JEE Physics Wave Optics MCQs Set 19

Answer: (c) 9 : 1

Explanation:
1. Intensities are proportional to widths, so amplitudes are in the ratio 2 : 1.
2. \( \dfrac{I_{max}}{I_{min}}=\dfrac{(2+1)^2}{(2-1)^2}=9 \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (a) 5

Explanation:
1. Diffraction minima: \( a\sin\theta=\lambda,2\lambda \). Interference maxima: \( d\sin\theta=m\lambda \).
2. So \( m \) lies between \( \dfrac da=4.8 \) and \( 9.6 \): \( m = 5, 6, 7, 8, 9 \), i.e. 5 bright fringes.

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (c) 625 nm, 500 nm

Explanation:
1. Path difference \( d\theta=0.1\ \text{mm}\times\dfrac1{40}=2500 \) nm \( =m\lambda \).
2. Wavelengths in range: 625 nm \( (m = 4), 500 \) nm \( (m = 5), 416.7 \) nm \( (m = 6) \) and so on; of the pairs offered only 625 nm and 500 nm are both exact values.

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

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

Explanation:
1. Beyond \( A \) the intensity is \( I/2 \), and \( B \) passes all of it, so \( B \) is parallel to \( A \).
2. With \( C \) at angle \( \theta \): \( \dfrac I2\cos^2\theta\cos^2\theta=\dfrac I8\Rightarrow\cos^4\theta=\dfrac14\Rightarrow\cos\theta=\dfrac1{\sqrt2} \), \( \theta=45^\circ \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (d) 7.8 mm

Explanation:
1. Coincidence: \( m_1\times650=m_2\times520\Rightarrow m_1:m_2=4:5 \), so \( m_1=4 \).
2. \( y=\dfrac{m_1\lambda_1D}d=\dfrac{4\times650\times10^{-9}\times1.5}{0.5\times10^{-3}}=7.8 \) mm.

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (b) \( (2n+1)\dfrac\lambda4 \)

Explanation:
1. \( I=I_{max}\cos^2\dfrac\phi2=\dfrac{I_{max}}2\Rightarrow\phi=(2n+1)\dfrac\pi2 \).
2. Path difference \( =\dfrac{\lambda}{2\pi}\phi=(2n+1)\dfrac\lambda4 \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

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

Explanation:
1. \( \cos^2\dfrac\phi2=\dfrac14\Rightarrow\dfrac\phi2=60^\circ\Rightarrow\phi=120^\circ=\dfrac{2\pi}3 \).
2. Path difference \( =\dfrac\lambda3=d\sin\theta\Rightarrow\theta=\sin^{-1}\dfrac\lambda{3d} \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (c) \( 2\lambda \)

Explanation:
1. The extra path \( (\mu-1)t \) must be a whole number of wavelengths for the intensity to be unchanged.
2. \( 0.5\,t=\lambda\Rightarrow t=2\lambda \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (a) 4I

Explanation:
1. \( I_A=I+4I+2\sqrt{4I^2}\cos\dfrac\pi2=5I \).
2. \( I_B=5I+4I\cos\pi=I \). Difference \( =4I \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (d) 2.4 mm

Explanation:
1. First minima at \( y=\dfrac{\lambda D}a=\dfrac{6\times10^{-7}\times2}{10^{-3}}=1.2 \) mm on each side.
2. Distance between them \( =2.4 \) mm.

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (b) quadrupled

Explanation:
1. \( \beta=\dfrac{\lambda D}d \).
2. \( \beta'=\dfrac{\lambda(2D)}{d/2}=4\beta \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (d) \( 2.945\times10^{-4} \) m

Explanation:
1. Separation of first minima \( =\dfrac{2\lambda f}d \).
2. \( d=\dfrac{2\times5.89\times10^{-7}\times0.5}{2\times10^{-3}}=2.945\times10^{-4} m \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (c) 90 nm

Explanation:
1. At the air-layer surface the reflection has a phase change of \( \pi \) (rare to dense); at the layer-glass surface (1.8 to 1.5) there is none. The relative phase change is \( \pi \).
2. Constructive: \( 2\mu t=\left(n-\tfrac12\right)\lambda \); least \( t=\dfrac\lambda{4\mu}=\dfrac{648}{7.2}=90 \) nm.

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (a) (a) and (c) only

Explanation:
1. Path difference in front of one slit: \( \sqrt{D^2+b^2}-D\approx\dfrac{b^2}{2D} \). Darkness needs \( \dfrac{b^2}{2D}=\left(n+\tfrac12\right)\lambda \).
2. \( \lambda=\dfrac{b^2}{(2n+1)D} \): \( \dfrac{b^2}D (n = 0) \), \( \dfrac{b^2}{3D} (n = 1) \), and so on: \( (a) \) and \( (c) \).

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.

Answer: (c) true, because two independent sources are not coherent

Explanation:
1. Two independent lamps are not coherent: their phase difference changes randomly and rapidly.
2. With no fixed phase relation, the intensities simply add and no stable fringes form.

Teacher's Note:
1. Use path difference = \( d \sin(\theta) = d y / D \), with bright fringes at whole wavelengths and dark ones at odd half wavelengths; fringe width is \( \lambda D / d \).
2. For intensities add amplitudes: \( I_{\max} = (a1 + a2)^2 \) and \( I_{\min} = (a1 - a2)^2 \), and intensity at phase difference \( \phi \) is 4 I \( \cos^2(\phi/2) \) for equal sources.