂ = 4 × 1.5 = 6 cm.
3. Thickness of the slab = d₁ + d₂ = 9 + 6 = 15 cm.
4. Hence, the correct option is (d).

Teacher's Note:
a) The two real depths add up to the thickness of the slab.
b) The bubble is seen closer than it really is.

8. A ray of light is incident at one face of an equilateral prism of μ equal to 1.5. If the ray after refraction through the prism, travels along the grazing angle at the opposite face, the angle of incident ray, is
(a) 18.2º
(b) 36.4º
(c) 27.9º
(d) 29.3º

Answer: (c) 27.9º

1. For grazing emergence, the angle of emergence is 90°, so the angle of incidence on the second face is the critical angle: sin r₂ = 1/1.5, so r₂ = 41.8°.
2. For the prism, r₁ + r₂ = A = 60°, so r₁ = 60° - 41.8° = 18.2°.
3. sin i = μ sin r₁ = 1.5 × sin 18.2° = 1.5 × 0.312 = 0.468.
4. i = 27.9°.
5. Hence, the correct option is (c).

Teacher's Note:
a) Grazing emergence means that the ray hits the second face at the critical angle.
b) Do not confuse the answer with r₁ = 18.2°, which is a distractor.

9. A prism has two acute angles of 30º each and refractive index of 1.5. If two parallel light rays AB and CD are incident as shown in the figure, the angle between the emergent rays is sin^-1 (0.75) = 48.6º.
(a) 18.6°
(b) 37.2º
(c) 48.6º
(d) 52º

Answer: (b) 37.2º

1. The rays AB and CD fall normally on the vertical face, so they go undeviated into the prism and meet the slant faces at an angle of incidence of 30°.
2. For each ray, sin r = 1.5 sin 30° = 0.75, so r = 48.6°.
3. The deviation of each ray = 48.6° - 30° = 18.6°, and the two rays are deviated towards each other.
4. Angle between the emergent rays = 2 × 18.6° = 37.2°.
5. Hence, the correct option is (b).

Teacher's Note:
a) Each ray is bent through 18.6°, so the angle between them is double.
b) The angle of incidence at the slant face equals the base angle of the prism, 30°.

10. A crown glass prism of 5º has refractive index for violet and red colours as 1.523 and 1.514 respectively. It is combined with a flint glass prism of 4.2º which has got refractive indices of 1.632 and 1.613 for violet and red colours respectively. The flint glass prism is combined with crown glass prism in opposite direction. The net angular dispersion is
(a) 0.124º
(b) 0.348º
(c) 0.0348º
(d) 0.0124º

Answer: (c) 0.0348º

1. Angular dispersion of the crown prism = (μ_v - μ_r)A = (1.523 - 1.514) × 5 = 0.009 × 5 = 0.045°.
2. Angular dispersion of the flint prism = (1.632 - 1.613) × 4.2 = 0.019 × 4.2 = 0.0798°.
3. The prisms are in opposite directions, so the net angular dispersion = 0.0798 - 0.045 = 0.0348°.
4. Hence, the correct option is (c).

Teacher's Note:
a) For prisms in opposite positions subtract the dispersions.
b) Angular dispersion = (μ_v - μ_r)A for a thin prism.

11. A projector lens of focal length 10 cm magnifies the image of a 2 cm × 2 cm slide on a screen at a distance of 5 m from the lens. The size of the picture on the screen is
(a) 200 cm²
(b) 2401 cm²
(c) 4802 cm²
(d) 9604 cm²

Answer: (d) 9604 cm²

1. The image distance v = 5 m = 500 cm and f = 10 cm. From 1/u + 1/v = 1/f (magnitudes, real image): 1/u = 1/10 - 1/500 = 49/500, so u = 500/49 cm.
2. Linear magnification m = v/u = 500/(500/49) = 49.
3. Each side of the picture = 49 × 2 = 98 cm, so the area = 98 × 98 = 9604 cm².
4. Hence, the correct option is (d).

Teacher's Note:
a) The area magnification is the square of the linear magnification, 49² = 2401, times the slide area of 4 cm².
b) Take care to convert 5 m to 500 cm.

12. A thin prism P₁ with angle 4º and made from glass of refractive index 1.54 is combined with another thin prism P₂ made of another type of glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism P₂ is
(a) 2.6º
(b) 3º
(c) 4º
(d) 5.33º

Answer: (b) 3º

1. For no deviation, the deviations of the two prisms are equal and opposite: (μ₁ - 1)A₁ = (μ₂ - 1)A₂.
2. (1.54 - 1) × 4 = (1.72 - 1) × A₂, so 2.16 = 0.72 A₂.
3. A₂ = 3°.
4. Hence, the correct option is (b).

Teacher's Note:
a) A combination with no deviation but with dispersion is the direct vision prism.
b) The prism with the larger refractive index has the smaller angle.

