JEE Physics Ray Optics and Optical Instruments MCQs Set 21

Answer: (c) 0.5 cm and 0.84 m

Explanation:
1. Focal length: \( f=+0.6 \) m for a convex mirror, image distance \( v=+0.35 \) m behind the mirror.
2. Mirror formula \( \frac1v+\frac1u=\frac1f \): \( \frac1u=\frac1{0.6}-\frac1{0.35}=-1.19 \), so \( u=-0.84 \) m.
3. Magnification \( m=-\frac vu=\frac{0.35}{0.84}=0.417 \), so the image height is \( 1.2\times0.417=0.5 \) cm.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (a) 40 cm/s

Explanation:
1. Image distance: \( \frac1v=\frac1{40}-\frac1{60}\Rightarrow v=120 \) cm.
2. Differentiating the mirror formula, \( \frac{dv}{dt}=-\frac{v^2}{u^2}\frac{du}{dt} \), so the speed is \( \left(\frac{120}{60}\right)^2\times10=40 \) cm/s.
3. The image moves in the opposite sense to the object (away from the mirror as the object approaches).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (b) 20 cm/s

Explanation:
1. Image distance \( v=120 \) cm, so the magnification is \( m=-\frac vu=-2 \).
2. Transverse velocity of the image: \( v_{i}=m\,v_{o} \) for the transverse part, since \( m \) is constant at that instant for this motion.
3. Speed \( 2\times10=20 \) cm/s, and the image moves opposite to the object (inverted).

Teacher's Note:
1. Transverse speeds scale with the lateral magnification; longitudinal speeds scale with its square.

Answer: (d) \( 6.7\times10^{-9} \) s

Explanation:
1. Angle of incidence \( =90^\circ-60^\circ=30^\circ \), so \( n=\frac{\sin30^\circ}{\sin15^\circ}=1.93 \).
2. Speed in ice \( =\frac{3\times10^8}{1.93}=1.55\times10^8 \) m/s; path length \( =\frac{1}{\cos15^\circ}=1.035 \) m.
3. Time \( =\frac{1.035}{1.55\times10^8}=6.7\times10^{-9} \) s.

Teacher's Note:
1. The ray travels obliquely, so the path in the slab is longer than its thickness.

Answer: (b) 9 cm/s

Explanation:
1. The height of an air object seen from water is magnified by the refractive index, so the bird's apparent speed (relative to the water surface) is \( \frac43v_b \).
2. The fish moves up at 4 cm/s, so the relative apparent speed is \( \frac43v_b+4=16 \).
3. Hence \( v_b=9 \) cm/s.

Teacher's Note:
1. Add the two speeds because the fish and the apparent bird approach each other.

Answer: (c) \( \sqrt3 \) cm

Explanation:
1. The ray to the rim: \( \tan\theta_1=\frac{3}{3\sqrt3}=\frac1{\sqrt3}\Rightarrow\theta_1=30^\circ \).
2. Snell's law: \( \sin\theta_2=\sqrt3\sin30^\circ=\frac{\sqrt3}{2}\Rightarrow\theta_2=60^\circ \).
3. Traced back to the axis, the ray is at 3 cm horizontal distance and \( 60^\circ \) from the vertical: apparent depth \( \frac{3}{\tan60^\circ}=\sqrt3 \) cm.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (a) 45 degrees

Explanation:
1. For refraction, the deviation is \( r-i \) and would have to equal 90 degrees, which is impossible for the glass-to-air direction because \( r<90^\circ \) and \( i>0 \).
2. So the ray must be totally reflected, with deviation \( 180^\circ-2i=90^\circ \), giving \( i=45^\circ \).
3. Check: the critical angle is \( \sin^{-1}\frac23=41.8^\circ \) and \( 45^\circ \) is greater, so total internal reflection does occur.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (c) 80 cm

Explanation:
1. Refraction from glass to air at a spherical surface of radius 20 cm: \( \frac{1}{v}+\frac{2}{x}=\frac{2-1}{20} \) where x is the real distance from the surface (sign convention: R behind the surface).
2. Nearest surface: \( -\frac15+\frac2x=\frac1{20} \), so \( x=8 \) cm. The farthest surface is \( 40-8=32 \) cm away.
3. Then \( \frac1v=\frac1{20}-\frac{2}{32}=-\frac1{80} \), so the image is 80 cm from that surface (virtual).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (d) 2

