JEE Physics Ray Optics and Optical Instruments MCQs Set 09

Answer: (d) 120°

Explanation:
1. The ray makes 60° with the surface, so the angle of incidence (from the normal) is \( 90°-60°=30° \).
2. Deviation by a plane mirror is \( \delta=180°-2i=180°-60°=120° \).

Teacher's Note:
1. Angles of incidence and reflection are always measured from the normal, not from the mirror surface.
2. A plane mirror deviates a ray by 180° minus twice the angle of incidence.

Answer: (a) 5

Explanation:
1. Here \( \frac{360°}{\theta}=\frac{360°}{60°}=6 \), which is an even integer.
2. Number of images \( n=\frac{360°}{\theta}-1=5 \) for any position of the object when 360°/θ is an even integer.

Teacher's Note:
1. If 360°/θ is an integer, the count is 360°/θ − 1 for an object on the bisector as well as for one placed unsymmetrically (when it is even).
2. If 360°/θ is not an integer, the count depends on where the object is placed.

Answer: (b) 60 m

Explanation:
1. The image of the tower top lies as far below the mirror as the top lies above it, so the top and its image are symmetric about the ground.
2. Both lines from B to the top and to its image are inclined at \( \frac{90°}{2}=45° \) to the ground.
3. Height \( h=60\tan45°=60\ \text{m} \).

Teacher's Note:
1. The image of a point in a plane mirror is as far behind the mirror as the object is in front.
2. A 90° angle split equally by the mirror plane gives two 45° angles, so height equals horizontal distance.

Answer: (b) f/2

Explanation:
1. For a convex mirror, f is positive and the object distance is \( u=-f \).
2. \( \frac1v=\frac1f-\frac1u=\frac1f+\frac1f=\frac2f \), so \( v=\frac f2 \) behind the mirror.

Teacher's Note:
1. Use the mirror formula 1/v + 1/u = 1/f with sign convention: distances measured along the incident light are positive.
2. A convex mirror always gives a virtual, erect and diminished image of a real object, between the pole and the focus.

Answer: (d) \( \frac{\pi}{18} \)

Explanation:
1. The sun is effectively at infinity, so the image forms at the focus: \( f=\frac R2=10\ \text{cm} \).
2. Angle subtended: \( 1°=\frac{\pi}{180}\ \text{rad} \).
3. Image diameter \( d=f\theta=10\times\frac{\pi}{180}=\frac{\pi}{18}\ \text{cm} \).

Teacher's Note:
1. An object at infinity forms its image in the focal plane, and the image size is focal length times the angle subtended.
2. Always convert the angle to radians before multiplying.

Answer: (a) infinity

Explanation:
1. The beam would converge 20 cm behind the mirror, so the mirror sees a virtual object: \( u=+20\ \text{cm} \).
2. \( \frac1v=\frac1f-\frac1u=\frac1{20}-\frac1{20}=0 \).
3. So \( v=\infty \): the reflected beam is parallel to the axis.

Teacher's Note:
1. A virtual object for a mirror is at a point behind the mirror towards which the incident rays are converging.
2. A beam converging at the focus of a convex mirror comes out parallel.

Answer: (b) \( \frac13 \)

Explanation:
1. Take the focal length as f (positive) and \( u=-2f \).
2. \( \frac1v=\frac1f+\frac1{2f}=\frac{3}{2f}\Rightarrow v=\frac{2f}{3} \).
3. \( m=-\frac vu=\frac{2f/3}{2f}=+\frac13 \).

Teacher's Note:
1. A convex mirror always forms an erect (positive magnification) and diminished image of a real object.
2. Magnification is m = −v/u, so check the sign against the image type.

Answer: (b) 15 cm

Explanation:
1. A real image is inverted, so \( m=-3 \) and \( v=3u=-60\ \text{cm} \) (with \( u=-20\ \text{cm} \)).
2. \( \frac1f=\frac1v+\frac1u=-\frac1{60}-\frac1{20}=-\frac{4}{60} \), so \( f=-15\ \text{cm} \).
3. The magnitude of the focal length is 15 cm.

