JEE Physics Ray Optics and Optical Instruments MCQs Set 23

Answer: (b) \( \sin^{-1}\left(\frac18\right) \)

Explanation:
1. For parallel boundaries, \( n\sin\theta \) is the same in every region.
2. The beam just misses region IV when it grazes the III-IV boundary, i.e. the angle in region III is the critical angle for III to IV.
3. Then \( n_0\sin\theta=\frac{n_0}{8}\sin90^\circ \) and \( \sin\theta=\frac18 \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) (66, 33) and (78, 39)

Explanation:
1. Use \( \frac1v=\frac1f-\frac1u \) with magnitudes for each u and compare with the recorded v.
2. (42): v = 56.0, (48): v = 48.0, (60): v = 40.0, all matching. For u = 66, v = 37.7, but 33 was recorded (error 4.7 cm); for u = 78, v = 34.7, but 39 was recorded (error 4.3 cm).
3. Both errors far exceed 0.2 cm, so (66, 33) and (78, 39) are incorrect.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) 6

Explanation:
1. Magnitude of magnification \( |m|=\frac{f}{u-f} \) for a real object beyond the focus.
2. At 25 cm: \( \frac{20}{5}=4 \); at 50 cm: \( \frac{20}{30}=\frac23 \).
3. Ratio \( 4\div\frac23=6 \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) 3

Explanation:
1. Convex mirror, \( f=+10 \) m: \( \frac1u=\frac1f-\frac1v \) in magnitudes of a virtual image gives \( \frac1u=\frac1v-\frac1f \) with a minus sign: \( u=\frac{vf}{f-v} \).
2. For \( v=\frac{25}3 \): \( u=50 \) m. For \( v=\frac{50}7 \): \( u=25 \) m.
3. The object moves 25 m in 30 s, i.e. \( \frac{25}{30}\text{ m/s}=3 \) km/h.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) (i) and (iii) only

Explanation:
1. A very thin film has no effect on the focal lengths; only the air-glass surface matters.
2. Air to glass: \( \frac{1.5}{f_1}=\frac{1.5-1}{R} \) gives \( f_1=3R \).
3. Glass to air (parallel rays inside the glass): \( f_2=\frac{1\times R}{1.5-1}=2R \). So (i) and (iii) are correct.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) 7

Explanation:
1. Mirror: \( \frac1v=-\frac1{10}+\frac1{15} \) gives \( v=-30 \) cm (real, 30 cm in front), \( m=-2 \).
2. In air the lens is 20 cm from this image: \( v=20 \) cm, \( m=-1 \), so \( M_1=+2 \) (erect).
3. In the medium the mirror is unchanged, but the lens focal length becomes \( f'=10\times\frac{0.5}{\frac{1.5\times6}{7}-1}=17.5 \) cm. Then \( \frac1v=\frac1{17.5}-\frac1{20} \) gives \( v=140 \) cm and \( m=\frac{140}{-20}=-7 \), so \( M_2=14 \).
4. The ratio is \( \frac{14}{2}=7 \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) 2

Explanation:
1. With \( \sin r_1=\frac{\sin60^\circ}{n} \), \( r_2=60^\circ-r_1 \) and \( \sin\theta=n\sin r_2 \).
2. Differentiating \( \sin\theta=n\sin r_2 \): \( \cos\theta\frac{d\theta}{dn}=\sin r_2+n\cos r_2\frac{dr_2}{dn} \), and from \( n\sin r_1=\sin60^\circ \) we get \( \frac{dr_2}{dn}=-\frac{dr_1}{dn}=\frac{\tan r_1}{n} \).
3. At \( n=\sqrt3 \): \( r_1=r_2=30^\circ \), \( \frac{dr_2}{dn}=\frac13 \), so \( \frac12\frac{d\theta}{dn}=\frac12+\sqrt3\cdot\frac{\sqrt3}2\cdot\frac13=1 \) and \( \frac{d\theta}{dn}=2 \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) \( NA_2 \)

Explanation:
1. Light must be guided in both sections. Rays accepted by the structure with the larger NA but outside the acceptance cone of the smaller one escape in the second.
2. Hence the combination accepts only the smaller cone, so its numerical aperture is \( NA_2 \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

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

Explanation:
1. Real magnified image: \( v=60 \) cm for \( u=30 \) cm, so \( \frac1f=\frac1{60}+\frac1{30}=\frac1{20} \): \( f=20 \) cm.
2. The faint image is virtual and erect (convex mirror), 10 cm behind the surface: \( \frac1{10}-\frac1{30}=\frac1{f_m} \) so \( f_m=15 \) cm and \( R=30 \) cm.
3. Then \( f=\frac{R}{n-1}=20 \) gives \( n=2.5 \). So (i) and (iv) are true; R = 30 cm (not 45) and the faint image is virtual (not real).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) (i), (iii) and (iv)

Explanation:
1. At minimum deviation \( i_1=\frac{A+\delta_m}{2}=A \) and \( r_1=\frac A2 \), so \( r_1=\frac{i_1}{2} \) (i).
2. Then \( \mu=\frac{\sin A}{\sin(A/2)}=2\cos\frac A2 \), so \( A=2\cos^{-1}\frac\mu2 \): statement (ii) (with a factor 1/2) is wrong, (iv) is right.
3. For \( i_1=A \) we have the minimum-deviation case, in which the ray inside is parallel to the base (iii).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) 130

