JEE Physics Ray Optics and Optical Instruments MCQs Set 18

Answer: (b) f1 > f and f2 becomes negative

Explanation:
1. \( \frac1{f_l}=\left(\frac{\mu_g}{\mu_l}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \).
2. In the liquid of index 4/3 the factor is 1/8, smaller than 1/2 in air, so f1 > f. In the liquid of index 5/3 the factor is negative (-1/10), so f2 is negative: the lens becomes diverging.

Teacher's Note:
1. A lens is converging only when its material is optically denser than the surrounding medium.

Answer: (b) L/I

Explanation:
1. In normal adjustment the objective and eyepiece are separated by \( f_o+f_e \), so the line lies at \( u=f_o+f_e \) from the eyepiece.
2. Magnification of the eyepiece for this object: \( \frac IL=\frac{f_e}{u-f_e}=\frac{f_e}{f_o} \).
3. Therefore \( M=\frac{f_o}{f_e}=\frac LI \).

Teacher's Note:
1. The magnifying power in normal setting is fo/fe, and the line on the objective gives that ratio directly.

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

Explanation:
1. Each layer of thickness h and index \( \mu \) shifts the image by \( h\left(1-\frac1\mu\right) \).
2. Adding the water and kerosene layers gives option (c).

Teacher's Note:
1. For near normal viewing the shifts of successive layers simply add.

Answer: (a) -50 cm

Explanation:
1. Each glass lens: \( P=\frac{0.5}{20}=\frac1{40}\ \text{cm}^{-1} \). Two of them: \( \frac1{20} \).
2. The oil forms a biconcave lens with both radii 20 cm: \( P=(1.7-1)\left(-\frac1{20}-\frac1{20}\right)=-0.07 \).
3. Total power \( =0.05-0.07=-0.02\ \text{cm}^{-1}\Rightarrow f=-50\ \text{cm} \)

Teacher's Note:
1. Lenses in contact add their powers; the oil between two convex surfaces acts as a diverging lens.

Answer: (d) \( 180^\circ-2A \)

Explanation:
1. \( \mu=\frac{\sin\frac{A+\delta}{2}}{\sin\frac A2}=\cot\frac A2\Rightarrow\sin\frac{A+\delta}2=\cos\frac A2=\sin\left(90^\circ-\frac A2\right) \)
2. \( \frac{A+\delta}{2}=90^\circ-\frac A2\Rightarrow\delta=180^\circ-2A \)

Teacher's Note:
1. Use the prism formula at minimum deviation and match the sine to a cosine.

Answer: (a) 45 degrees

Explanation:
1. \( |\vec A|=\sqrt{108+192+100}=20,\ \cos i=\frac{10}{20}\Rightarrow i=60^\circ \).
2. \( \sqrt2\sin60^\circ=\sqrt3\sin r\Rightarrow\sin r=\frac1{\sqrt2}\Rightarrow r=45^\circ \)

Teacher's Note:
1. The angle of incidence is measured from the normal (the z axis), so use the z component of the direction vector.

Answer: (d) The microscope will decrease but that of the telescope will increase

Explanation:
1. Microscope: \( M\propto\frac1{f_o} \) (roughly), so it decreases.
2. Telescope: \( M=\frac{f_o}{f_e} \), so it increases.

Teacher's Note:
1. A microscope wants a short objective focal length, a telescope wants a long one.

Answer: (b) 2 cos A

Explanation:
1. The ray retraces its path only if it strikes the silvered face normally, so the angle of refraction at the first face is \( r=A \).
2. \( \mu=\frac{\sin2A}{\sin A}=2\cos A \)

Teacher's Note:
1. Retracing the path means normal incidence on the mirror face.

Answer: (a) \( \frac{R}{\mu_1-\mu_2} \)

Explanation:
1. Plano-convex lens: \( P_1=\frac{\mu_1-1}{R} \); plano-concave lens: \( P_2=-\frac{\mu_2-1}{R} \).
2. \( P=P_1+P_2=\frac{\mu_1-\mu_2}{R}\Rightarrow f=\frac{R}{\mu_1-\mu_2} \)

Teacher's Note:
1. The two curved surfaces cancel in the interface, leaving a net power proportional to the difference of indices.

