JEE Physics Wave Optics MCQs Set 06

Answer: (c) 5

Explanation:
1. The intensity in the overlap varies between \(I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2\) and \(I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2\).
2. The average is \(\dfrac{I_{max}+I_{min}}2=I_1+I_2\).
3. So the average intensity \(=2+3=5\) units.

Teacher's Note:
1. The interference term averages to zero over the whole pattern.
2. Energy is only redistributed, not created.

Answer: (c) 9I and I

Explanation:
1. \(I_{max}=(\sqrt I+\sqrt{4I})^2=(1+2)^2I=9I\).
2. \(I_{min}=(\sqrt{4I}-\sqrt I)^2=(2-1)^2I=I\).

Teacher's Note:
1. Add the amplitudes for the maximum and subtract for the minimum.
2. Intensity is proportional to amplitude squared.

Answer: (c) I = 4I₀

Explanation:
1. With one slit the intensity at the point is \(I_0\).
2. With both identical slits the amplitudes add at the central fringe, so the amplitude is doubled and \(I=(2A)^2\), i.e. \(I=4I_0\).

Teacher's Note:
1. At the central maximum the two waves are in phase.
2. Doubling the amplitude gives four times the intensity.

Answer: (a) 3

Explanation:
1. Fringe width \(\beta=\dfrac{\lambda D}{d}\) is inversely proportional to d.
2. If d becomes \(d/3\), the fringe width becomes 3 times, so \(n=3\).

Teacher's Note:
1. Closer slits give wider fringes.
2. The fringe width is also proportional to λ and D.

Answer: (b) \(\dfrac Dd(\mu-1)t\)

Explanation:
1. The mica sheet adds an extra optical path \((\mu-1)t\).
2. The central fringe moves to the point where the path difference due to the slit separation, \(\dfrac{yd}D\), cancels this.
3. \(y=\dfrac Dd(\mu-1)t\), towards the side of the sheet.

Teacher's Note:
1. The shift is independent of the wavelength.
2. Shift in fringes \(=\dfrac{(\mu-1)t}{\lambda}\).

Answer: (a) 225 nm

Explanation:
1. Light reflected from the top and the bottom of the film both suffer a phase change of \(\pi\) (air to film and film to glass, each going to a denser medium).
2. So the condition for strong reflection is \(2\mu t=n\lambda\).
3. For the least thickness \(n=1\): \(t=\dfrac{600}{2\times1.33}\approx225\) nm.

Teacher's Note:
1. Count the phase changes at both surfaces before choosing the condition.
2. Equal phase changes cancel out.

Answer: (c) 1.4 mm

Explanation:
1. Width of the central maximum in single slit diffraction \(=\dfrac{2\lambda D}{a}\).
2. \(=\dfrac{2\times5\times10^{-7}\times2.1}{1.5\times10^{-3}}=1.4\times10^{-3}\) m \(=1.4\) mm.

Teacher's Note:
1. Convert all lengths to metres first.
2. The central maximum is twice as wide as the other maxima.

Answer: (d) 25 \(\mu\)m

Explanation:
1. The first minimum is at \(30^\circ\) from the axis: \(a\sin30^\circ=\lambda\), so \(\lambda=0.5\) \(\mu\)m.
2. Fringe width \(\beta=\dfrac{\lambda D}{d}\), so \(d=\dfrac{\lambda D}\beta=\dfrac{0.5\times10^{-6}\times0.5}{10^{-2}}=2.5\times10^{-5}\) m.
3. \(d=25\) \(\mu\)m.

Teacher's Note:
1. The angular width of the central maximum is twice the angle of the first minimum.
2. Separate the diffraction part from the interference part.

Answer: (b) \(45^\circ\)

Explanation:
1. A passes \(I/2\) of the unpolarised light. Since B also passes \(I/2\), B must be parallel to A.
2. With C at angle \(\theta\) to A (and to B), the intensity beyond B is \(\dfrac I2\cos^2\theta\cos^2\theta=\dfrac I2\cos^4\theta\).
3. Setting this equal to \(I/8\) gives \(\cos^4\theta=\dfrac14\), \(\cos\theta=\dfrac1{\sqrt2}\), \(\theta=45^\circ\).

Teacher's Note:
1. Malus's law: \(I=I_0\cos^2\theta\).
2. Apply it twice, once at C and once at B.

Answer: (a) 17.3 GHz

Explanation:
1. For light the relativistic Doppler formula for an observer approaching the source is \(f'=f\sqrt{\dfrac{1+v/c}{1-v/c}}\).
2. With \(v=c/2\): \(f'=10\sqrt{\dfrac{1.5}{0.5}}=10\sqrt3=17.3\) GHz.

Teacher's Note:
1. The classical formula would give only 15 GHz.
2. Use the relativistic form when v is comparable with c.

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

Explanation:
1. After P₁: \(I_1=\dfrac{I_0}2\).
2. After P₃ (at \(45^\circ\)): \(I_2=\dfrac{I_0}2\cos^245^\circ=\dfrac{I_0}4\).
3. P₂ is at \(45^\circ\) to P₃: \(I_3=\dfrac{I_0}4\cos^245^\circ=\dfrac{I_0}8\).

Teacher's Note:
1. Without P₃ no light would pass through crossed polaroids.
2. Unpolarised light loses half its intensity at the first polaroid.

Answer: (b) 16 : 1

Explanation:
1. \(I_{max}\propto(3+5)^2=64\) and \(I_{min}\propto(5-3)^2=4\).
2. The ratio is \(64:4=16:1\).

Teacher's Note:
1. Intensity is proportional to the square of the amplitude.
2. The ratio \(\dfrac{I_{max}}{I_{min}}=\left(\dfrac{a_1+a_2}{a_1-a_2}\right)^2\).

Answer: (d) \(\sin^{-1}\dfrac1{\sqrt3}\)

Explanation:
1. By Brewster's law \(\mu=\tan\theta_p=\tan60^\circ=\sqrt3\).
2. The critical angle satisfies \(\sin C=\dfrac1\mu=\dfrac1{\sqrt3}\).
3. So \(C=\sin^{-1}\dfrac1{\sqrt3}\).

Teacher's Note:
1. Brewster's law connects the refractive index and the polarising angle.
2. The sine of the critical angle is 1 over the index.