Explanation:
1. The beam converges to the focus at 20 cm and then diverges. At 22 cm the beam radius equals \( \frac{22-20}{20}=\frac1{10} \) of the lens radius.
2. Area ratio \( \frac1{100} \), so intensity is \( 100\times1.3=130 \) kW/m\( ^2 \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) \( (2f,\ 2f) \)

Explanation:
1. The 45 degree line is \( v=u \).
2. Substituting in \( \frac1u+\frac1v=\frac1f \) gives \( \frac2u=\frac1f \), so \( u=v=2f \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) \( \left(1-\frac1{\mu_1}\right)h_1+\left(1-\frac1{\mu_2}\right)h_2 \)

Explanation:
1. Each layer of thickness t and index \( \mu \) lifts the image by \( t\left(1-\frac1\mu\right) \).
2. Adding both layers gives the expression shown.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) 5.6 m

Explanation:
1. Focal length: \( \frac1f=\frac1{12}+\frac1{240}=\frac{21}{240} \) cm.
2. The plate shifts the image away by \( t\left(1-\frac1\mu\right)=\frac13 \) cm, so the lens must now form its image at \( 12-\frac13=\frac{35}3 \) cm.
3. Then \( \frac1u=\frac{21}{240}-\frac3{35} \) gives \( u=560 \) cm = 5.6 m.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) \( f_1>f \) and \( f_2 \) becomes negative

Explanation:
1. Use \( \frac1{f}\propto\frac{\mu_g}{\mu_l}-1 \).
2. Air: \( 0.5 \). In the liquid of index 4/3: \( \frac98-1=0.125 \), so \( f_1=4f \) (greater than f).
3. In the liquid of index 5/3: \( \frac{9}{10}-1=-0.1 \), so \( f_2=-5f \): negative, the lens becomes diverging.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) \( \theta>\sin^{-1}\left[\mu\sin\left(A-\sin^{-1}\frac1\mu\right)\right] \)

Explanation:
1. Transmission at AC needs \( r_2<C \) with \( \sin C=\frac1\mu \).
2. Since \( r_1=A-r_2 \), this means \( r_1>A-C \).
3. Hence \( \sin\theta=\mu\sin r_1>\mu\sin\left(A-\sin^{-1}\frac1\mu\right) \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) real, at 40 cm from the convergent lens

Explanation:
1. The diverging lens makes a virtual image at its own focus, 25 cm on the incoming side of it.
2. This point is 40 cm from the converging lens, so for it \( u=-40 \) cm, \( f=20 \) cm.
3. Then \( \frac1v=\frac1{20}-\frac1{40}=\frac1{40} \), \( v=40 \) cm: real, 40 cm from the convergent lens.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) \( \frac43 \)

Explanation:
1. Without the liquid the pin is at the focus: \( f=18 \) cm, so with \( n=1.5 \): \( \frac1{18}=0.5\times\frac2R \) and \( R=18 \) cm.
2. With the liquid, the lens and a plano-concave liquid lens are in contact: \( \frac1{27}=\frac1{18}+\frac1{f_l} \) gives \( f_l=-54 \) cm.
3. For the liquid lens \( \frac1{f_l}=-\frac{\mu_l-1}{18} \), so \( \mu_l-1=\frac13 \) and \( \mu_l=\frac43 \).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) 3 m

Explanation:
1. Image distance 8 m and magnification magnitude \( \frac13 \) give object distance 24 m, so \( \frac1f=\frac18+\frac1{24}=\frac16 \) and \( f=6 \) m.
2. Index: \( n=\frac{\lambda_0}{\lambda}=\frac32 \).
3. For a plano-convex lens \( f=\frac R{n-1} \), so \( R=6\times0.5=3 \) m.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) 2.5 cm

Explanation:
1. Relaxed eye: object at infinity, image on the retina.
2. So \( f_{max}=2.5 \) cm.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (a) -1 D

Explanation:
1. The lens must form a virtual image at 100 cm of a distant object, so \( f=-100 \) cm.
2. Power \( P=\frac{1}{-1\,\text{m}}=-1 \) D.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (b) +3 D

Explanation:
1. The lens must form a virtual image at 100 cm of an object at 25 cm.
2. \( \frac1f=\frac1{0.25}-\frac1{1} \) in metres gives \( P=4-1=+3 \) D (converging lens).

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (c) 4 rad/s

Explanation:
1. When the plane face turns through a small angle \( \theta \), the angle of incidence is \( \theta \) and the refraction angle is \( r\approx n\theta \).
2. The refracted ray deviates from the fixed x-axis by \( r-\theta=(n-1)\theta \).
3. So the angular velocity is \( \left(\frac53-1\right)\times6=4 \) rad/s.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.

Answer: (d) 80 cm

Explanation:
1. The spot disappears at total internal reflection: \( \sin\theta_c=\frac35 \), \( \theta_c=37^\circ \), at which the refracted ray grazes the plane face (\( r=90^\circ \)).
2. The deviation from the x-axis is \( 90^\circ-37^\circ=53^\circ \).
3. Distance \( =60\tan53^\circ=60\times\frac43=80 \) cm.

Teacher's Note:
1. Write the sign convention before substituting numbers, and check that the result is physically sensible.