JEE Physics Ray Optics and Optical Instruments MCQs Set 24

Answer: (b) \( 5\hat i+4\hat j+5\hat k \)

Explanation:
1. Component along the normal: \( v_{I} = 2v_m - v_O = 2(4)-3 = 5 \).
2. Components parallel to the mirror are unchanged: 4 and 5.

Teacher's Note:
1. Image velocity along the normal is 2v_m − v_O.
2. Tangential components of the image equal those of the object.

Answer: (a) 145 cm from M₂

Explanation:
1. Images in M₂ lie behind it at distances 5, 65, 75, 135, 145, ... cm (alternate gaps of 60 and 10 cm).
2. The fifth is at 145 cm.

Teacher's Note:
1. Image of an image: keep reflecting alternately in the two mirrors.
2. Mirror separation is 35 cm.

Answer: (c) \( \frac{7\pi}{12} \)

Explanation:
1. Deviation after two reflections \( = 2\pi - 2\alpha = \frac{5\pi}6 \).
2. \( \alpha = \frac{7\pi}{12} \).

Teacher's Note:
1. Deviation by two mirrors is 360° − 2α.
2. The result does not depend on the angle of incidence.

Answer: (c) away from the object, 8 m/s

Explanation:
1. With the object moving towards the mirror at +10, the image velocity is \( 2v_m - 10 = +6 \), so \( v_m = 8 \) m/s in the same direction as the object's motion, i.e. away from the object.

Teacher's Note:
1. Use v_image = 2v_mirror − v_object along the normal.
2. Take the object's direction as positive.

Answer: (d) 75 cm

Explanation:
1. Inverted and magnified means real, \( m = -2 = -\frac vu \), so |v| = 2|u|.
2. \( \frac1{2x}+\frac1x = \frac1{50} \Rightarrow x = 75 \) cm.

Teacher's Note:
1. Real images from concave mirrors are inverted.
2. Use the mirror formula with sign convention.

Answer: (b) 15 cm

Explanation:
1. Real image: \( v = 3u \) in magnitude, so \( \frac1{60}+\frac1{20} = \frac1f \Rightarrow \frac1f = \frac{4}{60} \).
2. f = 15 cm.

Teacher's Note:
1. For a real image both distances have the same sign.
2. Object between f and 2f gives an enlarged real image.

Answer: (b) virtual at \( \frac{20}{3} \) cm from the mirror

Explanation:
1. \( \frac1v = \frac1f-\frac1u = \frac1{10}+\frac1{20} = \frac3{20} \), so \( v = \frac{20}{3} \) cm behind the mirror.

Teacher's Note:
1. A convex mirror always forms a virtual, erect, diminished image.
2. Take f positive for a convex mirror.

Answer: (a) 54 cm, 5 cm

Explanation:
1. \( f = 18 \) cm: \( \frac1v = -\frac1{18}+\frac1{27} = -\frac1{54} \), v = 54 cm in front of the mirror.
2. \( |m| = \frac{54}{27} = 2 \), image size = 5 cm.

Teacher's Note:
1. Real image can be received on a screen in front of the mirror.
2. Size = |m| × 2.5 cm.

Answer: (d) 6.7 cm behind the mirror, 5/9

Explanation:
1. \( \frac1v = \frac1{15}+\frac1{12} = \frac{9}{60} \Rightarrow v = 6.67 \) cm.
2. \( m = \frac{v}{|u|} = \frac{6.67}{12} = \frac59 \).

Teacher's Note:
1. Convex mirror: f is positive, u is negative.
2. Magnification is less than 1 for a convex mirror.

Answer: (b) Aberration occurs only for real images

Explanation:
1. Spherical aberration occurs for real and virtual images alike.
2. The other three statements are correct: a virtual image from a concave mirror is erect, a virtual image cannot be caught on a screen, and reflection does not depend on wavelength.

Teacher's Note:
1. Chromatic aberration is a refraction effect.
2. Aberration is a defect of the optics, not of the image type.

Answer: (a) 66.67 cm

Explanation:
1. Virtual object at u = +0.2 m; real image at v = −0.5 m (in front). \( \frac1f = \frac1v+\frac1u = -2+5 = 3 \), f = 1/3 m = 33.3 cm.
2. R = 2f = 66.67 cm.

Teacher's Note:
1. A converging beam gives a virtual object for the mirror.
2. Take distances in front of the mirror as negative.

Answer: (c) may be at infinity

Explanation:
1. For a concave mirror an object at F gives an image at infinity. For a convex mirror the same condition gives a virtual image at f/2, not at infinity.
2. So the image may be at infinity (depends on the type of mirror).

Teacher's Note:
1. Concave and convex mirrors behave differently.
2. Object at F is the standard way to get a parallel beam.

Answer: (a) the refractive index fluctuations in the earth's atmosphere

Explanation:
1. Air layers of varying temperature and density continuously change the refractive index, bending the starlight by varying amounts.
2. The apparent position and brightness therefore flicker.

Teacher's Note:
1. Planets, being extended sources, twinkle much less.
2. Twinkling is an atmospheric refraction effect.