Answer: (a) 1/15 m/s

Explanation:
1. \( u=-280\ \text{cm},\ f=20\ \text{cm}\Rightarrow v=\frac{uf}{u-f}=\frac{-280\times20}{-300}=18.67\ \text{cm} \)
2. Longitudinal magnification \( \frac{v_I}{v_O}=\left(\frac{f}{f-u}\right)^2=\left(\frac{20}{300}\right)^2=\frac1{225} \), so \( v_I=\frac{15}{225}=\frac1{15}\ \text{m/s} \)

Teacher's Note:
1. Image speed equals the square of the lateral magnification times the object speed.

Answer: (a) \( \mu A \)

Explanation:
1. Normal emergence means \( r_2=0 \), so \( r_1=A \).
2. \( \sin i=\mu\sin A\approx\mu A\Rightarrow i\approx\mu A \)

Teacher's Note:
1. For small angles the sines can be replaced by the angles.

Answer: (a) \( f_1+f_2 \)

Explanation:
1. The lens focuses the parallel beam at distance \( f_2 \) from it.
2. For the beam to return parallel, this point must be the focus of the mirror, which is \( f_1 \) in front of the mirror: \( d-f_1=f_2 \), so \( d=f_1+f_2 \).

Teacher's Note:
1. A parallel beam returns parallel only if the point where it is focused is the focus of the mirror.

Answer: (a) (2f, 2f)

Explanation:
1. The line makes 135 degrees with the axis, so it represents \( |u|=|v| \).
2. \( \frac1v-\frac1u=\frac1f\Rightarrow\frac2{|u|}=\frac1f\Rightarrow|u|=|v|=2f \)

Teacher's Note:
1. Equal object and image distances for a thin lens occur when the object is at 2f.

Answer: (c) Real, inverted, height = 4 cm

Explanation:
1. For a biconvex lens of index 1.5 with both radii 20 cm, \( f=20\ \text{cm} \).
2. \( \frac1v=\frac1{20}-\frac1{30}\Rightarrow v=60\ \text{cm} \); \( m=\frac vu=-2 \), so the image is real, inverted and 4 cm tall.

Teacher's Note:
1. The object lies between f and 2f, giving a real, inverted, enlarged image beyond 2f.

Answer: (d) \( \sin^{-1}\frac1{\sqrt3} \)

Explanation:
1. Grazing the wall means the angle of incidence on the wall equals the critical angle: \( \sin C=\frac{\sqrt3}2\Rightarrow C=60^\circ \).
2. Refraction angle at the end face \( r=90^\circ-60^\circ=30^\circ \).
3. \( \sin\theta=\mu\sin r=\frac2{\sqrt3}\times\frac12=\frac1{\sqrt3} \)

Teacher's Note:
1. The refracted ray inside the rod makes angle (90 - C) with the axis.

Answer: (c) 10 degrees

Explanation:
1. No deviation: \( (\mu_1-1)A_1=(\mu_2-1)A_2 \).
2. \( 0.5\times15=0.75\,A_2\Rightarrow A_2=10^\circ \)

Teacher's Note:
1. For thin prisms the deviations are (mu - 1)A and they must cancel.

Answer: (c) \( \mu=1.50 \)

Explanation:
1. Total internal reflection needs \( \sin45^\circ\ge\frac1\mu\Rightarrow\mu\ge\sqrt2=1.414 \).
2. Only 1.50 satisfies this.

Teacher's Note:
1. The critical angle must be smaller than the angle of incidence.

Answer: (d) \( 1.8\times10^8 \) m/s

Explanation:
1. The ray path is 5 cm long (3-4-5 triangle), so the angle with the normal is the critical angle: \( \sin C=\frac35 \).
2. \( \mu=\frac1{\sin C}=\frac53,\ v=\frac c\mu=\frac{3\times10^8}{5/3}=1.8\times10^8\ \text{m/s} \)

Teacher's Note:
1. A ray that just skims the surface is incident at the critical angle.

Answer: (b) \( \frac{36}{\sqrt7} \)

Explanation:
1. \( r=\frac{h}{\sqrt{\mu^2-1}}=\frac{12}{\sqrt{16/9-1}}=\frac{12\times3}{\sqrt7}=\frac{36}{\sqrt7}\ \text{cm} \)

Teacher's Note:
1. The circle is bounded by rays reaching the fish at the critical angle.

Answer: (a) 45 degrees

Explanation:
1. Retracing requires normal incidence on the mirror, so \( r=A=30^\circ \).
2. \( \sin i=\sqrt2\sin30^\circ=\frac1{\sqrt2}\Rightarrow i=45^\circ \)

Teacher's Note:
1. Normal incidence on the mirrored face is the retracing condition.