JEE Physics Ray Optics and Optical Instruments MCQs Set 12

Answer: (c) 0.15 mm

Explanation:
1. The slit and its image are \( d=2\ \text{mm} \) apart and the reflection adds a phase change π, so the fringe pattern is shifted by half a fringe.
2. Fringe width \( \beta=\frac{\lambda D}{d}=\frac{600\times10^{-9}\times1}{2\times10^{-3}}=0.3\ \text{mm} \), so the first maximum is at \( \frac\beta2=0.15\ \text{mm} \).

Teacher's Note:
1. The mirror surface itself is a dark fringe.
2. The first maximum lies half a fringe width above the mirror.

Answer: (a) 5000 Å

Explanation:
1. The 9th bright fringe is at \( 9\beta \) and the 2nd dark fringe at \( 1.5\beta \).
2. So \( 7.5\beta=7.5\ \text{mm} \) gives \( \beta=1\ \text{mm} \).
3. \( \lambda=\frac{\beta d}{D}=\frac{10^{-3}\times0.5\times10^{-3}}{1}=5\times10^{-7}\ \text{m} \).

Teacher's Note:
1. Bright fringes lie at nβ and dark fringes at (n − ½)β.
2. Then λ = βd/D.

Answer: (c) \( 1.8\times10^{-2}\ \text{cm} \)

Explanation:
1. \( \beta=\frac{\lambda D}{d}=\frac{5.4\times10^{-5}\times80}{0.24}=1.8\times10^{-2}\ \text{cm} \).

Teacher's Note:
1. The two images act as coherent sources of separation 0.24 cm.
2. Use consistent units (cm).

Answer: (a) 1 mm

Explanation:
1. Image distance \( \frac1v=\frac1{0.1}-\frac1{0.15} \) gives \( v=0.30\ \text{m} \).
2. Each half has its optical centre 0.25 mm from the axis. The ray through that centre shows the image is at \( 0.25\times\frac{0.45}{0.15}=0.75\ \text{mm} \) from the axis, so \( d=1.5\ \text{mm} \).
3. Distance from the images to the screen \( D=1.30-0.30=1.0\ \text{m} \), so \( \beta=\frac{\lambda D}{d}=\frac{500\times10^{-9}\times1}{1.5\times10^{-3}}=0.33\ \text{mm} \) and the third maximum is at \( 3\beta=1\ \text{mm} \).

Teacher's Note:
1. Locate the two real images first; they are the coherent sources.
2. Measure D from the images, not from the lenses.

Answer: (d) increase

Explanation:
1. A smaller gap brings the optical centres closer to the axis, so the image separation d decreases.
2. Since \( \beta=\frac{\lambda D}{d} \), a smaller d gives a larger fringe width, so the distance OA increases.

Teacher's Note:
1. Image separation is proportional to the gap.
2. Fringe width is inversely proportional to d.

Answer: (c) \( \frac{\lambda}{2(\mu-1)\alpha} \)

Explanation:
1. Each half of the biprism deviates the beam by \( \delta=(\mu-1)\alpha \), so the two beams cross at an angle \( 2\delta \).
2. Two plane waves crossing at angle 2δ give fringe width \( \beta=\frac{\lambda}{2\delta}=\frac{\lambda}{2(\mu-1)\alpha} \), independent of D.

Teacher's Note:
1. For a parallel beam the sources are at infinity, so D does not appear.
2. Two plane waves at angle θ give β = λ/θ.

Answer: (a) \( \frac\pi6 \)

Explanation:
1. Write the second wave as a cosine: \( \sin\left(\theta+\frac\pi3\right)=\cos\left(\theta+\frac\pi3-\frac\pi2\right)=\cos\left(\theta-\frac\pi6\right) \).
2. Compared with \( \cos\theta \), the phase difference is \( \frac\pi6 \).

Teacher's Note:
1. Convert sine to cosine by subtracting π/2 from the argument.
2. Compare arguments of the two cosines.

