JEE Physics Ray Optics and Optical Instruments MCQs Set 14

Answer: (d) 6 cm

Explanation:
1. \( \sin\theta_c=\frac35 \), so \( \tan\theta_c=\frac34 \).
2. Radius \( R=8\tan\theta_c=8\times\frac34=6\ \text{cm} \).

Teacher's Note:
1. Only rays within the critical angle emerge from the top surface.
2. R = thickness × tan(critical angle).

Answer: (c) 3

Explanation:
1. Convex mirror, \( f=+10\ \text{m} \). First image \( v_1=\frac{25}3 \): \( \frac1{u_1}=\frac1{10}-\frac3{25}=-\frac1{50} \), so \( u_1=-50\ \text{m} \).
2. Second image \( v_2=\frac{50}7 \): \( \frac1{u_2}=\frac1{10}-\frac7{50}=-\frac1{25} \), so \( u_2=-25\ \text{m} \).
3. The object moved 25 m in 30 s: \( \frac{25}{30}=\frac56\ \text{m/s}=3\ \text{km/h} \).

Teacher's Note:
1. Find the object positions from the two image positions.
2. Convert m/s to km/h by multiplying by 3.6.

Answer: (a) planar

Explanation:
1. The beam is initially parallel and cylindrical, which means all its rays are parallel and the wavefront is perpendicular to them everywhere.
2. A wavefront perpendicular to a parallel set of rays is a plane.

Teacher's Note:
1. Wavefronts are perpendicular to the rays.
2. Parallel rays always come from a plane wavefront.

Answer: (c) 2

Explanation:
1. Treat the oil layer as a thin lens between air and water: \( \frac{4/3}{v}-\frac1{-24}=\frac{7/4-1}{6}+\frac{4/3-7/4}{\infty}=\frac18 \).
2. So \( \frac{4/3}{v}=\frac18-\frac1{24}=\frac1{12} \) and \( v=16\ \text{cm} \) below the oil lens.
3. The tank is 18 cm deep, so the image is \( 18-16=2\ \text{cm} \) above the bottom.

Teacher's Note:
1. The oil-water boundary is nearly plane, so only the convex oil-air surface contributes power.
2. The image distance is measured from the lens, in water.

Answer: (a) the refracted ray lies on the same side of the normal as the incident ray, and the speed of light in the medium is c/|n|

Explanation:
1. For a negative index Snell's law \( \sin\theta_1=n\sin\theta_2 \) gives a negative \( \sin\theta_2 \), so the refracted ray is on the same side of the normal as the incident ray (negative refraction).
2. The speed is \( v=\frac{c}{|n|} \), and the wavelength is \( \lambda_m=\frac{\lambda_{air}}{|n|} \) (frequency is unchanged).

Teacher's Note:
1. A negative refractive index reverses the sign of the angle of refraction.
2. Speed and wavelength depend on the magnitude of n.

Answer: (c) 30°

Explanation:
1. Only the j-component of the direction changes sign, so the normal to the mirror is along the y-axis.
2. The angle between the incident ray and the normal: \( \cos i=\frac{\sqrt3}2 \), so \( i=30° \).

Teacher's Note:
1. The component of the direction along the normal reverses on reflection.
2. Angle of incidence is the angle with the normal, not with the mirror.

Answer: (b) 1.36

Explanation:
1. Spot radius \( r=\frac{11.54}2=5.77\ \text{mm} \) and \( \tan\theta_c=\frac{5.77}{10}=0.577 \), so \( \theta_c=30° \).
2. Then \( \sin\theta_c=\frac{n_l}{2.72}=\frac12 \), so \( n_l=1.36 \).

Teacher's Note:
1. The edge of the bright spot is set by the critical angle.
2. sin θc = (index of liquid)/(index of block).

