JEE Physics Ray Optics and Optical Instruments MCQs Set 22

Answer: (c) 20 cm

Explanation:
1. For a convex mirror \( \frac1v=\frac1f-\frac1u \) with real \( u<0 \): \( v=\frac{f|u|}{|u|+f} \).
2. This increases with \( |u| \) and approaches \( f=20 \) cm as the object goes to infinity, so the maximum is 20 cm.

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

Answer: (a) 1.5

Explanation:
1. Frequency stays constant, so \( n=\frac{\lambda_{vac}}{\lambda_{med}}=\frac{6000}{4000}=1.5 \).

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

Answer: (c) closer by a distance t/3

Explanation:
1. Shift \( t\left(1-\frac1\mu\right)=t\left(1-\frac23\right)=\frac t3 \).
2. The image shifts towards the observer, so the source appears nearer.

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

Answer: (d) \( \frac{v}{\sin\theta} \)

Explanation:
1. At the critical angle \( \sin\theta=\frac{n_B}{n_A}=\frac{v_A}{v_B} \).
2. So \( v_B=\frac{v}{\sin\theta} \).

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

Answer: (b) at C for all values of the refractive index

Explanation:
1. Rays from the centre travel along radii and meet the surface normally.
2. They are not bent at all, so the apparent position is C itself, irrespective of the refractive index.

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

Answer: (d) 1.7

Explanation:
1. For a symmetric biconvex lens \( \frac1f=\frac{2(n-1)}{R} \).
2. Equal f gives \( \frac{n_A-1}{R_A}=\frac{n_B-1}{R_B} \), so \( n_B-1=0.63\times\frac{R_B}{0.9R_B}=0.70 \).
3. Hence \( n_B=1.7 \).

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

Answer: (a) between 1 and 1.33

Explanation:
1. It converges in air, so its index exceeds 1 (assuming a convex-shaped lens).
2. It diverges in water, so its index is less than that of water, 1.33.
3. Hence the index lies between 1 and 1.33.

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

Answer: (c) 4 mm

Explanation:
1. Image distance \( v=\frac{15\times10}{5}=30 \) cm.
2. Longitudinal magnification \( \left(\frac vu\right)^2=\left(\frac{30}{15}\right)^2=4 \).
3. The image is 4 mm long.

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

Answer: (a) 4f

Explanation:
1. With \( u+v=D \) and \( \frac1u+\frac1v=\frac1f \): \( D=\frac{u^2}{u-f} \).
2. Minimising gives \( u=2f \), so \( D_{min}=4f \).

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

Answer: (b) 6 cm

Explanation:
1. Apparent depth of the fish seen from above: \( \frac{4}{1.33}=3.0 \) cm, so the object distance from the lens is \( 2+3.0=5.0 \) cm.
2. Lens formula: \( \frac1v=\frac1{30}-\frac1{5.0} \), so \( |v|=6.0 \) cm (virtual image, same side as the fish).

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

Answer: (d) \( \frac{280}{9} \) cm

Explanation:
1. From the same analysis, \( R=20\mu \) with \( \mu=\frac{14}{9} \).
2. So \( R=\frac{280}{9}=31.1 \) cm.

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

Answer: (b) the refractive index of the medium is different for different wavelengths

Explanation:
1. Dispersion means that the refractive index depends on wavelength, so different colours travel at different speeds and bend by different amounts.

Teacher's Note:
1. Red light has the least refractive index and violet the greatest in normal materials.

Answer: (c) red

Explanation:
1. Apparent rise \( t\left(1-\frac1\mu\right) \) increases with the refractive index.
2. Red has the smallest \( \mu \) (violet the largest), so red is raised the least.

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

Answer: (a) 0.08 degrees

Explanation:
1. For a thin prism \( \delta_v-\delta_r=(n_v-n_r)A \).
2. So \( (1.5-1.48)\times4^\circ=0.08^\circ \).

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

Answer: (c) 98 cm

Explanation:
1. Magnifying power \( \frac{f_o}{f_e}=50 \), so \( f_e=2 \) cm.
2. For a Galilean telescope the separation is \( f_o-f_e=100-2=98 \) cm.

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

Answer: (d) 12

Explanation:
1. Normal adjustment: \( \frac{f_o}{f_e}=10 \), so \( f_o=50 \) cm.
2. Near-point adjustment: \( M=\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=10\left(1+\frac{5}{25}\right)=12 \).

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

Answer: (b) 200

Explanation:
1. Objective: \( \frac1{v_o}=\frac1{1.2}-\frac1{1.25}=0.0333 \), so \( v_o=30 \) cm and \( m_o=\frac{30}{1.25}=24 \).
2. Eyepiece for image at infinity: \( \frac{25}{3}=8.33 \).
3. Total magnification \( 24\times8.33=200 \).