Answer: (c) \( \frac{2\pi\mu x}{\lambda} \)

Explanation:
1. Wavelength in the medium is \( \lambda'=\frac\lambda\mu \).
2. Phase difference \( =\frac{2\pi x}{\lambda'}=\frac{2\pi\mu x}{\lambda} \).

Teacher's Note:
1. Equivalent optical path is μx.
2. Phase = 2π × (path)/λ.

Answer: (a) zero and 1.5 mm

Explanation:
1. Extra optical path \( (\mu-1)t=0.5\times1.5\times10^{-6}=0.75\ \mu m=1.5\lambda \), a phase of 3π, which is a minimum, so the intensity is zero.
2. Shift \( =\frac{(\mu-1)tD}{d}=\frac{0.75\times10^{-6}\times1}{5\times10^{-4}}=1.5\ \text{mm} \).

Teacher's Note:
1. An odd number of half wavelengths of extra path gives darkness.
2. Shift = (μ − 1)tD/d.

Answer: (d) 0.63 mm

Explanation:
1. Wavelength in the liquid \( \lambda'=\frac{6300}{1.33}\ \text{Å} \).
2. \( \beta=\frac{\lambda' D}{d}=\frac{6300\times10^{-10}}{1.33}\times\frac{1.33}{10^{-3}}=0.63\ \text{mm} \).

Teacher's Note:
1. The factor 1.33 cancels.
2. Use the wavelength in the liquid.

Answer: (b) 0.085

Explanation:
1. Extra optical path \( (\mu-1)t=0.17\times1.5\times10^{-7}=2.55\times10^{-8}\ \text{m} \).
2. The central maximum lies where the geometric path difference cancels it: \( d\sin\theta=2.55\times10^{-8} \), so \( \sin\theta=\frac{2.55\times10^{-8}}{3\times10^{-7}}=0.085 \) (towards the covered slit).

Teacher's Note:
1. The central maximum has zero total optical path difference.
2. It moves towards the slit carrying the sheet.

Answer: (d) \( \frac\lambda4 \)

Explanation:
1. With index between air and glass, both reflections are from a denser medium, so the phase changes cancel.
2. Destructive interference needs \( 2t=\frac\lambda2 \), so \( t=\frac\lambda4 \).

Teacher's Note:
1. Minimum reflection needs a path difference of half a wavelength in the film.
2. λ here is the wavelength inside the coating.

Answer: (b) \( 4I_0 \)

Explanation:
1. With the same incident intensity per unit area, doubling the slit width doubles the amplitude at the centre of the pattern.
2. Intensity is proportional to amplitude squared, so it becomes \( 4I_0 \).

Teacher's Note:
1. At the central maximum all secondary wavelets are in phase.
2. Amplitude is proportional to slit width.

Answer: (d) \( \frac34 \)

Explanation:
1. Phase difference \( \phi=\frac{2\pi}{\lambda}\cdot\frac\lambda6=\frac\pi3 \).
2. \( \frac I{I_0}=\cos^2\frac\phi2=\cos^2\frac\pi6=\frac34 \).

Teacher's Note:
1. I = I0 cos²(φ/2).
2. φ/2 = 30°.

Answer: (c) 442.5 nm

Explanation:
1. Coincidence: \( 3\times590=4\lambda \).
2. \( \lambda=\frac{1770}{4}=442.5\ \text{nm} \).

Teacher's Note:
1. Position of the nth bright fringe is nλD/d.
2. Equate positions.

Answer: (b) 2

Explanation:
1. Coherent: amplitude \( 2A \), so \( I_1\propto4A^2 \).
2. Incoherent: intensities add, \( I_2\propto A^2+A^2=2A^2 \).
3. \( \frac{I_1}{I_2}=2 \).

Teacher's Note:
1. Coherent waves add amplitudes, incoherent waves add intensities.
2. The central point has zero path difference.

