JEE Physics Ray Optics and Optical Instruments MCQs Set 27

Answer: (c) 40.0 cm

Explanation:
1. \( \dfrac1f=\dfrac{1.5-1}{14}+\dfrac{1.2-1}{14}=\dfrac{0.7}{14}=\dfrac1{20} \), so \( f=20 \) cm.
2. \( u=-40 \): \( \dfrac1v=\dfrac1{20}-\dfrac1{40}=\dfrac1{40}\Rightarrow v=40 \) cm.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (a) \( -15 \) and 10

Explanation:
1. Let the concave lens have focal length \( -3x \) and the convex lens \( 2x \) (positive combination of 30 cm needs the convex lens to be stronger). \( \dfrac1{2x}-\dfrac1{3x}=\dfrac1{6x}=\dfrac1{30}\Rightarrow x=5 \).
2. So the convex lens is 10 cm and the concave lens is \( -15 \) cm; combination \( \dfrac1{10}-\dfrac1{15}=\dfrac1{30} \).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (b) 2.5 cm

Explanation:
1. The convex lens alone would form the image at 30 cm. The concave lens at 26 cm sees it as a virtual object at \( u=+4 \) cm.
2. \( \dfrac1v=-\dfrac1{20}+\dfrac14=\dfrac15\Rightarrow v=5 \) cm; magnification \( \dfrac vu=\dfrac54 \), so the new size is \( 2\times1.25=2.5 \) cm.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (d) convergent lens of focal length 3.5 R

Explanation:
1. \( \dfrac1f=\left(\dfrac{1.5}{1.75}-1\right)\left(-\dfrac1R-\dfrac1R\right)=\left(-\dfrac17\right)\left(-\dfrac2R\right)=\dfrac2{7R} \).
2. \( f=+3.5R \): a convergent lens, because the lens material has a lower index than the surrounding medium.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (a) is smaller if the curved surface of the lens faces the object

Explanation:
1. Spherical aberration of a plano-convex lens depends on how the total deviation is shared between the two surfaces.
2. For a distant object it is smaller when the curved face is towards the object, since the deviation is then shared more evenly between refractions at the two faces.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (c) (a) and (d) only

Explanation:
1. Real image of size twice: \( m=-2 \), \( u=-30 \), \( v=60 \); \( \dfrac1f=\dfrac1{60}+\dfrac1{30}=\dfrac1{20} \), \( f=20 \) cm (d true).
2. Convex mirror: \( v=+10 \), \( u=-30 \): \( \dfrac2R=\dfrac1{10}-\dfrac1{30}=\dfrac1{15} \), \( R=30 \) cm, so (b) is false and (c) is false (the image is virtual). Then \( n-1=\dfrac Rf=1.5 \), \( n=2.5 \) (a true).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (a) 60 cm

Explanation:
1. \( \dfrac{f_m}{f_a}=\dfrac{\mu-1}{\mu/\mu_m-1}=\dfrac{0.5}{9/8-1}=4 \).
2. \( f_m=4\times15=60 \) cm.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (b) 15 cm

Explanation:
1. A parallel beam is focused by A at 20 cm; B must have this point as its virtual focus on the far side.
2. So \( d=f_A-|f_B|=20-5=15 \) cm (Galilean arrangement).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (d) 6

Explanation:
1. \( m=\dfrac f{f+u} \): \( m_{25}=\dfrac{20}{20-25}=-4 \) and \( m_{50}=\dfrac{20}{20-50}=-\dfrac23 \).
2. Ratio \( =\dfrac{-4}{-2/3}=6 \).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (a) 0.09 m/s and 0.3 s\(^{-1}\)

Explanation:
1. \( v=\dfrac{uf}{u-f}=1.2 \) m and \( \dfrac{dv}{dt}=\left(\dfrac f{u-f}\right)^2\left|\dfrac{du}{dt}\right|=9\times0.01=0.09 \) m/s.
2. \( |m|=\dfrac f{u-f} \), so \( \left|\dfrac{dm}{dt}\right|=\dfrac f{(u-f)^2}\left|\dfrac{du}{dt}\right|=\dfrac{0.3}{0.01}\times0.01=0.3 \) s\(^{-1}\).

