JEE Physics Wave Optics MCQs Set 12

Answer: (b) Statement I is true but Statement II is false

Explanation:
1. Air to glass: \( \tan\theta_B=\mu_g \).
2. Glass to air: \( \tan\theta_B'=\frac1{\mu_g}=\cot\theta_B \), so \( \theta_B'=\frac\pi2-\theta_B \): Statement I is true.
3. Statement II would require \( \tan\theta=\mu_g \), which is the air-to-glass angle, so it is false.

Teacher's Note:
1. Brewster's law: tan θ = μ2/μ1 (second medium over first).
2. At Brewster's angle the reflected and refracted rays are perpendicular, so the two angles in the reverse path are complementary.

Answer: (a) 60 \( \mu \)m

Explanation:
1. Position of the nth bright fringe: \( y_n=\frac{n\lambda D}{d} \).
2. \( d=\frac{5\times600\times10^{-9}\times1}{0.05}=6\times10^{-5}\ \text{m}=60\ \mu\text{m} \).

Teacher's Note:
1. Convert 5 cm to 0.05 m and 600 nm to 6 x 10⁻⁷ m.
2. The fringe width is β = λD/d.

Answer: (a) 18 km/s

Explanation:
1. Path difference \( \Delta x=\frac{\lambda}{2\pi}\Delta\phi=\frac\lambda6=6\ \text{m} \), so \( \lambda=36\ \text{m} \).
2. \( v=f\lambda=500\times36=18000\ \text{m/s}=18\ \text{km/s} \).

Teacher's Note:
1. A phase difference of 60° is 1/6 of a full cycle, i.e. λ/6 of path.
2. Speed = frequency x wavelength.

Answer: (a) 24 W/m²

Explanation:
1. After P1 (unpolarised light): \( I_1=\frac{256}{2}=128\ \text{W/m}^2 \).
2. After P2 (60° to P1): \( I_2=128\cos^260^\circ=32\ \text{W/m}^2 \).
3. P3 is at 90° - 60° = 30° to P2: \( I_3=32\cos^230^\circ=24\ \text{W/m}^2 \).

Teacher's Note:
1. Malus's law applies to the angle between successive pass axes.
2. The first polaroid halves the unpolarised intensity.

Answer: (c) 0.2 mm

Explanation:
1. Path difference at P: \( A_2P-A_1P=\sqrt{D^2+a^2}-D=\frac\lambda2 \).
2. Squaring: \( a^2=D\lambda+\frac{\lambda^2}{4}\approx0.05\times800\times10^{-9}=4\times10^{-8} \).
3. \( a=2\times10^{-4}\ \text{m}=0.2\ \text{mm} \).

Teacher's Note:
1. The first minimum needs a path difference of half a wavelength.
2. The λ²/4 term is negligible compared with Dλ.

Answer: (a) 10

Explanation:
1. Extra optical path difference \( =(1.55-1.51)\times0.1\ \text{mm}=4\times10^{-6}\ \text{m} \).
2. Number of fringes shifted \( =\frac{\Delta}{\lambda}=\frac{4\times10^{-6}}{4\times10^{-7}}=10 \).

Teacher's Note:
1. A path difference of one wavelength shifts the pattern by one fringe.
2. The shift is towards the slit with the slab of larger refractive index.

Answer: (a) 3

Explanation:
1. Intensity \( I=4I_0\cos^2\frac\phi2 \) with \( \phi=\frac{2\pi}\lambda\Delta x \).
2. Path difference \( \frac\lambda4 \): \( \phi=\frac\pi2,\ I_1=4I_0\cos^2\frac\pi4=2I_0 \).
3. Path difference \( \frac\lambda3 \): \( \phi=\frac{2\pi}3,\ I_2=4I_0\cos^2\frac\pi3=I_0 \).
4. \( \frac{I_1+I_2}{I_0}=3 \).

Teacher's Note:
1. Convert path difference to phase difference first.
2. Resultant intensity of two equal coherent sources: 4I0 cos²(φ/2).

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

Explanation:
1. De Broglie proposed that moving particles have wave nature, so electron beams can show interference and diffraction: A is true.
2. Davisson and Germer observed diffraction of electrons from a nickel crystal, confirming the wavelength predicted by de Broglie: R is true and it supports A directly.

Teacher's Note:
1. Diffraction needs a grating spacing comparable to the wavelength; crystal planes serve this purpose for electrons.
2. This experiment is the key evidence for matter waves.

Answer: (c) \( \frac{I_0}8 \)

Explanation:
1. After A: \( \frac{I_0}2 \).
2. After C (45° to A): \( \frac{I_0}2\cos^245^\circ=\frac{I_0}4 \).
3. After B (45° to C): \( \frac{I_0}4\cos^245^\circ=\frac{I_0}8 \).

Teacher's Note:
1. Without C no light would pass through crossed polaroids.
2. Each 45° polaroid halves the polarised intensity.

Answer: (a) 48 mm

Explanation:
1. Bright fringes coincide when \( n_1\times800=n_2\times600\Rightarrow n_1:n_2=3:4 \).
2. Smallest case: \( y=\frac{3\times800\times10^{-9}\times7}{0.35\times10^{-3}}=4.8\times10^{-2}\ \text{m}=48\ \text{mm} \).

Teacher's Note:
1. Find the lowest integers with n1 λ1 = n2 λ2.
2. Use y = nλD/d with the matching n and λ.

Answer: (b) 6

Explanation:
1. After the first sheet the intensity is \( \frac I2 \).
2. Each further sheet multiplies it by \( \cos^245^\circ=\frac12 \): \( I_n=\frac I2\left(\frac12\right)^{n-1}=\frac I{2^n} \).
3. \( 2^n=64\Rightarrow n=6 \).

Teacher's Note:
1. Count the first sheet separately: it only halves the unpolarised light.
2. Then apply Malus's law n - 1 times.

Answer: (a) 4

Explanation:
1. Shift of the central maximum \( =\frac{(\mu-1)tD}{d} \) and fringe width \( \beta_0=\frac{\lambda D}{d} \).
2. \( x=\frac{(\mu-1)t}{\lambda}=\frac{0.2\times10\times10^{-6}}{500\times10^{-9}}=4 \).

Teacher's Note:
1. The plate adds an extra optical path (μ - 1)t.
2. Express the shift as a multiple of the fringe width.

Answer: (a) 7

Explanation:
1. Refractive index \( n=\sqrt{\mu_r\varepsilon_r}=\sqrt{\varepsilon_r} \).
2. Brewster's law: \( \tan i_B=\frac{n_2}{n_1}=\sqrt{\frac{6.8}{2.8}}=\sqrt{1+\frac{4}{2.8}}=\sqrt{1+\frac{10}{7}} \).
3. Comparing with the given form, \( \theta=7 \).

Teacher's Note:
1. The reflected and refracted rays are perpendicular at Brewster's angle.
2. 4/2.8 = 10/7.