13. A ray of light passes from vacuum into a medium of refractive index μ. If the angle of incidence is twice the angle of refraction, then the angle of incidence is
(a) cos⁻¹ \( \left( \frac{\mu}{2} \right) \)
(b) sin⁻¹ \( \left( \frac{\mu}{2} \right) \)
(c) 2 cos⁻¹ \( \left( \frac{\mu}{2} \right) \)
(d) 2sin⁻¹ \( \left( \frac{\mu}{2} \right) \)

Answer: (c) 2 cos⁻¹ \( \left( \frac{\mu}{2} \right) \)

1. Let the angle of refraction be r, so the angle of incidence is 2r. By Snell's law, sin 2r = μ sin r.
2. 2 sin r cos r = μ sin r, so cos r = μ/2 and r = cos⁻¹(μ/2).
3. The angle of incidence = 2r = 2 cos⁻¹(μ/2).
4. Hence, the correct option is (c).

Teacher's Note:
a) Use the identity sin 2r = 2 sin r cos r to eliminate sin r.
b) The answer must be double the angle of refraction.

14. A ray of light entering a plane surface from air to glass of refractive index 1.5 is partly reflected and partly refracted. If the incident and the reflected lights are at right angles to each other, the angle of refraction is
(a) sin⁻¹ \( \left( \sqrt{\frac{2}{3}} \right) \)
(b) sin⁻¹ \( \left( \frac{\sqrt{2}}{3} \right) \)
(c) sin⁻¹ \( \left( \frac{2}{\sqrt{3}} \right) \)
(d) sin⁻¹ \( \left( \frac{1}{\sqrt{3}} \right) \)

Answer: (b) sin⁻¹ \( \left( \frac{\sqrt{2}}{3} \right) \)

1. The angle between the incident and reflected rays = 2i = 90°, so the angle of incidence i = 45°.
2. By Snell's law, sin 45° = 1.5 sin r, so sin r = (1/√2)/1.5 = √2/3.
3. r = sin⁻¹(√2/3).
4. Hence, the correct option is (b).

Teacher's Note:
a) The angle between incident and reflected rays is twice the angle of incidence.
b) Do not confuse this with the Brewster condition, where the reflected and refracted rays are at 90°.

15. A concave mirror of 30 cm focal length is used to obtain an image thrice the size of the object. The object should be placed in front of concave mirror at a distance of
(a) 10 cm
(b) 20 cm
(c) 30 cm
(d) 40 cm

Answer: (b) 20 cm and (d) 40 cm

1. For a real image, m = -3: with f = -30 cm, the relation u = f(m - 1)/m gives u = (-30)(-4)/(-3) = -40 cm, so the object is 40 cm from the mirror.
2. For a virtual (erect) image, m = +3: u = f(m - 1)/m = -30 × 2/3 = -20 cm, so the object is 20 cm from the mirror.
3. A concave mirror gives a magnified image of either kind, so both distances are possible.
4. Hence, the correct options are (b) 20 cm and (d) 40 cm.

Teacher's Note:
a) The question says only "thrice the size", so the image can be real and inverted or virtual and erect.
b) The object is between the pole and the focus for the virtual image, and beyond the focus for the real image.

16. A concave mirror of focal length f produces an image m times the linear size of object. If the image formed is real, the distance of the object from the mirror is
(a) \( \frac{m - 1}{f} \)
(b) \( \frac{(m - 1) f}{m} \)
(c) \( \frac{(m + 1) f}{m} \)
(d) (m + 1) f

Answer: (c) \( \frac{(m + 1) f}{m} \)

1. Let the object distance be u and the image distance v (both magnitudes). For a real image, v = mu.
2. The mirror formula for a real image: 1/u + 1/v = 1/f, so 1/u + 1/(mu) = 1/f.
3. (m + 1)/(mu) = 1/f, so u = (m + 1)f/m.
4. Hence, the correct option is (c).

Teacher's Note:
a) A real image formed by a concave mirror is inverted, with |m| = v/u.
b) Check with m = 1: u = 2f, as expected.

17. A short linear object of length l lies on the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. The size of the image is approximately
(a) l \( \left[ \frac{u - f}{f} \right]^{1/2} \)
(b) l \( \left[ \frac{f}{u - f} \right]^{1/2} \)
(c) l \( \left[ \frac{u - f}{f} \right] \)
(d) l \( \left[ \frac{f}{u - f} \right]^2 \)

Answer: (d) l \( \left[ \frac{f}{u - f} \right]^2 \)

1. From the mirror formula, v = uf/(u - f) (magnitudes).
2. Differentiating, dv/du = -f²/(u - f)². The longitudinal magnification has the magnitude f²/(u - f)².
3. The image size for a short object of length l along the axis = l × [f/(u - f)]².
4. Hence, the correct option is (d).

Teacher's Note:
a) The longitudinal magnification is the square of the transverse magnification m = f/(u - f).
b) The approximation holds when the object is small, so that differentiation is valid.

