JEE Physics Wave Optics MCQs Set 07

Answer: (c) 5460 Å

Explanation:
1. Phase difference and path difference are related by \( \phi = \dfrac{2\pi}{\lambda}\,\Delta \).
2. So \( \lambda = \dfrac{2\pi\Delta}{\phi} = \dfrac{2\pi \times 2.1\ \mu m}{7.692\pi} = 0.546\ \mu m \).
3. \( 0.546\ \mu m = 5460\ \text{Å} \).

Teacher's Note:
1. A path difference of one wavelength corresponds to a phase difference of \(2\pi\).
2. Convert micrometres to angstrom using \(1\ \mu m = 10^{4}\ \text{Å}\).

Answer: (d) I₀/4

Explanation:
1. Malus' law gives \( I = I_0\cos^2\theta \).
2. With \( \theta = 60^\circ \), \( \cos^2 60^\circ = \tfrac14 \), so \( I = I_0/4 \).

Teacher's Note:
1. Malus' law applies when the incident light is already plane polarised.
2. For unpolarised light the first polarizer halves the intensity.

Answer: (a) Ultrasonic waves

Explanation:
1. Only transverse waves can be polarised.
2. Radio waves, ultraviolet rays and X-rays are electromagnetic and transverse.
3. Ultrasonic waves in a fluid are longitudinal, so they cannot be polarised.

Teacher's Note:
1. Polarisation is a test for transverse nature.
2. Sound and ultrasound in air or liquids are longitudinal.

Answer: (b) concentric circles

Explanation:
1. The path difference to a point on the screen depends only on its angle from the line S₁S₂, which is the axis of the screen.
2. Points at the same distance from the axis lie at the same angle, so they have the same path difference.
3. Hence the locus of equal intensity is a circle, and the fringes are concentric circles.

Teacher's Note:
1. The fringe shape comes from the locus of constant path difference.
2. Straight fringes appear only when the screen is parallel to the line joining the sources.

Answer: (c) \( \dfrac{4\lambda}{\mu-1} \)

Explanation:
1. A plate of thickness t shifts the pattern by \( \dfrac{(\mu-1)tD}{d} \).
2. The fourth bright fringe is at \( \dfrac{4\lambda D}{d} \) from the centre.
3. Equating, \( (\mu-1)t = 4\lambda \), so \( t = \dfrac{4\lambda}{\mu-1} \).

Teacher's Note:
1. The shift equals the number of fringes shifted times the fringe width.
2. One fringe shift needs an extra optical path of one wavelength.

Answer: (b) 4500 Å

Explanation:
1. The nth bright fringe is at \( y = \dfrac{n\lambda D}{d} \).
2. Coincidence gives \( 3\lambda_A = 4\lambda_B \).
3. So \( \lambda_B = \dfrac{3\times6000}{4} = 4500\ \text{Å} \).

Teacher's Note:
1. Coinciding bright fringes mean equal path differences, \( n_A\lambda_A = n_B\lambda_B \).
2. The central bright band is counted as the zeroth order.

Answer: (b) 7

Explanation:
1. Maxima occur where \( d\sin\theta = n\lambda \), and \( \sin\theta \le 1 \).
2. So \( n \le d/\lambda = 7000/2000 = 3.5 \), giving \( n = 0, \pm1, \pm2, \pm3 \).
3. Total maxima = 7.

Teacher's Note:
1. The largest possible order is the integer part of \( d/\lambda \).
2. Count both sides of the central maximum.

Answer: (a) three

Explanation:
1. The path difference at a point (x, 0) is \( \Delta = \sqrt{x^2+16\lambda^2} - x \).
2. It decreases from \( 4\lambda \) at x = 0 to 0 at x = ∞.
3. Maxima need \( \Delta = n\lambda \) with n an integer, so n = 1, 2, 3 are the values lying strictly between 0 and 4λ.
4. That gives three points.

Teacher's Note:
1. Find the range of the path difference first, then count integer multiples of λ inside it.
2. The end values 4λ and 0 correspond to the excluded points.