Teacher's Note:
1. For a real image from a concave mirror, magnification is negative.
2. With u = −20 and m = −3, the image distance is v = −60 cm; the negative focal length simply indicates a concave mirror.

Answer: (d) 90°

Explanation:
1. At the critical angle in glass: \( \sin C=\frac{1}{3/2}=\frac23 \).
2. Glass to water: \( \frac32\cdot\frac23=\frac43\sin r\Rightarrow \sin r=\frac34 \).
3. Water to air: \( \frac43\cdot\frac34=1\cdot\sin e\Rightarrow\sin e=1 \), so \( e=90° \).

Teacher's Note:
1. Snell's law across several parallel layers means μ sin θ is the same in every layer.
2. So the emergent angle is fixed by the first medium alone: sin e = (3/2)(2/3) = 1, whatever the intermediate layer.

Answer: (b) 90°

Explanation:
1. At the first surface: \( \sin r=\frac{\sin60°}{\sqrt3}=\frac12 \), so \( r=30° \).
2. By symmetry the angle of incidence at the farther surface is also 30°, so the reflected ray makes 30° with the normal.
3. Refraction out: \( \sin e=\sqrt3\sin30°=\frac{\sqrt3}{2} \), so \( e=60° \).
4. Angle between the reflected and refracted rays \( =180°-30°-60°=90° \).

Teacher's Note:
1. Inside a sphere the angle of refraction at the entry point equals the angle of incidence at the exit point.
2. The reflected and refracted rays lie on opposite sides of the normal, and the normal makes a straight line with its extension.

Answer: (c) 1.732

Explanation:
1. Lateral shift \( d=\frac{t\sin(i-r)}{\cos r} \).
2. \( 5\sqrt3=\frac{15\,(\sin60°\cos r-\cos60°\sin r)}{\cos r}=15\left(\frac{\sqrt3}{2}-\frac{\tan r}{2}\right) \).
3. So \( \frac{\sqrt3}{3}=\frac{\sqrt3}{2}-\frac{\tan r}{2}\Rightarrow\tan r=\frac{1}{\sqrt3} \), giving \( r=30° \).
4. \( \mu=\frac{\sin60°}{\sin30°}=\sqrt3=1.732 \).

Teacher's Note:
1. The lateral displacement of a ray through a parallel slab is t sin(i − r)/cos r.
2. After finding r, apply Snell's law to get μ.

Answer: (a) 270°

Explanation:
1. At entry: \( \sin r=\frac{\sin45°}{\sqrt2}=\frac12 \), so \( r=30° \).
2. Deviation at each refraction is \( i-r=15° \) and at each internal reflection is \( 180°-2r=120° \).
3. Total deviation \( =2(15°)+2(120°)=270° \).

Teacher's Note:
1. Total deviation is the sum of the deviations at every surface interaction, all measured in the same rotational sense.
2. For k internal reflections: D = 2(i − r) + k(180° − 2r).

Answer: (c) \( \frac{\sqrt{43}}{5} \)

Explanation:
1. For equal deviation, the two incidence angles satisfy \( i_1+i_2=A+\delta=60°+23°=83° \) and \( i_1-i_2=23° \).
2. So \( i_1=53°,\ i_2=30° \). By reversibility, the ray entering at 53° leaves at 30°.
3. Let the second refraction angle be \( r_2 \): \( \sin30°=\mu\sin r_2 \), \( \sin53°=\mu\sin(60°-r_2) \).
4. Expanding: \( \mu\cos r_2=\frac{\sin53°+\sin30°/2}{\sin60°}=1.211 \) and \( \mu\sin r_2=0.5 \).
5. \( \mu^2=1.211^2+0.5^2=1.72=\frac{43}{25} \), so \( \mu=\frac{\sqrt{43}}{5} \).

Teacher's Note:
1. A prism gives the same deviation for two incidence angles that are interchanged on the way out; they add up to A + δ.
2. Use r1 + r2 = A together with Snell's law at both faces.