18. An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm and there is no parallax between the images formed by the two mirrors. The radius of curvature of the convex mirror is
(a) 25 cm
(b) 30 cm
(c) 50 cm
(d) 60 cm

Answer: (a) 25 cm

1. The plane mirror forms an image 30 cm behind it, which is 30 + 30 = 60 cm from the object.
2. No parallax means the image in the convex mirror is at the same place, so it lies 60 cm from the object, that is, 60 - 50 = 10 cm behind the convex mirror.
3. For the convex mirror: u = -50 cm and v = +10 cm. 1/f = 1/v + 1/u = 1/10 - 1/50 = 4/50, so f = 12.5 cm.
4. R = 2f = 25 cm.
5. Hence, the correct option is (a).

Teacher's Note:
a) Two images with no parallax are at the same position.
b) The image in a convex mirror is virtual and behind the mirror.

19. For a concave mirror, the magnification of a real image was found to be twice as great when the object was 15 cm from the mirror as it was when the object was 20 cm from the mirror. The focal length of the mirror is
(a) 5.0 cm
(b) 7.5 cm
(c) 10 cm
(d) 12.5 cm

Answer: (c) 10 cm

1. The magnification of a real image (magnitude) m = f/(u - f).
2. At u = 15 cm: m₁ = f/(15 - f). At u = 20 cm: m₂ = f/(20 - f). The condition is m₁ = 2m₂.
3. 1/(15 - f) = 2/(20 - f), so 20 - f = 30 - 2f.
4. f = 10 cm.
5. Hence, the correct option is (c).

Teacher's Note:
a) The magnification formula in terms of u and f is the quickest to use.
b) Check: m₁ = 10/5 = 2 and m₂ = 10/10 = 1.

20. A slab of glass of refractive index 1.5 and thickness 3 cm is placed with the faces perpendicular to the principal axis of a concave mirror. If the radius of curvature of the mirror is 10 cm, the distance at which an object must be placed from the mirror so that the image coincides with the object is
(a) 9 cm
(b) 10 cm
(c) 11 cm
(d) 12 cm

Answer: (c) 11 cm

1. The image coincides with the object when the rays fall normally on the mirror, that is, when the object appears to be at the centre of curvature, 10 cm from the mirror.
2. The slab shifts the apparent position of the object by t(1 - 1/μ) = 3(1 - 1/1.5) = 1 cm towards the mirror.
3. So the real object must be 10 + 1 = 11 cm from the mirror.
4. Hence, the correct option is (c).

Teacher's Note:
a) The slab is taken to lie between the object and the mirror.
b) Rays retrace their path only when they strike the mirror along the normal.

21. A concave lens of glass of refractive index 1.5 has both the surfaces of the same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a
(a) convergent lens of focal length 3.5 R
(b) convergent lens of focal length 3.0 R
(c) divergent lens of focal length 3.5 R
(d) divergent lens of focal length 3.0 R

Answer: (a) convergent lens of focal length 3.5 R

1. By the lens maker's formula, 1/f = (μ_g/μ_m - 1)(1/R₁ - 1/R₂).
2. For a biconcave lens, R₁ = -R and R₂ = +R, so (1/R₁ - 1/R₂) = -2/R.
3. μ_g/μ_m - 1 = 1.5/1.75 - 1 = -1/7.
4. 1/f = (-1/7)(-2/R) = 2/(7R), so f = 3.5R, which is positive.
5. Hence, the lens is convergent with focal length 3.5R and the correct option is (a).

Teacher's Note:
a) A lens whose refractive index is less than that of the surrounding medium behaves in the opposite way.
b) The sign of f changes when the medium is denser than the glass.

22. A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm, the power of the combination is
(a) -1.5 dioptres
(b) -6.5 dioptres
(c) +6.5 dioptres
(d) +6.67 dioptres

Answer: (a) -1.5 dioptres

1. Power of the convex lens P₁ = 100/40 = +2.5 D.
2. Power of the concave lens P₂ = -100/25 = -4 D.
3. Power of the combination P = P₁ + P₂ = 2.5 - 4 = -1.5 D.
4. Hence, the correct option is (a).

Teacher's Note:
a) Powers add for lenses in contact.
b) A concave lens has negative power; convert the focal length to metres first.

23. A spherical surface of radius of curvature R separates air μ = 1.0 from glass of μ= 1.5. The centre of curvature is in glass. A point object P placed in air is found to have a real image Q in glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to
(a) 5 R
(b) 3 R
(c) 2 R
(d) 1.5 R

Answer: (a) 5 R

1. For refraction at a spherical surface, μ₂/v - μ₁/u = (μ₂ - μ₁)/R.
2. Let PO = OQ = x. Then u = -x and v = +x.
3. 1.5/x + 1/x = 0.5/R, so 2.5/x = 0.5/R.
4. x = 5R.
5. Hence, the correct option is (a).

Teacher's Note:
a) The object distance is negative and the image distance positive in the sign convention used.
b) Put u and v as -x and +x to use the equal-distance condition.