Answer: (d) point 5

Explanation:
1. In single-slit diffraction the minima occur where the path difference between the extreme rays is \( n\lambda \).
2. Path difference \( \lambda \) gives the first minimum (point 4) and \( 2\lambda \) gives the second minimum (point 5).
3. Points 1, 2 and 3 are maxima.

Teacher's Note:
1. Minima are at whole numbers of wavelengths; secondary maxima lie roughly midway between them.
2. Count the dark points from the centre to find the order.

Answer: (a) Width of fringe increases

Explanation:
1. The fringe width is \( \beta = \dfrac{\lambda D}{d} \).
2. Moving the screen away increases D, so the fringe width increases in proportion.

Teacher's Note:
1. Fringe width is directly proportional to the screen distance.
2. The pattern is simply magnified.

Answer: (b) \( \dfrac{2b\lambda}{e} \)

Explanation:
1. The first minima are at \( y = \pm\dfrac{\lambda b}{e} \) from the centre.
2. The central maximum lies between them, so its size is \( \dfrac{2b\lambda}{e} \).

Teacher's Note:
1. Central maximum width equals twice the distance to the first minimum.
2. It grows when the slit is made narrower.

Answer: (b) 0% to 50%

Explanation:
1. The lamp gives unpolarised light, so the first polaroid transmits 50%.
2. The second transmits \( 50\%\times\cos^2\theta \), which varies from 50% (axes parallel) to 0 (axes crossed).

Teacher's Note:
1. The first polaroid always cuts unpolarised light to half.
2. Rotating the second one only changes the Malus factor from 1 to 0.

Answer: (c) \( \sin^{-1}\left(\dfrac{3}{4}\right) \)

Explanation:
1. The first minimum gives \( b\sin30^\circ = \lambda \), so \( b = 2\lambda \).
2. The first secondary maximum is at \( b\sin\theta = \dfrac{3\lambda}{2} \).
3. So \( \sin\theta = \dfrac{3\lambda}{2\times2\lambda} = \dfrac34 \), that is \( \theta = \sin^{-1}(3/4) \).

Teacher's Note:
1. Secondary maxima lie near path differences \( (n+\tfrac12)\lambda \).
2. Use the first minimum to find b in terms of λ.

Answer: (d) \( 1 : \dfrac{4}{9\pi^2} : \dfrac{4}{25\pi^2} \)

Explanation:
1. The intensity at the secondary maxima follows \( I_n \approx \dfrac{I_0}{(n+\tfrac12)^2\pi^2} \) for n = 1, 2, 3, ...
2. For n = 1: \( \dfrac{4}{9\pi^2}I_0 \); for n = 2: \( \dfrac{4}{25\pi^2}I_0 \).
3. So the ratio is \( 1 : \dfrac{4}{9\pi^2} : \dfrac{4}{25\pi^2} \).

Teacher's Note:
1. Secondary maxima are very weak compared with the central one.
2. Their intensities fall off as \( 1/(2n+1)^2 \).

Answer: (a) depends on wavelength and is different for different colours

Explanation:
1. Brewster's law is \( \tan i_p = \mu \).
2. The refractive index \( \mu \) of a material depends on wavelength (dispersion).
3. So the polarising angle depends on wavelength and differs for different colours.

Teacher's Note:
1. Anything that depends on μ depends on colour through dispersion.
2. This is why the Brewster angle is quoted for a given wavelength.

Answer: (c) 2A²

Explanation:
1. An ideal polaroid transmits half of the intensity of unpolarised light.
2. So \( \dfrac{I}{2} = A^2 \) and \( I = 2A^2 \).

Teacher's Note:
1. Half of unpolarised light is blocked even by an ideal polaroid.
2. The transmitted fraction for unpolarised light is always 1/2.

Answer: (a) \( \left(\dfrac{I_0}{8}\right)\sin^2(2\theta) \)

Explanation:
1. The first and last polaroids are crossed.
2. After the first polaroid the intensity is \( I_0/2 \); after the middle one \( \dfrac{I_0}{2}\cos^2\theta \); after the last (axis at \( 90^\circ-\theta \) to the middle one) \( \dfrac{I_0}{2}\cos^2\theta\sin^2\theta \).
3. Since \( \sin\theta\cos\theta = \tfrac12\sin2\theta \), this equals \( \dfrac{I_0}{8}\sin^2 2\theta \).