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (c) 12 cm

Explanation:
1. \( f_{lens}=\dfrac R{\mu-1}=24 \) cm. With the plane face silvered the system acts as a concave mirror of focal length \( \dfrac{f_{lens}}2 \).
2. The rays converge at \( 12 \) cm from the lens.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (d) 30 cm in front of the lens (real)

Explanation:
1. Equivalent concave mirror of focal length \( F=12 \) cm.
2. \( \dfrac1v=\dfrac1F-\dfrac1u\) with signs: \( \dfrac1v=-\dfrac1{12}+\dfrac1{20}=-\dfrac1{30}\Rightarrow v=-30 \) cm: a real image 30 cm in front of the lens.

Teacher's Note:
1. For thin lenses use 1/v - 1/u = 1/f with Cartesian signs and 1/f = (mu - 1)(1/R1 - 1/R2); powers add for lenses in contact.
2. In a medium of index mu_m replace mu by mu/mu_m in the lens-maker formula; a lens can change from converging to diverging.

Answer: (b) \( \theta\gt \sin^{-1}\left[\mu\sin\left(A-\sin^{-1}\dfrac1\mu\right)\right] \)

Explanation:
1. Emergence at AC needs \( r_2\lt C \) where \( \sin C=\dfrac1\mu \). Since \( r_1=A-r_2 \), this means \( r_1\gt A-C \).
2. \( \sin\theta=\mu\sin r_1\gt \mu\sin\left(A-\sin^{-1}\dfrac1\mu\right) \), so \( \theta\gt \sin^{-1}\left[\mu\sin\left(A-\sin^{-1}\dfrac1\mu\right)\right] \).

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (d) \( 30^\circ \) for both the colours

Explanation:
1. At minimum deviation \( r_1=r_2=\dfrac A2 \), whatever the colour.
2. With \( A=60^\circ \), \( r=30^\circ \) for both red and violet (only the angle of incidence differs).

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (a) the ray inside the prism (QR) is horizontal, parallel to the base

Explanation:
1. At minimum deviation the ray passes symmetrically, so inside the prism it is parallel to the base (the table).
2. The incident and emergent rays are inclined to the horizontal by equal and opposite angles, so neither is horizontal.

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (c) \( 3^\circ \)

Explanation:
1. No deviation for the mean ray: \( A_1(\mu_1-1)=A_2(\mu_2-1) \).
2. \( 4\times0.54=A_2\times0.72\Rightarrow A_2=3^\circ \).

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (d) separate the red colour from the green and blue colours

Explanation:
1. Critical angles: \( \sin C=\dfrac1\mu \) gives red \( 46.0^\circ \), green \( 44.0^\circ \), blue \( 42.9^\circ \).
2. At \( 45^\circ \): red (\( C\gt 45^\circ \)) is transmitted; green and blue (\( C\lt 45^\circ \)) are totally reflected. So red is separated from green and blue.

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (b) \( 15^\circ \)

Explanation:
1. Normal incidence gives \( r_1=0 \), so \( r_2=A=30^\circ \) at the second face.
2. \( \sin e=\sqrt2\sin30^\circ=\dfrac1{\sqrt2}\Rightarrow e=45^\circ \); \( \delta=e-A=15^\circ \).

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (c) \( 0^\circ \)

Explanation:
1. Minimum deviation for \( \mu=\sqrt2 \), \( A=60^\circ \): \( \sin\dfrac{A+\delta_m}2=\sqrt2\sin30^\circ=\dfrac1{\sqrt2}\Rightarrow\dfrac{A+\delta_m}2=45^\circ\Rightarrow\delta_m=30^\circ \).
2. So the given deviation is the minimum one and the ray inside is parallel to the base: angle \( 0^\circ \).

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (a) 2

Explanation:
1. \( \sin r_1=\dfrac{\sin60^\circ}n \), \( r_2=60^\circ-r_1 \), \( \sin\theta=n\sin r_2 \). At \( n=\sqrt3 \): \( r_1=r_2=30^\circ \).
2. \( \dfrac{dr_1}{dn}=-\dfrac13 \) (from \( \cos r_1\,dr_1=-\dfrac{\sin60^\circ}{n^2}dn \)). Then \( \cos\theta\,d\theta=\sin r_2\,dn+n\cos r_2\cdot\dfrac13dn=\tfrac12+\tfrac12=1 \), so \( \dfrac{d\theta}{dn}=\dfrac1{\cos60^\circ}=2 \).

