JEE Physics Ray Optics and Optical Instruments MCQs Set 04

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

Explanation:
1. The angle of incidence is measured from the normal, and it is 0 for normal incidence.
2. By the law of reflection the angle of reflection is also 0.

Teacher's Note:
1. Angles are always measured from the normal.

Answer: (a) 3

Explanation:
1. Number of images \( =360^\circ/\theta-1=360/90-1=3 \).
2. This formula applies when 360/θ is an integer.

Teacher's Note:
1. n = 360/θ - 1.

Answer: (c) 5, 25, 35

Explanation:
1. The first image in M2 is 5 cm behind M2.
2. The object is 10 cm from M1, so its image in M1 is 10 cm behind M1, which is 15 + 10 = 25 cm from M2.
3. The image of the first image in M1 (which is 15 + 5 = 20 cm from M1) is 20 cm behind M1, at 35 cm from M2.

Teacher's Note:
1. Images in parallel mirrors form an infinite series.
2. Measure every distance from the same mirror.

Answer: (b) 3 feet

Explanation:
1. The minimum length of a plane mirror needed to see the full image is half the height of the person.
2. So it is 3 feet.

Teacher's Note:
1. Independent of the distance from the mirror.

Answer: (c) 7

Explanation:
1. \( n=360/45-1=8-1=7 \).
2. The formula gives integer values here.

Teacher's Note:
1. Use n = 360/θ - 1.

Answer: (b) 1.78 cm

Explanation:
1. \( f=12.5 \) cm, \( u=-40 \): \( 1/v=1/12.5+1/40=0.105 \), so \( v=9.52 \) cm.
2. \( m=-v/u=0.238 \) and \( h'=7.5\times0.238=1.78 \) cm.

Teacher's Note:
1. Convex mirror gives a small virtual erect image.

Answer: (c) 60 cm

Explanation:
1. Real image: \( m=-1/4 \) so \( v=u/4 \).
2. \( \dfrac4u+\dfrac1u=-\dfrac1{12} \) gives \( u=-60 \) cm.

Teacher's Note:
1. Real image is inverted so m is negative.

Answer: (a) -10 cm

Explanation:
1. \( m=-2 \) so \( v=-30 \) cm.
2. \( 1/f=1/v+1/u=-1/30-1/15=-1/10 \), so \( f=-10 \) cm.

Teacher's Note:
1. Concave mirror has negative focal length in this convention.

Answer: (c) 4 cm/s

Explanation:
1. \( 1/v=-1/24+1/60 \) gives \( v=-40 \) cm.
2. \( v_I=(v/u)^2v_O=(40/60)^2\times9=4 \) cm/s.

Teacher's Note:
1. Longitudinal magnification is (v/u) squared.

Answer: (d) Zero

Explanation:
1. When the object is at the centre of curvature, the real image is formed at the same point.
2. So the distance between object and image can be zero.

Teacher's Note:
1. u = v = 2f gives a zero separation.

Answer: (d) Same for all lights

Explanation:
1. For a mirror \( f=R/2 \), which does not depend on the refractive index.
2. So it is the same for all colours.

Teacher's Note:
1. Mirrors have no chromatic aberration.

Answer: (d) 40 cm

Explanation:
1. \( R=2f=40 \) cm.
2. The relation holds for spherical mirrors of small aperture.

Teacher's Note:
1. R = 2f.

Answer: (a) Frequency

Explanation:
1. Frequency does not change when light passes from one medium to another, while wavelength and velocity do.
2. So colour is characterised by frequency.

Teacher's Note:
1. Wavelength changes in a medium.

Answer: (b) 2 cm

Explanation:
1. Apparent shift \( =t(1-1/\mu)=6\times(1/3)=2 \) cm.
2. The microscope must be raised by 2 cm.

Teacher's Note:
1. Apparent depth is real depth over mu.

Answer: (b) 6 m

Explanation:
1. \( \sin C=3/5 \), so \( \tan C=3/4 \).
2. Radius \( =h\tan C=4\times3/4=3 \) m and diameter 6 m.

Teacher's Note:
1. Light beyond the critical angle is totally reflected.

Answer: (a) Yellow, orange, red

Explanation:
1. Colours of larger wavelength (yellow, orange, red) have smaller refractive index and larger critical angle, so they are not totally reflected and emerge.
2. Violet, indigo and blue have smaller critical angles and are totally reflected.

Teacher's Note:
1. Violet has the smallest critical angle.

Answer: (c) 10.5 cm

Explanation:
1. The apparent depths add to 7 cm.
2. Real thickness = 7 x 1.5 = 10.5 cm.

Teacher's Note:
1. Real depth = apparent depth x mu.

Answer: (c) \( \dfrac{nt}{c} \)

Explanation:
1. Speed in glass \( v=c/n \).
2. \( T=t/v=nt/c \).

Teacher's Note:
1. Time = optical path / c.

Answer: (a) Total internal reflection

Explanation:
1. Light is guided along the fibre by repeated total internal reflection at the core-cladding boundary.
2. The other phenomena do not guide light.

Teacher's Note:
1. Core has a higher index than cladding.

Answer: (c) Total internal reflection of light

Explanation:
1. Hot air near the ground has a lower refractive index and light from the sky bends and is totally internally reflected.
2. This makes the ground appear to have water.

Teacher's Note:
1. Seen in deserts.

Answer: (d) Have a refractive index exactly matching that of the surrounding fluid

Explanation:
1. When the refractive indices match, there is no reflection or refraction at the boundary.
2. So the object cannot be seen.

Teacher's Note:
1. Glass rod disappears in a liquid of the same index.

Answer: (b) \( 2.3^\circ \)

Explanation:
1. For a thin prism \( \delta=(\mu-1)A=0.52\times4.5^\circ \).
2. \( \delta=2.34^\circ \approx2.3^\circ \).

Teacher's Note:
1. Thin prism deviation (mu - 1) A.