Answer: (d) \( \frac{I_m}{9}\left(1+8\cos^2\frac\phi2\right) \)

Explanation:
1. With amplitudes a and 2a: \( I_m=(a+2a)^2=9a^2 \).
2. \( I=a^2+4a^2+4a^2\cos\phi=a^2(5+4\cos\phi) \).
3. Using \( \cos\phi=2\cos^2\frac\phi2-1 \): \( I=a^2\left(1+8\cos^2\frac\phi2\right)=\frac{I_m}{9}\left(1+8\cos^2\frac\phi2\right) \).

Teacher's Note:
1. Maximum intensity is the square of the sum of amplitudes.
2. Express cosφ in terms of cos²(φ/2).

Answer: (b) concentric circles

Explanation:
1. Each point on the screen at a given distance from the axis has the same path difference to S1 and S2, since the path difference depends only on the angle from the axis.
2. So the loci of constant path difference are circles about the axis: concentric circular fringes.

Teacher's Note:
1. The pattern has rotational symmetry about the line joining the sources.
2. Fringes are loci of constant path difference.

Answer: (d) 30 cm

Explanation:
1. Refractive index \( \mu=\frac{3\times10^8}{2\times10^8}=1.5 \).
2. Radius of curvature from the sag: \( R=\frac{h^2+s^2}{2s}=\frac{30^2+3^2}{6}\ \text{mm}\approx151\ \text{mm}\approx15\ \text{cm} \).
3. \( f=\frac{R}{\mu-1}=\frac{15}{0.5}=30\ \text{cm} \).

Teacher's Note:
1. Find μ from c/v.
2. Use the chord-sagitta relation for R.

Answer: (c) \( 30\ \mu m \)

Explanation:
1. Limit of resolution \( \theta=\frac{1.22\lambda}{D}=\frac{1.22\times500\times10^{-9}}{5\times10^{-3}}=1.22\times10^{-4}\ \text{rad} \).
2. Separation \( =\theta\times0.25\ \text{m}=3.05\times10^{-5}\ \text{m}\approx30\ \mu m \).

Teacher's Note:
1. The aperture diameter is 0.5 cm.
2. Separation = θ × viewing distance.

Answer: (a) bends upwards

Explanation:
1. Near the ground the refractive index is smaller, so light travels faster there.
2. The lower part of each wavefront moves ahead of the upper part, so the wavefront tilts upwards and the beam bends upwards.

Teacher's Note:
1. Speed is c/μ.
2. A beam bends towards the region of higher refractive index.

Answer: (c) 45°

Explanation:
1. Beyond A the intensity is \( \frac I2 \). Since B transmits all of it, B is parallel to A.
2. With C inserted at angle θ to A (and so to B): \( \frac I2\cos^2\theta\cos^2\theta=\frac I8 \), so \( \cos^4\theta=\frac14 \) and \( \cos^2\theta=\frac12 \).
3. Hence \( \theta=45^\circ \).

Teacher's Note:
1. B is parallel to A.
2. Apply Malus' law twice.

Answer: (a) \( \sin^{-1}\left(\frac{\lambda}{3d}\right) \)

Explanation:
1. \( I\cos^2\frac\phi2=\frac I4 \) gives \( \cos\frac\phi2=\frac12 \), so \( \phi=\frac{2\pi}{3} \).
2. Path difference \( d\sin\theta=\frac\lambda3 \), so \( \theta=\sin^{-1}\frac{\lambda}{3d} \).

Teacher's Note:
1. I = I_max cos²(φ/2).
2. φ = 2π/3 corresponds to a path difference λ/3.

Answer: (c) \( \pi r^2\propto f^2 \)

Explanation:
1. The sun subtends a fixed angle, so the image radius is \( r=f\theta \) and the area \( \pi r^2\propto f^2 \).
2. Covering half the lens does not change the image size, only its brightness, and doubling f increases the image area four times, so the intensity decreases.

Teacher's Note:
1. Image size of a distant object is f × (angle subtended).
2. Blocking part of a lens dims the image.