JEE Physics Ray Optics And Optical Instruments MCQs Set 16

Answer: (b) 45°

Explanation:
1. Number of images \( n=\frac{360^\circ}{\theta}-1 \) (when 360/θ is an even integer).
2. \( 7=\frac{360}{\theta}-1\Rightarrow\theta=45^\circ \).

Teacher's Note:
1. Use the formula for two inclined mirrors.
2. Check that 360/45 = 8 is an integer.

Answer: (a) 6 rad/s

Explanation:
1. When a mirror turns through an angle θ, the reflected ray turns through 2θ.
2. So the angular velocity of the reflected beam is \( 2\times3=6\ \text{rad/s} \).

Teacher's Note:
1. The angle between the incident and reflected rays changes twice as fast as the mirror's rotation.
2. Both angles i and r change by θ.

Answer: (a) the object can be real and the image virtual

Explanation:
1. A real object always gives a virtual image in a plane mirror, so (a) is true.
2. A real image would need converging rays reflected by the mirror, which only happens for a virtual object (converging incident beam). Both being real, or both being virtual, cannot occur in the same formation.

Teacher's Note:
1. A plane mirror gives a virtual image of a real object.
2. A virtual object gives a real image in a plane mirror.

Answer: (d) 5, 25, 35

Explanation:
1. First image in M2: 5 cm behind M2, so at 5 cm.
2. Object is 10 cm from M1, so its image in M1 is 10 cm behind M1: distance from M2 \( 15+10=25\ \text{cm} \).
3. The first image (5 cm behind M2, so 20 cm from M1) forms an image 20 cm behind M1: distance from M2 \( 15+20=35\ \text{cm} \).
4. So the three nearest images are at 5, 25, 35 cm.

Teacher's Note:
1. Images from mirror M2 are on its far side; images in M1 are on M1's far side.
2. Repeat reflections to find the farther images.

Answer: (c) \( 2v\cos\theta \)

Explanation:
1. Component of the man's velocity normal to the mirror: \( v\cos\theta \) (towards the mirror).
2. The image moves towards the mirror with the same normal speed, so relative to the man the normal velocity is \( v\cos\theta+v\cos\theta=2v\cos\theta \).

Teacher's Note:
1. Only the normal components change sign for the image.
2. Parallel components are the same for the man and the image.

Answer: (b) 100

Explanation:
1. At 45° incidence the ray advances \( 1\ \text{cm}\times\tan45^\circ=1\ \text{cm} \) along the mirrors per reflection (the gap is 1 cm).
2. Length of mirrors 100 cm, so the number of reflections is \( \frac{100}{1}=100 \).

Teacher's Note:
1. Each crossing of the gap advances the ray by gap × tan i.
2. Divide the mirror length by that advance.

Answer: (a) f

Explanation:
1. For a mirror \( f=\frac R2 \), which depends only on the geometry of the surface and not on the refractive index of the surrounding medium.
2. So the focal length remains f.

Teacher's Note:
1. A mirror works by reflection, not refraction.
2. Only lenses change focal length with the medium.

Answer: (d) 60 cm and 30 cm

Explanation:
1. Focal length \( f=\frac{40}{2}=20\ \text{cm} \). Magnification magnitude \( \frac vu=\frac12\Rightarrow v=\frac u2 \).
2. Mirror formula (magnitudes, real image): \( \frac1v+\frac1u=\frac1f\Rightarrow\frac2u+\frac1u=\frac1{20} \).
3. \( u=60\ \text{cm},\ v=30\ \text{cm} \).

Teacher's Note:
1. A diminished image from a concave mirror is real and beyond f but between f and 2f.
2. Use v = u/2 in the mirror formula.

Answer: (c) 3 : 1

Explanation:
1. Focal length \( f=5\ \text{cm} \); object at \( u=-20\ \text{cm} \): \( \frac1v=\frac1f-\frac1u\ \Rightarrow\ \frac1v=-\frac15+\frac1{20}=-\frac{3}{20} \), so \( v=-\frac{20}{3}\ \text{cm} \).
2. Lateral magnification \( |m|=\frac{|v|}{|u|}=\frac13 \).
3. Longitudinal magnification (small length along the axis) \( =m^2=\frac19 \).
4. Ratio of lengths: \( \frac13:\frac19=3:1 \).

Teacher's Note:
1. Transverse magnification is m; longitudinal magnification is m².
2. Both portions have the same object length.