Answer: (d) 22.5 cm

Explanation:
1. For an observer in water, apparent distance of an object in air \( = d\times\frac{\mu_w}{\mu_a} \).
2. \( 30 = d\times\frac43 \Rightarrow d = 22.5 \) cm.

Teacher's Note:
1. The bird appears farther than it is, because the fish looks into a rarer medium.
2. Apply apparent depth with the observer in the denser medium.

Answer: (b) 1.5

Explanation:
1. Real depths x and 15 − x: \( \frac x\mu = 6 \) and \( \frac{15-x}{\mu} = 4 \).
2. Adding: \( \frac{15}{\mu} = 10 \Rightarrow \mu = 1.5 \).

Teacher's Note:
1. The two real depths add up to the edge of the cube.
2. Apparent depth = real depth / μ.

Answer: (c) 5 cm

Explanation:
1. Normal shift \( = t\left(1-\frac1\mu\right) = 15\times\frac13 = 5 \) cm.

Teacher's Note:
1. The shift does not depend on the distance of the observer.
2. Shift is towards the observer.

Answer: (a) \( \sin^{-1}\frac{11}{12} \)

Explanation:
1. \( \sin C = \frac{n_{II}}{n_I} = \frac{v_I}{v_{II}} = \frac{2.2}{2.4} = \frac{11}{12} \).

Teacher's Note:
1. Refractive index is inversely proportional to speed.
2. Sine of critical angle cannot exceed 1.

Answer: (d) 6 cm

Explanation:
1. \( n_q = 2 \), \( n_g = \frac43 \), relative index \( \mu = 1.5 \).
2. Shift \( = t\left(1-\frac1\mu\right) = 18\times\frac13 = 6 \) cm.

Teacher's Note:
1. Use the refractive index of the slab relative to the surrounding medium.
2. The shift is towards the slab.

Answer: (d) 12 cm

Explanation:
1. Let the water depth be x. Apparent depth of water \( = \frac{3x}{4} \). Air column \( = 21-x \).
2. Half-filled appearance: \( 21-x = \frac{3x}{4} \Rightarrow x = 12 \) cm.

Teacher's Note:
1. Apparent depth of a liquid column is real depth divided by μ.
2. Compare apparent water depth with the air height.

Answer: (a) 90°

Explanation:
1. The ray refracts in, reflects from the silvered face and refracts out, leaving at 45° to the normal on the same side as the incident ray.
2. Initial direction (sin45°, −cos45°); final direction (sin45°, +cos45°): the angle between them is 90°.

Teacher's Note:
1. The slab's refractive index does not change the final direction.
2. The reflection reverses the normal component only.

Answer: (c) 30°

Explanation:
1. Snell's law across parallel layers: \( \sin45^\circ = \sqrt2\sin\theta \).
2. \( \sin\theta = \frac12 \Rightarrow \theta = 30^\circ \).

Teacher's Note:
1. In stacked parallel layers n sin θ is constant.
2. Only the first and the required medium matter.

Answer: (a) 1.84

Explanation:
1. \( 1.5\sin60^\circ = n\sin45^\circ \Rightarrow n = 1.5\times\frac{0.866}{0.707} = 1.84 \).

Teacher's Note:
1. Use n₁ sin θ₁ = n₂ sin θ₂ with both absolute indices.
2. Do not treat 1.5 as relative index.

Answer: (d) \( \sqrt3 \)

Explanation:
1. Perpendicular reflected and refracted rays mean \( i + r = 90^\circ \), so \( r = 30^\circ \).
2. \( \mu = \frac{\sin60^\circ}{\sin30^\circ} = \sqrt3 \) (Brewster's law: μ = tan i).

Teacher's Note:
1. Angle between reflected and refracted rays is 180° − i − r.
2. Brewster angle gives μ = tan i_B.

Answer: (b) \( \sin^{-1}\frac45 \)

Explanation:
1. \( \sin C = \frac{n_{water}}{n_{glass}} = \frac{4/3}{5/3} = \frac45 \).

Teacher's Note:
1. TIR occurs when going from the denser to the rarer medium.
2. Critical angle uses the ratio of indices.

Answer: (c) less than θ

Explanation:
1. Shorter wavelength means larger refractive index (normal dispersion), and \( \sin C = \frac1n \) falls as n rises.
2. So the critical angle for yellow is smaller.

Teacher's Note:
1. Violet has the smallest critical angle among visible colours.
2. n is greater for shorter wavelength.

Answer: (a) 2.6 m²

Explanation:
1. \( \tan C = \frac{1}{\sqrt{\mu^2-1}} = \frac{1}{\sqrt{0.769}} = 1.14 \). Radius \( r = 0.8\times1.14 = 0.91 \) m.
2. Area \( = \pi r^2 = 3.14\times0.83 \approx 2.6 \) m².

Teacher's Note:
1. Light escapes only within the critical cone.
2. Radius = depth × tan C.

Answer: (b) 90°

Explanation:
1. \( \sin C = \frac1{\sqrt2} \), C = 45°, so r ≤ 45° with the axis.
2. \( \sin i = \sqrt2\sin45^\circ = 1 \Rightarrow i = 90^\circ \).

Teacher's Note:
1. Every ray entering the end face is guided.
2. This is the case of an uncladded fibre in air.