Answer: (b) A and C only

Explanation:
1. The film is thin, so the two curved surfaces act like a single surface between air and glass: from air to glass \( \frac{n_2}{f_1}=\frac{n_1-1}{R}+\frac{n_2-n_1}{R}=\frac{n_2-1}{R} \), so \( f_1=\frac{1.5R}{0.5}=3R \) (A).
2. From glass to air \( \frac1{f_2}=\frac{1-n_2}{-R}=\frac{0.5}{R} \), so \( |f_2|=2R \) (C).

Teacher's Note:
1. A thin film does not change the focal length: only the outer media matter.
2. Use n2/f = (n2 − n1)/R for refraction at a spherical surface.

Answer: (c) 70 cm

Explanation:
1. Refraction at the curved face of S1 (glass to air): \( \frac1v-\frac{1.5}{-50}=\frac{1-1.5}{-10} \), so \( \frac1v=0.05-0.03=0.02 \) and \( v=50\ \text{cm} \) in air.
2. For the beam to become parallel inside S2, this image must be at the first focal point of the curved face of S2 (air to glass): \( -\frac1u=\frac{1.5-1}{10} \), so \( u=-20\ \text{cm} \).
3. So \( d=50+20=70\ \text{cm} \).

Teacher's Note:
1. Find the image formed by S1's face, then use it as the object for S2's face.
2. A parallel beam inside S2 means the object for S2 is at the focus of its curved face.

Answer: (b) A and C only

Explanation:
1. For a fibre in a medium of refractive index m: \( NA=\frac{\sqrt{n_1^2-n_2^2}}{m} \).
2. S1: \( \sqrt{n_1^2-n_2^2}=\sqrt{\frac{45}{16}-\frac94}=\frac34 \). S2: \( \sqrt{\frac{64}{25}-\frac{49}{25}}=\frac{\sqrt{15}}5 \).
3. (A) S1 in water: \( \frac{3/4}{4/3}=\frac9{16} \); S2 in liquid \( \frac{16}{3\sqrt{15}} \): \( \frac{\sqrt{15}/5}{16/(3\sqrt{15})}=\frac{45}{80}=\frac9{16} \): equal.
4. (C) S1 in air: \( \frac34 \); S2 in liquid \( \frac4{\sqrt{15}} \): \( \frac{\sqrt{15}/5}{4/\sqrt{15}}=\frac{15}{20}=\frac34 \): equal.
5. (B) gives 0.18 and 0.58; (D) gives 0.75 and 0.58: not equal.

Teacher's Note:
1. The numerical aperture is √(n1² − n2²) divided by the index of the medium in front of the end face.
2. Compare the values numerically for each option.

Answer: (c) 15°

Explanation:
1. At PQ: \( \sin45°=\sqrt2\sin r \), so \( r=30° \).
2. The critical angle: \( \sin C=\frac1{\sqrt2} \), so \( C=45° \).
3. The angle of incidence at PR is \( \theta+r \) (PR makes angle θ with PQ and the ray goes down towards PR). The minimum α corresponds to the limit \( \theta+30°=45° \), so \( \theta=15° \).

Teacher's Note:
1. Minimum α corresponds to minimum r, which makes the incidence angle on PR just equal to the critical angle.
2. Express the angle of incidence at PR in terms of θ and r.

Answer: (c) A and D only

Explanation:
1. Real image of double size: \( m=-2 \), \( u=-30 \), \( v=60\ \text{cm} \): \( \frac1f=\frac1{60}+\frac1{30}=\frac1{20} \), so \( f=20\ \text{cm} \) (D is correct).
2. Faint image from the convex mirror surface: \( u=-30,\ v=+10 \) (virtual, erect, behind the surface): \( \frac1{f_m}=\frac1{10}-\frac1{30}=\frac1{15} \), so \( f_m=15 \) and \( R=30\ \text{cm} \) (B is wrong, C is wrong because the image is virtual).
3. Lens: \( f=\frac R{n-1} \), so \( n-1=\frac{30}{20}=1.5 \) and \( n=2.5 \) (A is correct).

Teacher's Note:
1. A convex mirror always gives a virtual, erect image of a real object.
2. Use the mirror image to find R, and the lens image to find f.