Explanation:
1. Refraction at the first surface with the object at infinity: \( \frac{n}{v}=\frac{n-1}{r} \), so \( v=\frac{nr}{n-1} \).
2. The farther surface is at distance \( 2r \): \( \frac{n}{n-1}=2 \).
3. So \( n=2 \).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (b) 1.4

Explanation:
1. Same radii give \( f\propto\frac{1}{n-1} \).
2. Ratio \( \frac{f_{1.5}}{f_{1.7}}=\frac{1.7-1}{1.5-1}=\frac{0.7}{0.5}=1.4 \).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (d) 4 cm

Explanation:
1. Focal length \( f=\frac1{5}=0.2 \) m = 20 cm.
2. Lens formula: \( \frac1v=\frac1{20}-\frac1{25}\Rightarrow v=100 \) cm.
3. Magnification \( -\frac{v}{u}=-4 \), so the image is 4 cm high (inverted).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (a) 30 cm

Explanation:
1. First position: \( v=3u \), so the candle-screen distance is \( D=4u \).
2. Second position: \( u'=u+0.8 \) and \( v'=3u-0.8 \) with \( v'=\frac{u'}{3} \), giving \( 9u-2.4=u+0.8 \), so \( u=0.4 \) m and \( v=1.2 \) m.
3. Focal length \( f=\frac{uv}{u+v}=\frac{0.48}{1.6}=0.3 \) m.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (c) 4 mm

Explanation:
1. Angle subtended by the sun's diameter: \( \theta=\frac{2\times0.75\times10^9}{1.5\times10^{11}}=0.01 \) rad.
2. Image diameter: \( f\theta=0.4\times0.01=4\times10^{-3} \) m = 4 mm.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (a) 60 cm

Explanation:
1. The final image coincides with the object when the rays strike the mirror perpendicularly, which happens when they appear to diverge from the mirror's centre of curvature (20 cm from the mirror, so 15 cm to the left of the lens).
2. So the lens must form a virtual image 15 cm to its left: \( v=-15 \).
3. Lens formula: \( \frac1u=\frac1v-\frac1f=-\frac1{15}+\frac1{20}=-\frac1{60} \), so the object is 60 cm from the lens.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (b) 13.3 cm

Explanation:
1. To emerge parallel, the reflected light must appear to diverge from the lens focus on the mirror side: 25 cm from the lens, i.e. 10 cm behind the mirror (virtual image).
2. Convex mirror \( f=+40 \) cm, image \( v=+10 \) cm (behind): \( \frac1u=\frac1f-\frac1v=\frac1{40}-\frac1{10}=-\frac{3}{40} \).
3. So \( u=-13.3 \) cm, a real source 13.3 cm in front of the mirror.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (d) 2.5 mm

Explanation:
1. The first lens would focus the beam at 10 cm; the diverging lens is 5 cm from it, so the beam is 5 cm before its focus there. The virtual object is 5 cm from the diverging lens, equal to its focal length, so the emergent beam is parallel.
2. The beam diameter at the second lens is reduced in the ratio \( \frac{10-5}{10}=\frac12 \), so it becomes \( 2.5 \) mm.
3. (The intensity of the beam is therefore 4 times larger.)

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (b) 10 cm and -60 cm

Explanation:
1. Convex lens alone: \( \frac1{f_1}=\frac1{30}+\frac1{15}=\frac1{10} \), so \( f_1=10 \) cm.
2. With the concave lens the image is at 60 cm: \( \frac1F=\frac1{60}+\frac1{15}=\frac1{12} \), so \( F=12 \) cm.
3. Then \( \frac1{f_2}=\frac1{12}-\frac1{10}=-\frac1{60} \), so \( f_2=-60 \) cm.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (c) 0.25 and 0.9 degrees

