JEE Physics Ray Optics and Optical Instruments MCQs Set 10

Answer: (c) 07 : 34 : 23

Explanation:
1. The mirror image of a clock reads the time that, added to the actual time, gives 12 : 00 : 00.
2. \( 12:00:00-04:25:37=07:34:23 \).

Teacher's Note:
1. A clock seen in a mirror is laterally inverted, so the displayed time is 12 h minus the actual time.
2. Borrow carefully: 12:00:00 − 4:25:37 = 7:34:23.

Answer: (a) 3L/2

Explanation:
1. Let the mirrors be at x = 0 and x = L and the object at x = L/3. Images are at \( x=2nL+\frac L3 \) and \( x=2nL-\frac L3 \).
2. The distance between two images of the same family is a multiple of 2L; between images of different families it is \( 2kL\pm\frac{2L}{3} \).
3. So possible separations include 2L/3, 4L/3, 2L, 8L/3, and so on. The value 3L/2 never occurs.

Teacher's Note:
1. Images in two parallel mirrors form two equally spaced families, each with spacing 2L.
2. The offset between the families is 2L/3, so only multiples of 2L and 2L ± 2L/3 are possible.

Answer: (d) 0.6 m

Explanation:
1. The mirror spans heights 0.8 m to 1.1 m. A point of the person at height h is seen only if the reflection point, at height \( \frac{h+1.5}{2} \), lies on the mirror.
2. \( 0.8\le\frac{h+1.5}{2}\le1.1\Rightarrow 0.1\le h\le0.7 \).
3. The visible part of the person has length \( 0.7-0.1=0.6\ \text{m} \), twice the mirror length.

Teacher's Note:
1. The portion of a body seen in a plane mirror is always twice the length of the mirror (when the eye and the object are the same distance away).
2. The reflection point is midway in height between the eye and the point seen.

Answer: (c) the boy cannot see his feet

Explanation:
1. The mirror covers heights 0.8 m to 1.55 m. The reflection point for a body point at height h is at \( \frac{h+1.4}{2} \).
2. \( 0.8\le\frac{h+1.4}{2}\le1.55\Rightarrow 0.2\le h\le1.7 \).
3. He sees heights from 0.2 m up to 1.7 m. His hair (1.5 m) is visible, but his feet (0 m) are below 0.2 m and are not.

Teacher's Note:
1. Find the range of heights of the object that reflect into the eye, then compare with the actual body extent.
2. The lower edge of the mirror is the limiting factor for the feet.

Answer: (a) 75 cm

Explanation:
1. For a bulb at distance d below the mirror, its image is d above the mirror. Reflected light appears to come from the image, through the mirror edge.
2. The lit radius on the floor is \( R=r\cdot\frac{d+3}{d} \).
3. \( 1=0.2\cdot\frac{d+3}{d}\Rightarrow d=0.2d+0.6\Rightarrow d=0.75\ \text{m} \).
4. Moving the bulb farther makes the lit circle smaller, so 75 cm is the maximum.

Teacher's Note:
1. Reflected rays from a plane mirror seem to diverge from the image, so draw the cone from the image through the mirror edge.
2. Similar triangles relate the mirror radius to the floor radius.

Answer: (b) \( 3\hat i+4\hat j+11\hat k \)

Explanation:
1. Components parallel to the mirror (\( \hat i,\hat j \)) of the image equal those of the object: 3 and 4.
2. Along the normal, the image position is \( z_I=2z_m-z_O \), so \( v_{Iz}=2v_{mz}-v_{Oz}=2(8)-5=11 \).
3. Velocity of the image \( =3\hat i+4\hat j+11\hat k \).

Teacher's Note:
1. The image mirrors the object's perpendicular position about the mirror, so its normal velocity is 2vm − vo.
2. Tangential components are unaffected by the mirror.