Answer: (d) 30 cm

Explanation:
1. Refractive index \( n=\frac c v=\frac{3\times10^8}{2\times10^8}=1.5 \).
2. The edge thickness is zero, so the centre thickness is the sagitta of the curved surface: \( s=\frac{a^2}{2R} \) with a = 3 cm = 30 mm and s = 3 mm: \( R=\frac{900}{6}=150\ \text{mm}=15\ \text{cm} \).
3. Focal length \( f=\frac R{n-1}=\frac{15}{0.5}=30\ \text{cm} \).

Teacher's Note:
1. The plano-convex lens has zero thickness at the edge, so the centre thickness equals the sagitta.
2. f = R/(n − 1) for a plano-convex lens.

Answer: (b) 8

Explanation:
1. For parallel layers \( n\sin\theta \) is invariant: \( 1.6\sin30°=n_m\sin90° \).
2. So \( n_m=0.8=1.6-0.1m \), giving \( m=8 \).

Teacher's Note:
1. Emerging parallel to the interface means a refraction angle of 90° in the mth slab.
2. Use n sin θ = constant across the stack.

Answer: (d) \( \text{greater than or equal to }\sqrt2 \)

Explanation:
1. The ray enters normally and meets the hypotenuse at 45°. Total internal reflection needs \( \sin45°\ge\frac1n \).
2. So \( n\ge\sqrt2 \).

Teacher's Note:
1. TIR occurs when the angle of incidence is at least the critical angle.
2. sin θc = 1/n.

Answer: (b) 40 cm from the pole in the medium of refractive index 1, virtual, erect and 4 cm in size

Explanation:
1. Sign convention: \( u=-10,\ R=+20,\ n_1=1,\ n_2=2 \).
2. Refraction formula: \( \frac2v-\frac1{-10}=\frac{2-1}{20} \), so \( \frac2v=0.05-0.1=-0.05 \) and \( v=-40\ \text{cm} \) (virtual, in the first medium).
3. Magnification \( m=\frac{n_1v}{n_2u}=\frac{1\times(-40)}{2\times(-10)}=+2 \): erect, size \( 2\times2=4\ \text{cm} \).

Teacher's Note:
1. A negative image distance means the image is on the object side, so it is virtual.
2. For refraction at a spherical surface m = (n1 v)/(n2 u).

Answer: (a) 10 cm

Explanation:
1. Light goes from the paper through the glass and refracts at the upper (concave) surface into air. For light travelling upwards, the centre of curvature is above the surface, so \( R=+20 \); \( u=-20 \) (flower), \( n_1=1.5,\ n_2=1 \).
2. Refraction formula: \( \frac1v-\frac{1.5}{-20}=\frac{1-1.5}{20} \), so \( \frac1v=-0.025-0.075=-0.1 \) and \( v=-10\ \text{cm} \).
3. The image is 10 cm below the top vertex, that is 10 cm above the paper.

Teacher's Note:
1. The flat bottom surface of the lens does not refract, since the paper is in contact with it.
2. Only the top surface contributes.

Answer: (d) the image is always virtual

Explanation:
1. The centre lies on the incident side, so \( R=-10 \). \( \frac{3/2}{v}-\frac{4/3}{u}=\frac{3/2-4/3}{-10}=-\frac1{60} \).
2. For a real object (u < 0): \( \frac{3/2}{v}=-\frac1{60}-\frac{4/3}{|u|}<0 \), so v is always negative: the image is virtual for every object distance.

Teacher's Note:
1. Check the sign of each term in the refraction formula.
2. Both terms on the right are negative, so v cannot be positive.

Answer: (b) \( -15\ \text{cm} \)

Explanation:
1. L1: \( \frac1v=\frac1{15}-\frac1{40}=\frac{1}{24} \), so the image from L1 would be 24 cm beyond L1.
2. L2 is 14 cm from L1, so this is a virtual object for L2 at \( u=+10\ \text{cm} \).
3. The final image is at \( v=+30\ \text{cm} \): \( \frac1{f_2}=\frac1v-\frac1u=\frac1{30}-\frac1{10}=-\frac1{15} \), so \( f_2=-15\ \text{cm} \) (a diverging lens).

