JEE Physics Ray Optics and Optical Instruments MCQs Set 20

Answer: (d) \( 12\text{ cm}\times 8\text{ cm} \)

Explanation:
1. Light from the edge of the face reaches the eye after reflection at the mirror midway between the face and the eye.
2. By similar triangles, the mirror needs only half the extent of the face in each direction.
3. Height \( =24/2=12 \) cm, width \( =16/2=8 \) cm.

Teacher's Note:
1. The rule of half the object size comes from the image being as far behind the mirror as the object is in front.
2. It does not depend on how far the person stands from the mirror.

Answer: (a) \( \dfrac{b\sqrt{1-n^2\cos^2\theta}}{\sin\theta} \)

Explanation:
1. Angle of incidence in glass: \( i=90^\circ-\theta \). The length of face covered by the beam is \( \dfrac{b}{\cos i}=\dfrac{b}{\sin\theta} \).
2. Snell's law: \( n\sin i=\sin r\Rightarrow \sin r=n\cos\theta \), so \( \cos r=\sqrt{1-n^2\cos^2\theta} \).
3. Width in air \( =(\text{length on face})\times\cos r=\dfrac{b\sqrt{1-n^2\cos^2\theta}}{\sin\theta} \).

Teacher's Note:
1. A beam's width is measured perpendicular to its rays, so it changes when the angle with the surface changes.
2. The footprint on the surface is the same before and after refraction.

Answer: (d) \( \dfrac12-\dfrac{\sqrt{n^2-1}}{2n} \)

Explanation:
1. Light escapes only within a cone of half-angle equal to the critical angle \( \theta_c \), with \( \sin\theta_c=1/n \).
2. Fraction of the sphere of directions in the downward-and-upward hemisphere: the cone's share of all directions is \( \dfrac{1-\cos\theta_c}{2} \).
3. With \( \cos\theta_c=\dfrac{\sqrt{n^2-1}}{n} \): \( f=\dfrac12-\dfrac{\sqrt{n^2-1}}{2n} \), independent of h.

Teacher's Note:
1. The solid angle of a cone of half-angle theta is \( 2\pi(1-\cos\theta) \) out of \( 4\pi \) total.
2. The depth h does not appear because only directions matter for an isotropic source.

Answer: (a) \( \dfrac{4-\sqrt7}{8}\approx0.17 \)

Explanation:
1. \( \sqrt{n^2-1}=\sqrt{16/9-1}=\sqrt7/3 \).
2. \( f=\dfrac12-\dfrac{\sqrt7/3}{2\times4/3}=\dfrac12-\dfrac{\sqrt7}{8}=\dfrac{4-\sqrt7}{8}\approx0.169 \).

Teacher's Note:
1. About 17 percent of the light from a submerged source escapes straight up; the rest is reflected back inside.
2. Substitute carefully: the 3 in the denominator cancels.

Answer: (b) \( 49.7^\circ \)

Explanation:
1. At minimum deviation \( r_1=r_2=A/2=30^\circ \).
2. \( \sin i=1.52\sin30^\circ=0.760 \), so \( i=49.7^\circ \).

Teacher's Note:
1. At minimum deviation the ray passes symmetrically through the prism.
2. The same incident beam then meets the prism at a different deviation for violet.

Answer: (c) \( 6.3^\circ \)

Explanation:
1. Red: \( \delta_r=2i-A=2(49.7)-60=39.4^\circ \).
2. Violet: \( \sin r_1=\dfrac{0.760}{1.6}=0.475\Rightarrow r_1=28.4^\circ,\ r_2=60-28.4=31.6^\circ \); \( \sin e=1.6\times0.520=0.832\Rightarrow e=56^\circ \).
3. Violet deviation \( =49.7+56-60=45.7^\circ \). Width \( =45.7-39.4=6.3^\circ \).

Teacher's Note:
1. Violet bends more than red, so the spectrum spreads between the two deviations.
2. Trace the ray through both faces separately for violet.

Answer: (d) 11 cm

Explanation:
1. Length \( =f\,\Delta\delta=100\times6.3\times\dfrac{22}{7\times180} \) cm.
2. This equals \( \approx11\text{ cm} \).

Teacher's Note:
1. A lens focuses parallel beams at different angles to different points in its focal plane.
2. Convert the angle to radians before multiplying by the focal length.

Answer: (c) \( \sin^{-1}(0.75)=48.6^\circ \)

Explanation:
1. At 600 nm the interface BC is optically invisible, so the block acts as a single prism of angle \( 60^\circ \) and index 1.5.
2. At minimum deviation \( \sin i=n\sin(A/2)=1.5\sin30^\circ=0.75 \).
3. So \( i=\sin^{-1}(0.75)=48.59^\circ \).

Teacher's Note:
1. When two media have equal index, the boundary between them has no optical effect.
2. Use the minimum-deviation relation for the equivalent single prism.

Answer: (c) 2

Explanation:
1. A ray entering at distance x from the centre of the bend (R at most x at most R+d) meets the outer curved surface with \( \sin i=\dfrac{x}{R+d} \).
2. The innermost ray (x = R) has the smallest angle, so condition for TIR is \( \dfrac{R}{R+d}\ge\dfrac1n=\dfrac23 \).
3. This gives \( 3R\ge2R+2d\Rightarrow R/d\ge2 \).

Teacher's Note:
1. Total internal reflection needs the angle of incidence to exceed the critical angle, whose sine is 1/n.
2. The worst case is the ray closest to the inner edge.

Answer: (d) 0.34

Explanation:
1. In water the critical angle satisfies \( \sin\theta_c=\dfrac{1.33}{1.5}=0.8867 \).
2. With R = 2d, outer radius = 3d; a ray at distance x has \( \sin i=x/3d \). TIR needs \( x\ge3d(0.8867)=2.66d \).
3. Rays enter uniformly for x from 2d to 3d (width d), so the fraction is \( \dfrac{3d-2.66d}{d}=0.34 \).

Teacher's Note:
1. Because the beam is uniform, the flux fraction equals the fraction of the entrance width that satisfies the condition.
2. Reducing the index contrast (water around glass) makes total internal reflection harder.