JEE Physics Ray Optics and Optical Instruments MCQs Set 19

Answer: (a) \( D_1\lt D_2 \)

Explanation:
1. Minimum deviation increases with refractive index for a given prism angle.
2. Blue has the larger index, so \( D_2\gt D_1 \).

Teacher's Note:
1. Violet and blue deviate more than red.

Answer: (b) 3 m

Explanation:
1. \( r=\frac h{\sqrt{\mu^2-1}}=\frac4{\sqrt{25/9-1}}=\frac4{4/3}=3\ \text{m} \)

Teacher's Note:
1. Light striking the surface beyond the critical angle cannot leave, so the disc must cover the critical cone.

Answer: (a) 20 cm

Explanation:
1. Light enters the plane face, travels in glass and is reflected at the curved surface. The silvered lens acts as a mirror of focal length \( f=\frac R{2\mu}=\frac{30}{3}=10\ \text{cm} \).
2. A real image of the same size forms when the object is at the centre of curvature: \( u=2f=20\ \text{cm} \).

Teacher's Note:
1. A silvered plano-convex lens behaves like a concave mirror of focal length R/(2 mu).

Answer: (c) 10.5 cm

Explanation:
1. Real depth = apparent depth times \( \mu \): \( d_1=5\times1.5=7.5\ \text{cm} \), \( d_2=2\times1.5=3\ \text{cm} \).
2. Thickness \( =7.5+3=10.5\ \text{cm} \)

Teacher's Note:
1. The bubble lies at fixed real distances from the two faces, which add up to the thickness.

Answer: (b) 3 feet

Explanation:
1. Rays from the top of the head and from the feet reach the eye after reflection at points that are half way up and down from the eye level.
2. The mirror needs half the height of the man: 3 feet.

Teacher's Note:
1. A plane mirror needs to be only half as tall as the person, whatever the distance.

Answer: (c) \( n\gt \sqrt2 \)

Explanation:
1. Worst case: incidence 90 degrees, so \( \sin r_{max}=\frac1n \) and the angle at the lateral face is \( 90^\circ-r \).
2. Total reflection there needs \( \cos r\ge\frac1n\Rightarrow\sqrt{1-\frac1{n^2}}\ge\frac1n\Rightarrow n\ge\sqrt2 \).

Teacher's Note:
1. Use the largest possible refraction angle at the end face.

Answer: (c) Real and enlarged

Explanation:
1. The object is just outside the focus of the objective, so its image is real, inverted and enlarged.
2. This image acts as the object for the eyepiece.

Teacher's Note:
1. The objective of a microscope makes a real magnified intermediate image.

Answer: (b) \( \frac{\mu t}c \)

Explanation:
1. \( v=\frac c\mu\Rightarrow T=\frac tv=\frac{\mu t}c \)

Teacher's Note:
1. Light is slower in glass by the factor mu.

Answer: (c) 30 cm

Explanation:
1. \( \mu=\frac{c}{v}=1.5 \).
2. Radius of the curved face from the sag: \( R=\frac{r^2+s^2}{2s}=\frac{3^2+0.3^2}{0.6}\approx15.2\ \text{cm} \).
3. \( f=\frac R{\mu-1}\approx30\ \text{cm} \)

Teacher's Note:
1. Find mu from the speed, R from the chord and sag geometry, then use the lens maker's formula.

Answer: (c) \( 2\times10^{-11} \) s

Explanation:
1. \( T=\frac{\mu t}c=\frac{1.5\times4\times10^{-3}}{3\times10^8}=2\times10^{-11}\ \text{s} \)

Teacher's Note:
1. Time equals optical path divided by c.

Answer: (d) 20 times nearer

Explanation:
1. The telescope increases the angle subtended by the tree 20 times, which is exactly what would happen if the tree were 20 times nearer.
2. The actual size of the tree is unchanged.

Teacher's Note:
1. Angular magnification is equivalent to bringing the object closer.

Answer: (d) 30 cm below the flat surface

Explanation:
1. The silvered surface is a concave mirror of \( R=10 \) cm, so \( f=5\ \text{cm} \). Its pole is 10 cm below the flat surface.
2. The bubble is \( 10-6=4 \) cm from the pole: \( \frac1v+\frac1{-4}=\frac1{-5}\Rightarrow v=+20\ \text{cm} \) behind the pole (virtual).
3. That point is \( 10+20=30 \) cm below the flat surface.

Teacher's Note:
1. Treat the silvered face as a concave mirror inside the glass and take distances from its pole.

Answer: (b) Bends upwards

Explanation:
1. Near the ground the refractive index is smaller, so the wavelets travel faster there.
2. The lower part of the wavefront advances more, tilting the wavefront and the beam upwards.

Teacher's Note:
1. Light bends towards the region of higher refractive index; the lower part of the wavefront runs ahead.

Answer: (c) 13

Explanation:
1. Each crossing advances the ray along the mirrors by \( 20\tan30^\circ=11.55\ \text{cm} \).
2. Number of reflections \( =\left\lfloor\frac{160}{11.55}\right\rfloor=\lfloor13.86\rfloor=13 \)

Teacher's Note:
1. Count the whole number of 11.55 cm steps along the 160 cm mirror.

Answer: (b) 0.06

Explanation:
1. The image is 10 cm behind the mirror, so the insect line is 30 cm from the image. The mirror edges are 10 cm above and below the axis.
2. The visible stretch is bounded by the rays from the image through the mirror edges: half-width \( 10\times\frac{30}{10}=30\ \text{cm} \), so total 60 cm.
3. \( t=\frac{0.6}{10}=0.06\ \text{s} \)

Teacher's Note:
1. The field of view of a mirror is the cone from the image through the mirror edges.

Answer: (c) 90 cm

Explanation:
1. \( f=40\ \text{cm}:\ \frac1v+\frac1{-60}=\frac1{-40}\Rightarrow v=-120 \): the concave mirror alone would focus at 120 cm.
2. The plane mirror at distance x shifts this image to \( 2x-120 \) from the concave mirror; set it equal to 60: \( x=90 \) cm.

Teacher's Note:
1. A plane mirror reflects the converging beam to the mirror-image position of the focus.

Answer: (c) 14

Explanation:
1. Apparent depth of the silvered surface from the top \( =\frac3{1.5}=2\ \text{cm} \).
2. Object is 10 cm above the top, so 12 cm above this equivalent mirror; the image is 12 cm below it, i.e. \( 12+2=14 \) cm below the top surface.

Teacher's Note:
1. Replace the silvered face by its apparent position and use plane mirror symmetry.

Answer: (d) 90 degrees

Explanation:
1. The ray refracts in, reflects from the silvered face and comes out at 45 degrees on the other side of the normal, independent of the index.
2. The incoming and outgoing directions are 90 degrees apart, so the deviation is 90 degrees.

Teacher's Note:
1. A plane mirror turns a ray symmetrically, so the slab does not change the exit angle.

Answer: (c) 16 m/s

Explanation:
1. Speed after falling \( 20-12.8=7.2 \) m: \( v=\sqrt{2\times10\times7.2}=12\ \text{m/s} \).
2. The apparent height is \( \mu h \), so the apparent speed is \( \frac43\times12=16\ \text{m/s} \)

Teacher's Note:
1. The fish sees both the position and the velocity of the ball magnified by the refractive index.

Answer: (b) 20 cm

Explanation:
1. \( m=\frac f{f+u} \): for +2, \( u=-10 \) cm; for -2, \( u=-30 \) cm.
2. The object must be moved 20 cm away from the lens.

Teacher's Note:
1. Use the magnification formula m = f/(f + u) with sign convention.