Teacher's Note:
1. The image from L1 lies beyond L2, so it acts as a virtual object for L2.
2. A negative focal length means a concave lens.

Answer: (d) 33.75 cm from the mirror

Explanation:
1. The water layer acts as a lens in contact with the mirror: \( \frac1{f'}=\frac{2\mu}{R} \) with \( R=60\ \text{cm} \), so \( f'=\frac{R}{2\mu}=\frac{60}{8/3}=22.5\ \text{cm} \).
2. A real image on a screen has \( m=-2 \): \( v=2u \). Mirror formula: \( \frac1{2u}+\frac1u=\frac1{22.5} \), so \( u=1.5\times22.5=33.75\ \text{cm} \).

Teacher's Note:
1. A thin layer of liquid shortens the effective focal length to R/(2μ).
2. For a real image with m = −2 the object is at 1.5 times the focal length.

Answer: (b) diverging lens of focal length 0.25 m

Explanation:
1. Power of each surface \( =\frac{n_2-n_1}{R} \): the two outer glass-air concave surfaces: \( \frac{1.5-1}{-0.2}=-2.5\ \text{D} \) and \( \frac{1-1.5}{0.2}=-2.5\ \text{D} \).
2. The two inner glass-liquid surfaces bound a biconvex liquid lens: \( \frac{1.6-1.5}{0.2}=+0.5\ \text{D} \) each.
3. Total \( P=-2.5-2.5+0.5+0.5=-4\ \text{D} \), a diverging lens with \( f=-\frac1{4}=-0.25\ \text{m} \).

Teacher's Note:
1. Add the powers of all four surfaces for a thin combination.
2. The liquid lens is denser than the glass, so it converges; the glass shell is concave and diverges more strongly.

Answer: (a) 60°

Explanation:
1. For a ray reflected successively from two plane mirrors inclined at angle α (in a plane perpendicular to both), the incident and final rays are inclined at 2α to each other.
2. Here 2α = 120°, and the acute angle between the two lines is \( 180°-120°=60° \).

Teacher's Note:
1. Total deviation after two reflections is 360° − 2α, independent of the angle of incidence.
2. The angle between the rays (as lines) is 2α, or its supplement if this exceeds 90°.

Answer: (c) greater than 90 cm

Explanation:
1. The centre lies in X, the medium of the object, so \( R=-10 \). \( \frac{4/3}{v}-\frac{3/2}{u}=\frac{4/3-3/2}{-10}=\frac1{60} \).
2. With \( u=-|u| \): \( \frac{4/3}{v}=\frac1{60}-\frac{3/2}{|u|} \). The image is real (v > 0) only if \( \frac{3/2}{|u|}<\frac1{60} \), that is \( |u|>90\ \text{cm} \).

Teacher's Note:
1. A real image needs v positive.
2. Solve the inequality for the object distance.

Answer: (a) 5 cm

Explanation:
1. The beam enters the flat face normally and meets the curved face at incidence angle θ, where \( \sin\theta=\frac hR \). Total internal reflection occurs for \( \sin\theta\ge\frac1n=\frac{\sqrt3}2 \), that is \( \theta\ge60° \).
2. The limiting ray with θ = 60° emerges grazing the surface along the tangent, which meets the table at \( \frac{R}{\cos60°}=2R=10\ \text{cm} \) from the axis of the cylinder (the centre of the base).
3. The cylinder ends at R = 5 cm, so the unilluminated patch is \( 10-5=5\ \text{cm} \) long.

Teacher's Note:
1. Find the ray that just escapes (the critical ray); rays above it are totally reflected.
2. The tangent at angle φ from the horizontal meets the table at R/cos φ from the centre.