Answer: (c) 450 cm

Explanation:
1. Image is real and magnified 3 times: \( |m|=\frac93=3 \).
2. Mirror to wall = x (image distance), and mirror to candle = \( x-300 \) (object distance), so \( \frac{x}{x-300}=3 \).
3. \( x=3x-900\Rightarrow x=450\ \text{cm} \).
4. Check: \( \frac1v+\frac1u=\frac1{450}+\frac1{150}=\frac{4}{450}\Rightarrow f=112.5\ \text{cm} \), a positive focal length magnitude for a concave mirror.

Teacher's Note:
1. The image is on the wall, so image distance from the mirror equals the mirror-to-wall distance.
2. Real magnified image: v = 3u in magnitude.

Answer: (d) (n − 1)f

Explanation:
1. For a convex mirror, \( m=\frac{f}{f-u} \) with \( u=-d \): \( m=\frac{f}{f+d} \).
2. Given \( m=\frac1n \): \( f+d=nf\Rightarrow d=(n-1)f \).

Teacher's Note:
1. A convex mirror image is erect, so magnification is positive 1/n.
2. Use m = f/(f − u) which comes from combining the mirror formula with m = −v/u.

Answer: (b) 45°

Explanation:
1. A ray parallel to the axis hits the mirror at a point whose angular position (from the axis, at the centre of curvature) is \( \alpha \), and the angle of incidence is also \( \alpha \).
2. Inside the circle the reflected ray travels along a chord subtending \( 180°-2\alpha \) at the centre, landing at angular position \( |3\alpha-180°| \).
3. A second reflection needs \( |3\alpha-180°|\le\theta \) with the first point inside the mirror (\( \alpha\le\theta \)).
4. The best case is \( \alpha=\theta \): \( 180°-3\theta\le\theta\Rightarrow\theta\ge45° \).

Teacher's Note:
1. Use the circle geometry: the angle of incidence for a parallel ray equals its angular position from the axis.
2. A ray at the rim has the largest angle of incidence and therefore the best chance of hitting the mirror again.

Answer: (c) 10/9

Explanation:
1. Apparent depth \( =1-0.1=0.9\ \text{m} \).
2. \( \mu=\frac{\text{real depth}}{\text{apparent depth}}=\frac{1}{0.9}=\frac{10}{9} \).

Teacher's Note:
1. The apparent shift is real depth × (1 − 1/μ).
2. Refractive index is always greater than 1, which rules out 9/10.

Answer: (d) \( \tan^{-1}n \)

Explanation:
1. In air the normal component of velocity is \( c\cos i \); in glass it is \( \frac cn\cos r \).
2. Equal components: \( n\cos i=\cos r \) with \( \sin r=\frac{\sin i}{n} \).
3. Square and add: \( n^2\cos^2i=1-\frac{\sin^2i}{n^2} \). Substituting \( \tan i=n \) gives \( \frac{n^2}{1+n^2}=1-\frac{1}{1+n^2} \), which is true.

Teacher's Note:
1. Use the normal component of the velocity vector, v cos θ, in each medium.
2. Check each option by substituting; tan i = n satisfies the condition exactly.

Answer: (a) \( \sin^{-1}\left(\frac45\right) \)

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

Teacher's Note:
1. Total internal reflection is possible only from the denser to the rarer medium.
2. Critical angle: sin C = n(rarer)/n(denser); the value must be less than 1.

Answer: (c) 30°

Explanation:
1. \( \mu=\frac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}\Rightarrow\sin\frac{90°+\delta_m}{2}=\sqrt{\tfrac32}\sin45°=\frac{\sqrt3}{2} \).
2. \( \frac{90°+\delta_m}{2}=60°\Rightarrow\delta_m=30° \).

Teacher's Note:
1. Use the prism formula with the angle of minimum deviation.
2. The value sin 60° = √3/2 identifies the angle immediately.

Answer: (c) 60°

Explanation:
1. \( \mu=\frac{\sin A}{\sin(A/2)}=2\cos\frac A2=\sqrt3 \).
2. \( \cos\frac A2=\frac{\sqrt3}{2}\Rightarrow\frac A2=30°\Rightarrow A=60° \).

