JEE Physics Ray Optics and Optical Instruments MCQs Set 25

Answer: (d) \( \sin^{-1}\frac1{\sqrt3} \)

Explanation:
1. Grazing the wall means the incidence there equals the critical angle: \( \sin C = \frac{\sqrt3}{2} \), C = 60°, so refraction at the end face is r = 30°.
2. \( \sin\theta = \frac{2}{\sqrt3}\sin30^\circ = \frac1{\sqrt3} \).

Teacher's Note:
1. At the wall the angle with the normal is 90° − r.
2. Use the critical-angle condition at the lateral surface.

Answer: (a) 30 cm

Explanation:
1. \( \frac1f = (\mu-1)\frac1R = \frac{0.6}{18} \Rightarrow f = 30 \) cm.

Teacher's Note:
1. For a plano-convex lens one radius is infinite.
2. Lens maker's formula.

Answer: (c) 60 cm, virtual

Explanation:
1. \( \frac1v = \frac1f+\frac1u = \frac1{12}-\frac1{10} = -\frac1{60} \), v = −60 cm.

Teacher's Note:
1. Object inside the focus gives a virtual, enlarged, erect image.
2. The image is on the same side as the object.

Answer: (a) +0.667

Explanation:
1. \( \frac1v = -\frac1{20}-\frac1{10} = -\frac3{20} \), v = −6.67 cm.
2. \( m = \frac vu = \frac{-6.67}{-10} = +0.667 \).

Teacher's Note:
1. A diverging lens always gives a virtual, erect, reduced image.
2. Magnification is positive for an erect image.

Answer: (d) 10 cm (concave)

Explanation:
1. Lens focal length \( f = \frac{10}{0.5} = 20 \) cm. Light passes through the lens twice: \( \frac1F = \frac2f = \frac1{10} \).
2. A converging system: concave mirror of focal length 10 cm.

Teacher's Note:
1. A silvered lens counts the lens power twice.
2. The plane silvered face acts as a mirror of zero power.

Answer: (b) −1.5 D

Explanation:
1. \( P = \frac{100}{40}-\frac{100}{25} = 2.5-4 = -1.5 \) D.

Teacher's Note:
1. Power of lenses in contact adds.
2. Concave lens has negative power.

Answer: (c) of the same size and inverted

Explanation:
1. \( F = 10 \) cm, and u = −20 cm = −2F.
2. The image forms at 2F on the other side, the same size and inverted.

Teacher's Note:
1. Object at 2F gives an image at 2F of the same size.
2. Lenses in contact: powers add.

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

Explanation:
1. \( \frac1f = \left(\frac{1.5}{1.75}-1\right)\left(-\frac2R\right) = \left(-\frac17\right)\left(-\frac2R\right) = \frac{2}{7R} \).
2. f = +3.5R, so the lens converges.

Teacher's Note:
1. A glass lens in a denser medium behaves oppositely.
2. A biconcave lens has R₁ = −R and R₂ = +R.

Answer: (b) of the mirror remains unchanged while that of the lens increases

Explanation:
1. A mirror's focal length is R/2 and does not depend on the medium.
2. The lens focal length depends on μ_lens/μ_medium − 1, which becomes smaller in glycerine, so f increases.

Teacher's Note:
1. Reflection is independent of the surrounding medium.
2. Refraction depends on the relative refractive index.

Answer: (d) \( \frac{(\mu+1)f}{\mu} \)

Explanation:
1. Real image: \( v = \mu x \) for object distance x: \( \frac1{\mu x}+\frac1x = \frac1f \).
2. \( x = \frac{(\mu+1)f}{\mu} \).

Teacher's Note:
1. Real image from a convex lens is inverted, so v and u have opposite signs.
2. Check μ = 1 gives x = 2f.

Answer: (c) −72 cm (diverging, 72 cm)

Explanation:
1. Glass lenses: \( \frac1{f_1} = -\frac{0.5}{20} \), \( \frac1{f_2} = -\frac{0.5}{30} \). Liquid lens (biconvex): \( \frac1{f_3} = \frac13\left(\frac1{20}+\frac1{30}\right) = \frac1{36} \).
2. \( \frac1F = -\frac1{40}-\frac1{60}+\frac1{36} = -\frac1{72} \), so F = −72 cm.

Teacher's Note:
1. The liquid between two concave surfaces forms a biconvex lens.
2. Add the three powers.

Answer: (c) 22 cm

Explanation:
1. \( \frac1f = (\mu-1)\frac2R \Rightarrow R = 2(0.55)(20) = 22 \) cm.

Teacher's Note:
1. For a symmetrical biconvex lens R₁ = R, R₂ = −R.
2. R is slightly more than f.

Answer: (a) 7.5 cm from the lens on the side of P

Explanation:
1. P acts as a virtual object at u = +12 cm. \( \frac1v = \frac1f+\frac1u = \frac1{20}+\frac1{12} = \frac2{15} \).
2. v = 7.5 cm, on the same side as P.