Explanation:
1. Dispersive power: \( \omega=\frac{n_v-n_r}{n_y-1}=\frac{0.15}{0.60}=0.25 \).
2. Angular dispersion of a thin prism: \( (n_v-n_r)A=0.15\times6^\circ=0.9^\circ \).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (a) 0.020

Explanation:
1. For a lens, \( \omega=\frac{f_r-f_v}{f_y} \).
2. So \( \omega=\frac{100-98}{99}=0.0202 \).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (c) 3.0 degrees and 0.225 degrees

Explanation:
1. Mean deviations add: \( 0.30\times5^\circ+0.50\times3^\circ=1.5^\circ+1.5^\circ=3.0^\circ \).
2. Dispersions add: \( 0.105^\circ+0.12^\circ=0.225^\circ \).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (d) 24 and 150 cm

Explanation:
1. Magnifying power \( =\frac{f_o}{f_e}=\frac{144}{6}=24 \).
2. Separation \( f_o+f_e=144+6=150 \) cm.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (b) 2.5 cm and 20

Explanation:
1. Eyepiece: image at 25 cm gives \( u_e=5 \) cm, so the objective image is \( 15-5=10 \) cm from the objective.
2. Objective: \( \frac1{10}-\frac1{u_o}=\frac12 \) gives \( u_o=-2.5 \) cm.
3. Magnification \( \frac{10}{2.5}\left(1+\frac{25}{6.25}\right)=4\times5=20 \).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (d) parallel to the first mirror

Explanation:
1. Take the first mirror along the x-axis and the second at 60 degrees. The incident ray travels along \( (-\cos60^\circ,-\sin60^\circ) \), parallel to the second mirror.
2. After the first reflection the y-component reverses: \( (-\cos60^\circ,+\sin60^\circ) \).
3. Reflecting this from the second mirror (direction \( (\cos60^\circ,\sin60^\circ) \)) gives \( (1,0) \): parallel to the first mirror.

Teacher's Note:
1. For two mirrors at angle theta, the net deviation is \( 360^\circ-2\theta \) for two reflections.

Answer: (a) 4 cm

Explanation:
1. For a mirror at position \( x_m \) and a fixed object at \( x_o \), the image is at \( 2x_m-x_o \).
2. So the image displacement is twice the mirror displacement: amplitude \( 2\times2=4 \) cm.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (c) 07:34:23

Explanation:
1. The mirror image of the clock reads the time \( 12:00:00-\text{(actual time)} \).
2. Compute \( 11:59:60-04:25:37=07:34:23 \).

Teacher's Note:
1. Subtract the displayed time from 12:00:00, borrowing as needed.

Answer: (a) 45 cm

Explanation:
1. Put A at x = 0, B at x = 20 and the object at x = 15. Images lie at \( x=15+40k \) and \( x=-15+40k \).
2. Those behind A (negative x, formed in mirror A) are at \( x=-15,-25,-55,-65,\ldots \), that is, at distances 15, 25, 55, 65 cm from A.
3. A distance of 45 cm does not appear in this list.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (b) 15 cm

Explanation:
1. Let the object be at x from the concave mirror, with image at v. The plane mirror (22.5 cm away) images that point at \( 45-v \).
2. The final image coincides with the object when \( 45-v=x \), so \( v=45-x \).
3. Mirror formula \( \frac1x+\frac1{45-x}=\frac1{10} \) gives \( x^2-45x+450=0 \), so \( x=15 \) or 30 cm. Only 15 cm lies between the mirrors.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (d) 4 mm/s

Explanation:
1. Image distance \( \frac1v=\frac1{10}-\frac1{15}\Rightarrow v=30 \) cm, so the lateral magnification magnitude is \( \frac{30}{15}=2 \).
2. Transverse image speed \( 2\times2=4 \) mm/s.

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.

Answer: (b) \( \frac{a^2}{4b} \)

Explanation:
1. Newton's relation for mirrors: \( xx'=f^2 \) where x and x' are the distances of object and image from the focus.
2. With \( f=\frac a2 \) and \( x=b \): \( x'=\frac{a^2}{4b} \).

Teacher's Note:
1. Use the sign convention consistently and check that the answer is physically reasonable.