JEE Physics Ray Optics and Optical Instruments MCQs Set 17

Answer: (d) the same for all colours of light

Explanation:
1. Focal length of a spherical mirror is \( \frac R2 \), which depends only on the geometry.
2. Reflection does not depend on wavelength, so the focal length is the same for all colours.

Teacher's Note:
1. A lens shows chromatic aberration because refraction depends on colour; a mirror does not.

Answer: (b) 10 cm away from the mirror

Explanation:
1. At \( u=-60 \): \( \frac1v=-\frac1{15}+\frac1{60}=-\frac3{60} \), so \( v=-20 \) cm.
2. After moving 30 cm closer, \( u=-30 \): \( \frac1v=-\frac1{15}+\frac1{30}=-\frac1{30} \), so \( v=-30 \) cm.
3. The image shifts from 20 cm to 30 cm in front of the mirror, which is 10 cm away from the mirror.

Teacher's Note:
1. As a real object approaches a concave mirror beyond the focus, the image moves away from the mirror.

Answer: (c) 15 cm

Explanation:
1. Apparent depth \( =\frac{\text{real depth}}{\mu} \). From one side the real depth is \( 6\times1.5=9 \) cm.
2. From the other side the real depth is \( 4\times1.5=6 \) cm.
3. Thickness \( =9+6=15 \) cm.

Teacher's Note:
1. The two real depths of the bubble add up to the thickness of the slab.

Answer: (d) 30°

Explanation:
1. Angle of prism \( A=60^\circ \), so \( i=e=\frac34\times60^\circ=45^\circ \).
2. \( \delta=i+e-A=45+45-60 \).
3. \( \delta=30^\circ \).

Teacher's Note:
1. For a prism, deviation is the sum of the angles of incidence and emergence minus the prism angle.

Answer: (c) scattering of light

Explanation:
1. Short wavelengths are scattered much more strongly by air molecules, in proportion to \( \frac1{\lambda^4} \) (Rayleigh scattering).
2. Blue light is therefore scattered in all directions and the sky looks blue.

Teacher's Note:
1. Red light scatters least, which is why sunsets look red.

Answer: (b) \( \sqrt2 \)

Explanation:
1. The ray enters the equal face normally, so it goes straight to the hypotenuse and meets it at \( 45^\circ \).
2. For TIR, \( 45^\circ\ge C \), so \( \sin45^\circ\ge\frac1\mu \).
3. \( \mu\ge\sqrt2 \).

Teacher's Note:
1. Minimum refractive index for a right-angled prism to act as a reflector is \( \sqrt2 \).

Answer: (a) separate the red part from the green and blue

Explanation:
1. Critical angle: \( \sin C=\frac1\mu \). Red \( 1.39 \): \( \sin C=0.719 \), \( C\approx46^\circ \). Green \( 1.44 \): \( C\approx44^\circ \). Blue \( 1.47 \): \( C\approx43^\circ \).
2. At the hypotenuse the angle of incidence is \( 45^\circ \). Red has \( 45^\circ\lt C \), so it is transmitted; green and blue have \( 45^\circ>C \), so they are totally reflected.
3. So red is separated from green and blue.

Teacher's Note:
1. Colours whose critical angle is less than the angle of incidence are totally reflected.

Answer: (b) \( 30^\circ,\ \sqrt2 \)

Explanation:
1. At minimum deviation the ray passes symmetrically, so \( r=\frac A2=30^\circ \) and \( i=45^\circ \).
2. \( \mu=\frac{\sin45^\circ}{\sin30^\circ}=\sqrt2 \).
3. \( \delta_{min}=2i-A=90-60=30^\circ \).

Teacher's Note:
1. Minimum deviation: \( i=e \) and \( r_1=r_2=\frac A2 \).

Answer: (c) -40 cm

Explanation:
1. Lens maker formula in a medium: \( \frac1f=\left(\frac{\mu_g}{\mu_m}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right) \).
2. \( \frac{\mu_g}{\mu_m}=\frac{1.5}2=0.75 \), and \( \frac1{R_1}-\frac1{R_2}=\frac2{20}=0.1 \).
3. \( \frac1f=(0.75-1)(0.1)=-0.025 \), so \( f=-40 \) cm.

Teacher's Note:
1. A convex lens of lower index than the surrounding liquid behaves as a diverging lens.

Answer: (d) -1 D

Explanation:
1. Power \( P=\frac1f \) in metres. Convex lens: \( P_1=\frac1{1\ \text{m}}=+1 \) D. Concave lens: \( P_2=-\frac1{0.5}=-2 \) D.
2. Powers add for lenses in contact: \( P=1-2 \).
3. \( P=-1 \) D.

Teacher's Note:
1. Convert focal lengths to metres before using power.

Answer: (b) 6 mm

Explanation:
1. Limit of resolution \( \Delta\theta=\frac{1.22\lambda}{D}=\frac{1.22\times5\times10^{-7}}{0.1}\approx6.1\times10^{-6} \) rad.
2. Minimum separation \( =\Delta\theta\times1000\ \text{m} \).
3. About \( 6\times10^{-3} \) m, that is 6 mm.

Teacher's Note:
1. Resolving power increases with larger aperture and shorter wavelength.

Answer: (d) 81 cm

Explanation:
1. Objective: \( \frac1v=\frac1{60}-\frac1{300}=\frac4{300} \), so \( v=75 \) cm.
2. The final image at infinity needs the eyepiece to be at distance \( f_e=6 \) cm from the intermediate image.
3. Separation \( =75+6=81 \) cm.

Teacher's Note:
1. For a nearby object, the objective forms its image beyond its focal point, so the tube length is larger than \( f_o+f_e \).

Answer: (d) 50 cm concave

Explanation:
1. The far point is 50 cm, so the lens must form images of distant objects at 50 cm on the same side as the object.
2. \( f=-50 \) cm, which is a concave lens.

Teacher's Note:
1. Myopia is corrected with a diverging lens of focal length equal to the far point distance.

Answer: (d) 2 cm

Explanation:
1. Total power \( P=35+15=50 \) D.
2. Focal length \( f=\frac1{50}\ \text{m}=2 \) cm.
3. The retina lies at the focus for distant objects, so the distance is 2 cm.

Teacher's Note:
1. For a normal eye relaxed on distant objects, the image forms at the focal plane.

Answer: (d) concave, -0.5 D

Explanation:
1. To see distant objects the lens must form an image at the far point 200 cm (virtual).
2. \( f=-200\ \text{cm}=-2\ \text{m} \).
3. \( P=\frac1f=-0.5 \) D, a concave lens.

Teacher's Note:
1. Only the far point matters for correcting distant vision.

Answer: (b) 30 cm, -45 cm

Explanation:
1. Achromatic condition: \( \frac{\omega_1}{f_1}+\frac{\omega_2}{f_2}=0 \), so \( \frac{f_1}{f_2}=-\frac{\omega_1}{\omega_2}=-\frac{0.024}{0.036}=-\frac23 \).
2. Only the pair 30 cm and -45 cm has this ratio.

Teacher's Note:
1. The two lenses must have opposite signs, one convex and one concave.