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

Answer: (d) \( \frac{200}{132} \)

Explanation:
1. Seen from the front, the object is 20 cm away; the block's optical effect on the double passage adds \( \frac{2t}{\mu} \) to the apparent distance.
2. Apparent image position from the front face: \( 20+\frac{2\times10}{\mu} \), which must equal \( 10+23.2=33.2 \) cm.
3. So \( \frac{20}{\mu}=13.2 \) and \( \mu=\frac{200}{132}=1.515 \).

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

Answer: (a) \( \frac{3f}{2} \)

Explanation:
1. Real image with \( m=-2 \): \( u=-\frac{3f}{2} \), \( v=-3f \) (distance between them \( \frac{3f}{2} \)).
2. Virtual image with \( m=+2 \): \( u=-\frac f2 \), \( v=+f \), giving the separation \( \frac{3f}{2} \) again.
3. Both cases give \( \frac{3f}{2} \).

Teacher's Note:
1. Check both the real and the virtual cases before choosing 'it depends'.

Answer: (c) a convex lens of focal length 27 cm

Explanation:
1. In the liquid, \( \frac{1}{f'}=\left(\frac{1.5}{1.35}-1\right)\left(\frac{1}{R_1}-\frac1{R_2}\right) \) and in air \( \frac{1}{12}=0.5\left(\frac{1}{R_1}-\frac1{R_2}\right) \).
2. So \( \frac1{f'}=\frac{0.111}{0.5}\times\frac1{12}=\frac1{54} \), a convex lens of focal length 54 cm each.
3. Two in contact: \( F=27 \) cm, convex.

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

Answer: (a) 45

Explanation:
1. The mirror spans heights 0.5 m to 1.0 m. A ray from the eye striking the mirror at height h descends by \( 1.8-h \) over 2 m, and after reflection it travels away from the wall with the same slope.
2. It reaches the floor at horizontal distance \( 2+\frac{2h}{1.8-h} \) from the eye: \( h=1.0 \) gives 4.5 m and \( h=0.5 \) gives \( 2+\frac{1}{1.3}=2.769 \) m.
3. Length visible \( 4.5-2.769=1.731=\frac{45}{26} \), so x = 45.

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

Answer: (b) 2

Explanation:
1. The source (at 2f) is at the centre of curvature, with image at 2f and magnification \( -1 \).
2. Each half has its own axis shifted by 0.5 cm, so the source is 0.5 cm off-axis for each half. The image appears on the other side of that axis at the same distance, 0.5 cm, so it is displaced by 1 cm from the original axis for each half, in opposite directions.
3. The images are 2 cm apart.

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

Answer: (d) 4

Explanation:
1. For a lens, \( \omega=\frac{f_r-f_v}{f} \).
2. So \( f_r-f_v=0.04\times10=0.4 \) cm = 4 mm.

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

Answer: (b) 0

Explanation:
1. The apparent shift of an object seen through a slab is \( t\left(1-\frac1\mu\right) \), which depends only on the thickness and index, not on where the slab is.
2. So as the slab moves, the image of a fixed object does not move: its velocity is zero.

Teacher's Note:
1. If the object itself moved, the image would move with the object's velocity.

Answer: (c) (i) or (ii)

Explanation:
1. Real image: \( m=-2 \), so \( v=2u \) with \( \frac1{2u}+\frac1u=\frac1{20} \) gives \( u=30 \) cm.
2. Virtual image: \( m=+2 \), so \( v=-2u \) and \( -\frac{1}{2u}+\frac1u=\frac1{20} \) gives \( u=10 \) cm.
3. The object distance is 10 cm or 30 cm.

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

Answer: (a) (i) and (iv)

Explanation:
1. With \( \delta=i+e-A \): \( i+e=40+60=100^\circ \).
2. Incidence angles differing by 20 degrees: \( i-e=20^\circ \), so \( i=60^\circ \) and \( e=40^\circ \).
3. The two angles of incidence are 40 degrees and 60 degrees (by the reversibility of the ray).

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

Answer: (c) (i) and (iv) only

Explanation:
1. Every part of the lens forms a complete image, so blocking some parts does not put streaks on the image: it still looks like the donkey.
2. Less light reaches the film, so the image is less intense.

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

Answer: (d) (i) and (iv) only

Explanation:
1. Apparent depths: \( \frac{d_1}{\mu}=5 \) and \( \frac{d_2}{\mu}=3 \) with \( d_1+d_2=12 \).
2. Adding: \( \frac{12}{\mu}=8 \), so \( \mu=1.5 \).
3. Then \( d_1=5\times1.5=7.5 \) cm.

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