Teacher's Note:
1. Insert the angle between each consecutive pair of axes in Malus' law.
2. The result is largest at 45° and zero at 0° and 90°.

Answer: (b) 10⁶

Explanation:
1. Resolving power is \( \dfrac{1}{\Delta\theta} = \dfrac{D}{1.22\lambda} \).
2. \( = \dfrac{0.61}{1.22\times5\times10^{-7}} = 10^{6} \).

Teacher's Note:
1. The factor 1.22 is for a circular aperture.
2. A larger lens or a shorter wavelength improves resolution.

Answer: (c) 0.31 × 10⁻⁶ rad

Explanation:
1. The limit of resolution is \( \Delta\theta = \dfrac{1.22\lambda}{D} \).
2. \( = \dfrac{1.22\times5\times10^{-7}}{2} = 3.05\times10^{-7}\ \text{rad} \approx 0.31\times10^{-6}\ \text{rad} \).

Teacher's Note:
1. Rayleigh's criterion uses 1.22λ/D for circular apertures.
2. A bigger mirror resolves smaller angles.

Answer: (d) 0.85

Explanation:
1. The phase difference is \( \phi = \dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{8} = \dfrac{\pi}{4} \).
2. \( \dfrac{I}{I_0} = \cos^2\dfrac{\phi}{2} = \cos^2\dfrac{\pi}{8} \).
3. \( \cos^2 22.5^\circ = 0.854 \approx 0.85 \).

Teacher's Note:
1. Half of the phase difference appears in the cosine.
2. Near the central maximum the intensity falls off slowly.

Answer: (b) 5 µm

Explanation:
1. A shift of five fringes needs an extra optical path of \( 5\lambda \): \( (\mu-1)t = 5\lambda \).
2. \( t = \dfrac{5\times5000\times10^{-10}}{0.5} = 5\times10^{-6}\ \text{m} = 5\ \mu m \).

Teacher's Note:
1. Each fringe of shift equals one wavelength of extra optical path.
2. Use \( (\mu-1) \) because the film replaces air of equal thickness.

Answer: (a) Both A and R are true and R is the correct explanation of A

Explanation:
1. By conservation of energy, the light that is not reflected is transmitted (neglecting absorption).
2. So the conditions for constructive interference in reflection are the conditions for destructive interference in transmission, and the reverse.
3. R states exactly this and therefore explains A.

Teacher's Note:
1. Reflected and transmitted intensities add up to the incident intensity.
2. A bright reflected film means little light is left to be transmitted.

Answer: (c) 7.8 mm

Explanation:
1. Bright fringes coincide when \( n_1\times650 = n_2\times520 \), so \( n_1 : n_2 = 4 : 5 \) and the least values are \( n_1 = 4 \), \( n_2 = 5 \).
2. \( y = \dfrac{n_1\lambda_1 D}{d} = \dfrac{4\times650\times10^{-9}\times1.5}{0.5\times10^{-3}} \).
3. \( y = 7.8\times10^{-3}\ \text{m} = 7.8\ \text{mm} \).

Teacher's Note:
1. Reduce the ratio of wavelengths to the smallest whole numbers to find the first coincidence.
2. Both orders give the same distance, which is a good check.

Answer: (d) Reflected light is completely polarised with intensity less than I₀/2

Explanation:
1. At the Brewster angle the reflected light contains only the component vibrating perpendicular to the plane of incidence, so it is completely polarised.
2. Only part of that component is reflected, so the reflected intensity is less than half of I₀.
3. The transmitted light is partially polarised.

Teacher's Note:
1. At the polarising angle the reflected and refracted rays are at 90° to each other.
2. The reflected beam is always weaker than I₀/2 because glass reflects only a small part.

Answer: (c) conservation of energy holds good and energy is redistributed

Explanation:
1. Interference does not create or destroy energy.
2. Energy is only redistributed: it is less than average at the dark fringes and more than average at the bright fringes.
3. The total (average) energy over the pattern is the same as the sum of the two separate beams.

Teacher's Note:
1. Interference is a redistribution of energy in space, so energy conservation always holds.
2. Average intensity with two equal sources is twice the single-source value.