Teacher's Note:
1. For a prism use A = r1 + r2 and delta = i + e - A; at minimum deviation r1 = r2 = A/2 and the ray inside is parallel to the base.
2. For emergence or total internal reflection compare the angle of incidence on the second face with the critical angle sin C = 1/mu.

Answer: (c) 0.24 \(\mu\)m

Explanation:
1. \( d=\dfrac{0.61\lambda}{NA} \).
2. \( d=\dfrac{0.61\times0.5\ \mu m}{1.25}=0.244\ \mu m\approx0.24\ \mu m \).

Teacher's Note:
1. Resolving limit of a telescope is 1.22 lambda / D and of a microscope 0.61 lambda / NA; larger aperture or smaller wavelength means better resolution.
2. For a telescope in normal adjustment the lens separation is fo + fe and the angular magnification is fo / fe.

Answer: (d) \( 3.0\times10^{-7} \) rad

Explanation:
1. \( \Delta\theta=\dfrac{1.22\lambda}D \).
2. \( =\dfrac{1.22\times6\times10^{-7}}{2.5}=2.93\times10^{-7}\approx3.0\times10^{-7} \) rad.

Teacher's Note:
1. Resolving limit of a telescope is 1.22 lambda / D and of a microscope 0.61 lambda / NA; larger aperture or smaller wavelength means better resolution.
2. For a telescope in normal adjustment the lens separation is fo + fe and the angular magnification is fo / fe.

Answer: (b) 20 times nearer

Explanation:
1. The telescope increases the visual angle subtended by the tree 20 times.
2. The apparent size is the same as if the tree were 20 times nearer (its actual height does not change).

Teacher's Note:
1. Resolving limit of a telescope is 1.22 lambda / D and of a microscope 0.61 lambda / NA; larger aperture or smaller wavelength means better resolution.
2. For a telescope in normal adjustment the lens separation is fo + fe and the angular magnification is fo / fe.

Answer: (c) \( a=\sqrt{\lambda L}\ \text{and}\ b_{min}=\sqrt{4\lambda L} \)

Explanation:
1. Spot size \( b=a+\dfrac{\lambda L}a \) (geometrical plus diffraction spread).
2. Minimum when \( a=\dfrac{\lambda L}a\Rightarrow a=\sqrt{\lambda L} \), giving \( b_{min}=2\sqrt{\lambda L}=\sqrt{4\lambda L} \).

Teacher's Note:
1. Resolving limit of a telescope is 1.22 lambda / D and of a microscope 0.61 lambda / NA; larger aperture or smaller wavelength means better resolution.
2. For a telescope in normal adjustment the lens separation is fo + fe and the angular magnification is fo / fe.

Answer: (a) 2.4 and 12.0

Explanation:
1. Final image at infinity means the intermediate image is at the eyepiece focus, 3 cm from the eyepiece, so \( v_o=15-3=12 \) cm.
2. \( \dfrac1{u}=\dfrac1{v}-\dfrac1{f}=\dfrac1{12}-\dfrac12=-\dfrac5{12}\Rightarrow u=-2.4 \) cm.

Teacher's Note:
1. Resolving limit of a telescope is 1.22 lambda / D and of a microscope 0.61 lambda / NA; larger aperture or smaller wavelength means better resolution.
2. For a telescope in normal adjustment the lens separation is fo + fe and the angular magnification is fo / fe.

Answer: (b) \( f_o=30 \) cm and \( f_e=6 \) cm

Explanation:
1. Normal adjustment: \( f_o+f_e=36 \) and \( \dfrac{f_o}{f_e}=5 \).
2. \( 6f_e=36\Rightarrow f_e=6 \) cm, \( f_o=30 \) cm.

Teacher's Note:
1. Resolving limit of a telescope is 1.22 lambda / D and of a microscope 0.61 lambda / NA; larger aperture or smaller wavelength means better resolution.
2. For a telescope in normal adjustment the lens separation is fo + fe and the angular magnification is fo / fe.