Teacher's Note:
1. A converging beam gives a virtual object for the lens.
2. The lens converges it further.

Answer: (b) virtual, erect and diminished

Explanation:
1. \( \frac1v = -\frac1{21}-\frac1{14} = -\frac5{42} \), v = −8.4 cm.
2. \( m = \frac{v}{u} = 0.6 \), so the image is 1.8 cm: virtual, erect, diminished.

Teacher's Note:
1. A concave lens always gives a virtual, erect, diminished image.
2. m between 0 and 1 means diminished.

Answer: (b) −60 cm

Explanation:
1. \( \frac1F = \frac1{30}-\frac1{20} = -\frac1{60} \).

Teacher's Note:
1. Add the powers (1/f) of lenses in contact.
2. Net negative power means diverging.

Answer: (c) a diverging lens

Explanation:
1. The net focal length is −60 cm, which is negative.
2. A negative focal length means a diverging system for all object positions.

Teacher's Note:
1. Nature of a lens does not depend on the object position.
2. Concave lens has the larger power here.

Answer: (d) 45°

Explanation:
1. \( i = \frac{A+\delta_m}{2} = \frac{60^\circ+30^\circ}{2} = 45^\circ \).

Teacher's Note:
1. At minimum deviation the ray passes symmetrically.
2. i = e = (A + δ)/2.

Answer: (b) 38°

Explanation:
1. \( \sin\frac{A+\delta_m}{2} = \mu\sin\frac A2 = 1.5\times0.5 = 0.75 \Rightarrow \frac{A+\delta_m}2 = 49^\circ \).
2. \( \delta_m = 98^\circ-60^\circ = 38^\circ \).

Teacher's Note:
1. Use the prism formula with A = 60°.
2. sin 49° = 0.75 is given.

Answer: (a) 90°

Explanation:
1. \( \delta = i+e-A \Rightarrow 30 = 60+e-30 \Rightarrow e = 0 \).
2. The emergent ray is along the normal to the second face, so it makes 90° with the face.

Teacher's Note:
1. Angle with the face is 90° minus angle with the normal.
2. Check δ = i + e − A.

Answer: (c) 0°

Explanation:
1. \( r_1+r_2 = A \), and with e = 0 we have \( r_2 = 0 \).

Teacher's Note:
1. The ray inside the prism meets the second face normally.
2. Then r₁ = A.

Answer: (c) 30°

Explanation:
1. \( r_1 = A - r_2 = 30^\circ - 0 = 30^\circ \).

Teacher's Note:
1. r₁ + r₂ = A.
2. Use r₂ = 0 from the earlier result.

Answer: (d) 2.9°

Explanation:
1. No net deviation: \( (\mu_1-1)A_1 = (\mu_2-1)A_2 \).
2. \( A_2 = \frac{0.54\times5}{0.92} = 2.9^\circ \).

Teacher's Note:
1. For thin prisms δ = (μ − 1)A.
2. The prisms are placed with their bases opposite.

Answer: (b) 70°

Explanation:
1. \( \delta = i+e-A \Rightarrow e = 55+60-45 = 70^\circ \).

Teacher's Note:
1. Deviation formula for a prism.
2. Emergence angle is measured from the normal.

Answer: (b) 10°

Explanation:
1. Symmetric passage: i = e = 35°. \( \delta = 2i - A = 70^\circ-60^\circ = 10^\circ \).

Teacher's Note:
1. Parallel to the base means symmetric.
2. Deviation here is the minimum deviation.

Answer: (a) 0.206

Explanation:
1. Mean index \( \mu = \frac{1.52+1.64}{2} = 1.58 \).
2. \( \omega = \frac{\mu_v-\mu_r}{\mu-1} = \frac{0.12}{0.58} = 0.206 \).

Teacher's Note:
1. Dispersive power has no units.
2. Use the mean refractive index in the denominator.

Answer: (d) 1.44°

Explanation:
1. \( \delta_v-\delta_r = (\mu_v-\mu_r)A = 0.12\times12^\circ = 1.44^\circ \).

Teacher's Note:
1. Angular dispersion = (μ_v − μ_r)A.
2. Thin prism: δ = (μ − 1)A.

Answer: (b) \( \sqrt2 \)

Explanation:
1. Retracing means the ray strikes the silvered face normally, so r₁ = A = 30°.
2. \( \mu = \frac{\sin45^\circ}{\sin30^\circ} = \sqrt2 \).

Teacher's Note:
1. Normal incidence on the silvered face reflects the ray back along itself.
2. Then r₁ equals the prism angle.

Answer: (a) 20°

Explanation:
1. At minimum deviation i = e, \( \delta_m = 2i - A = 80^\circ-60^\circ = 20^\circ \).

Teacher's Note:
1. Symmetric passage gives i = e.
2. δ = i + e − A.