Answer: (b) 49°

Explanation:
1. For the same deviation 44° at two incidence angles: \( i_1+i_2=A+\delta\Rightarrow 42°+62°=A+44° \), so \( A=60° \).
2. At minimum deviation: \( i=\frac{A+\delta_{min}}{2}=\frac{60°+38°}{2}=49° \).

Teacher's Note:
1. At minimum deviation the angle of incidence equals the angle of emergence, so each is (A + δmin)/2.
2. The two incidence angles for the same deviation add up to A + δ.

Answer: (b) \( \frac{6}{25} \)

Explanation:
1. Mean index \( n=\frac{1.56+1.44}{2}=1.50 \).
2. \( \omega=\frac{n_v-n_r}{n-1}=\frac{0.12}{0.50}=0.24=\frac{6}{25} \).

Teacher's Note:
1. Dispersive power is the angular dispersion divided by the mean deviation, which for a thin prism gives (nv − nr)/(n − 1).
2. It is a pure number and a property of the material only.

Answer: (a) \( \frac43 \)

Explanation:
1. For a sphere of radius R, paraxial parallel rays focus at a distance \( \frac{nR}{2(n-1)} \) from the centre.
2. Setting \( \frac{nR}{2(n-1)}=2R\Rightarrow n=4(n-1) \), so \( 3n=4,\ n=\frac43 \).

Teacher's Note:
1. Treat the sphere as two refracting surfaces in sequence and use n2/v − n1/u = (n2 − n1)/R twice.
2. The focal point lies outside the sphere only if n is less than 2.

Answer: (c) 3R

Explanation:
1. Light from the nose (in air) refracts at the spherical surface into water. Object distance from the surface \( u=-R \) (since the child is 2R from the centre and the surface is R from the centre).
2. Centre of curvature lies in water, so the radius is \( +R \): \( \frac{4/3}{v}-\frac{1}{-R}=\frac{4/3-1}{R} \).
3. \( \frac{4}{3v}=\frac{1}{3R}-\frac1R=-\frac{2}{3R}\Rightarrow v=-2R \).
4. The image is 2R from the surface on the air side, which is 3R from the centre.

Teacher's Note:
1. Measure object and image distances from the refracting surface, then convert to distances from the centre.
2. The negative v shows a virtual image on the same side as the object.

Answer: (b) \( \frac12(u_1+u_2) \)

Explanation:
1. Magnification of a convex lens image: real image \( |m|=\frac{f}{u_1-f} \); virtual image \( |m|=\frac{f}{f-u_2} \).
2. Equal sizes: \( u_1-f=f-u_2\Rightarrow f=\frac{u_1+u_2}{2} \).

Teacher's Note:
1. For a real image the object is beyond f (u1 > f), and for a virtual image it is within f (u2 < f).
2. Equating the magnitudes of m for the two positions gives the result directly.

Answer: (c) 75 cm, 150 cm

Explanation:
1. Let the radii be R and 2R. Lens maker's formula: \( \frac1f=(\mu-1)\left(\frac1{R}+\frac1{2R}\right) \).
2. \( \frac1{50}=(2-1)\cdot\frac{3}{2R}\Rightarrow R=75\ \text{cm} \).
3. The radii are 75 cm and 150 cm.

Teacher's Note:
1. For a double convex lens both surfaces contribute positive power, so 1/R1 and 1/R2 add.
2. Always check that the final radii are in the ratio given in the problem.

Answer: (c) 38.3 cm

Explanation:
1. Objective image: \( \frac1{v}=\frac1{30}-\frac1{200}=\frac{17}{600}\Rightarrow v=35.3\ \text{cm} \).
2. A relaxed eye needs the intermediate image at the focus of the eyepiece, so the lens separation is \( 35.3+3.0=38.3\ \text{cm} \).

Teacher's Note:
1. For a nearby object the objective image forms beyond its focus, so the tube must be longer than fo + fe.
2. For an object at infinity, the separation would be exactly 33 cm.