Teacher's Note:
1. With δm = A, the prism formula reduces to μ = 2 cos(A/2).
2. Such a prism of μ = √3 is a 60° prism with the symmetric ray path.

Answer: (b) 45°

Explanation:
1. Critical angle: \( \sin C=\frac1{\sqrt2}\Rightarrow r_2=45° \).
2. \( r_1=A-r_2=75°-45°=30° \).
3. \( \sin i=\sqrt2\sin30°=\frac{\sqrt2}{2}\Rightarrow i=45° \).

Teacher's Note:
1. Inside a prism r1 + r2 = A.
2. At the critical angle, the angle of incidence on the second face r2 equals C.

Answer: (a) 3°

Explanation:
1. No net deviation: \( (\mu_1-1)A_1=(\mu_2-1)A_2 \).
2. \( A_2=\frac{(1.54-1)\times4°}{1.72-1}=\frac{2.16°}{0.72}=3° \).

Teacher's Note:
1. Zero net deviation means the two prisms deviate by equal and opposite amounts.
2. The prisms must be placed with their bases in opposite directions.

Answer: (b) 0.08°

Explanation:
1. Angular dispersion of a thin prism: \( \theta=(n_v-n_r)A=(1.5-1.48)\times4°=0.08° \).

Teacher's Note:
1. The angle of dispersion is the angular separation between the violet and red rays.
2. For a thin prism each deviation is (n − 1)A, so the difference is (nv − nr)A.

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

Explanation:
1. The rays meet the hypotenuse at an angle of incidence of 45°. Total internal reflection occurs if \( \sin45°>\frac1\mu \), i.e. if \( \mu>\sqrt2=1.414 \).
2. Red: 1.39 is less than 1.414, so red is not totally reflected and passes out through the hypotenuse.
3. Green (1.44) and blue (1.47) exceed 1.414 and are totally internally reflected.
4. So red is separated from green and blue, while green and blue stay together.

Teacher's Note:
1. Compare each colour's critical angle with 45°: the colour has TIR if its critical angle is smaller than 45°.
2. Critical angles here are about 46.0° (red), 43.9° (green) and 42.8° (blue).

Answer: (c) \( \frac{\mu-\mu_0}{\mu-1}\cdot\frac{P}{\mu_0} \)

Explanation:
1. In air \( P=(\mu-1)\left(\frac1{R_1}-\frac1{R_2}\right) \).
2. In the liquid the relative index is \( \mu/\mu_0 \): \( P'=\left(\frac{\mu}{\mu_0}-1\right)\left(\frac1{R_1}-\frac1{R_2}\right)=\frac{\mu-\mu_0}{\mu_0}\cdot\frac{P}{\mu-1} \).

Teacher's Note:
1. The lens-maker formula uses the index of the lens relative to the surrounding medium.
2. Divide the two expressions for power to eliminate the radii.

Answer: (b) convex with focal length 10/3 cm

Explanation:
1. A real image from a single lens is formed only by a convex lens and is inverted, so \( m=-0.5 \) and \( v=+5\ \text{cm} \) with \( u=-10\ \text{cm} \).
2. \( \frac1f=\frac1v-\frac1u=\frac15+\frac1{10}=\frac{3}{10}\Rightarrow f=\frac{10}{3}\ \text{cm} \).

Teacher's Note:
1. A concave lens always gives a virtual image of a real object.
2. Magnitude of v follows from |m| = v/u.

Answer: (b) 4 mm

Explanation:
1. For the object position: \( \frac1v=\frac1{10}-\frac1{15}=\frac1{30}\Rightarrow v=30\ \text{cm} \).
2. Lateral magnification \( m=\frac vu=2 \).
3. For a small object along the axis, longitudinal magnification is \( m^2=4 \), so the image length is \( 4\times1=4\ \text{mm} \).

Teacher's Note:
1. Longitudinal magnification equals the square of lateral magnification for a short object along the axis.
2. Differentiating the lens formula gives dv/du = (v/u)².