Answer: (b) 25 cm

Explanation:
1. The object and image coincide only at the centre of curvature.
2. So \( R=50\ \text{cm} \) and \( f=\frac R2=25\ \text{cm} \).

Teacher's Note:
1. Object and image coincide at the centre of curvature (u = v = R).
2. The speed of the object is not needed.

Answer: (a) 150 cm from the pole in front of the reflecting surface

Explanation:
1. \( \frac1v+\frac1u=\frac1f \) with \( u=-30,\ f=-25 \): \( \frac1v=-\frac1{25}+\frac1{30}=-\frac{1}{150} \).
2. \( v=-150\ \text{cm} \): a real image 150 cm in front of the mirror.

Teacher's Note:
1. Take the distances measured against the direction of the incident light as negative.
2. The object is between f and 2R... here between f and 2f, so the image is real and beyond 2f.

Answer: (c) a straight line with negative slope, cutting both axes

Explanation:
1. Mirror formula for real images (magnitudes): \( \frac1v+\frac1u=\frac1f \), so \( \frac1v=\frac1f-\frac1u \).
2. This is a straight line of slope −1 with intercept 1/f on both axes.

Teacher's Note:
1. Rearrange into the form y = c − x.
2. Straight lines are obtained by plotting reciprocals.

Answer: (d) \( \frac h\mu \)

Explanation:
1. Apparent depth for near-normal viewing \( =\frac{\text{real depth}}{\mu}=\frac h\mu \).

Teacher's Note:
1. The pool looks shallower than it is.
2. Apparent depth = real depth / μ.

Answer: (c) \( \frac52 \) cm

Explanation:
1. Snell's law: \( \sin r=\frac{\sin60^\circ}{\sqrt3}=\frac12\Rightarrow r=30^\circ \).
2. Lateral shift \( d=\ell\sin(i-r)=5\sin30^\circ=\frac52\ \text{cm} \).

Teacher's Note:
1. The lateral shift uses the path length inside the slab and the angle i − r.
2. The emergent ray is parallel to the incident ray.

Answer: (b) 60°

Explanation:
1. At minimum deviation \( i=\frac{A+\delta_m}{2} \) with \( A=60^\circ \).
2. \( 60=\frac{60+\delta_m}{2}\Rightarrow\delta_m=60^\circ \).

Teacher's Note:
1. For an equilateral prism at minimum deviation, i = e and r = A/2 = 30°.
2. δ = 2i − A.

Answer: (a) 45°

Explanation:
1. At minimum deviation \( r=\frac A2=30^\circ \).
2. \( \sin i=\mu\sin r=\sqrt2\times\frac12=\frac1{\sqrt2}\Rightarrow i=45^\circ \).

Teacher's Note:
1. At minimum deviation the ray passes symmetrically through the prism.
2. Apply Snell's law at the first face.

Answer: (d) between unity and 1.33

Explanation:
1. A lens is converging in air only if its refractive index is greater than 1 (for a convex shape).
2. It is diverging in water if its refractive index is less than that of water, 1.33.
3. So the index lies between 1 and 1.33.

Teacher's Note:
1. Focal length depends on (μ_lens/μ_medium − 1).
2. The sign changes when μ_lens < μ_medium.

Answer: (c) −160 cm

Explanation:
1. In air: \( \frac1{20}=(1.5-1)\left(\frac1{R_1}-\frac1{R_2}\right)\Rightarrow\frac1{R_1}-\frac1{R_2}=0.1 \).
2. In the liquid: \( \frac1{f'}=\left(\frac{1.5}{1.6}-1\right)(0.1)=-0.00625 \).
3. \( f'=-160\ \text{cm} \).

Teacher's Note:
1. The lens becomes diverging because the liquid has a higher index than the lens.
2. Use the relative refractive index in the lens-maker formula.

Answer: (b) 10 cm

Explanation:
1. Power of the curved surface \( =\frac{\mu-1}{R}=\frac{0.5}{10}=0.05\ \text{cm}^{-1} \). The plane silvered surface adds no power.
2. Light passes through the curved surface twice: \( P=2\times0.05=0.1\ \text{cm}^{-1} \).
3. \( f=\frac1P=10\ \text{cm} \).

Teacher's Note:
1. The silvered plane acts as a plane mirror, adding no power.
2. Light refracts at the curved face on the way in and on the way out.