Practice MCQs for JEE Physics Wave Optics
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Multiple Choice Questions
(a) 4 : 1
(b) 10 : 8
(c) 9 : 1
(d) 2 : 1
Show Answer & Explanation
Answer: (a) 4 : 1
Explanation:
1. Amplitudes are in the ratio \( 3:1 \).
2. \( I_{max}/I_{min}=\left(\dfrac{3+1}{3-1}\right)^2=\dfrac{16}{4}=4:1 \).
Teacher's Note:
1. T
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124. .
(a) \( 6\pi \)
(b) \( 2\pi \)
(c) \( 4\pi \)
(d) zero
Show Answer & Explanation
Answer: (a) \( 6\pi \)
Explanation:
1. The third bright fringe has path difference \( 3\lambda \).
2. Phase difference \( =\dfrac{2\pi}{\lambda}\times3\lambda=6\pi \).
Teacher's Note:
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135. .
(a) 6000 \( \mathring{A} \)
(b) 4500 \( \mathring{A} \)
(c) 3000 \( \mathring{A} \)
(d) 5000 \( \mathring{A} \)
Show Answer & Explanation
Answer: (a) 6000 \( \mathring{A} \)
Explanation:
1. \( \beta=\lambda D/d \), so \( \Delta\beta=\lambda\,\Delta D/d \).
2. \( \lambda=\dfrac{\Delta\beta\,d}{\Delta D}=\dfrac{3\times10^{-5}\times10^{-3}}{5\times10^{-2}}=6\times10^{-7} \) m = 6000 Å.
Teacher's Note:
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124. .
(a) K/2
(b) K/4
(c) K
(d) zero
Show Answer & Explanation
Answer: (a) K/2
Explanation:
1. Path difference \( \lambda \) gives a maximum, so \( I_{max}=K \).
2. For path difference \( \lambda/4 \) the phase difference is \( \pi/2 \): \( I=I_{max}\cos^2(\pi/4)=K/2 \).
Teacher's Note:
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(a) \( \dfrac{D}{d}(\mu-1)t \)
(b) \( \dfrac{d}{D}(\mu-1)t \)
(c) \( \dfrac{d}{(\mu-1)D} \)
(d) \( (\mu-1)\dfrac{D}{d}\,\lambda \)
Show Answer & Explanation
Answer: (a) \( \dfrac{D}{d}(\mu-1)t \)
Explanation:
1. The extra optical path is \( (\mu-1)t \).
2. The central fringe shifts so that the path difference \( d\,y/D \) compensates it: \( y=\dfrac{D}{d}(\mu-1)t \).
Teacher's Note:
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109. .
(a) 1 cm
(b) 10 \( \mathring{A} \)
(c) 1000 \( \mathring{A} \)
(d) 100 \( \mathring{A} \)
Show Answer & Explanation
Answer: (c) 1000 \( \mathring{A} \)
Explanation:
1. Thin-film interference is seen when the film thickness is comparable to the wavelength of light.
2. Wavelengths are about 4000 to 7000 Å, so a thickness of the order of 1000 Å is suitable.
Teacher's Note:
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123. .
(a) 12
(b) 18
(c) 24
(d) 30
Show Answer & Explanation
Answer: (b) 18
Explanation:
1. Fringe width \( \beta\propto\lambda \), so the number of fringes in a fixed segment is proportional to \( 1/\lambda \).
2. \( N=12\times600/400=18 \).
Teacher's Note:
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85. .
(a) 0.16\(^\circ\)
(b) 0.08\(^\circ\)
(c) 0.20\(^\circ\)
(d) 0.32\(^\circ\)
Show Answer & Explanation
Answer: (a) 0.16\(^\circ\)
Explanation:
1. For the first dark fringe, \( d\sin\theta=\lambda/2 \).
2. \( \theta=\dfrac{\lambda}{2d}=\dfrac{5460\times10^{-10}}{2\times10^{-4}}=2.73\times10^{-3} \) rad.
3. \( \theta=2.73\times10^{-3}\times\dfrac{180}{\pi}\approx0.16^\circ \).
Teacher's Note:
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135. .
(a) 2I
(b) 4I
(c) 5I
(d) 7I
Show Answer & Explanation
Answer: (b) 4I
Explanation:
1. \( I_R=I_1+I_2+2\sqrt{I_1I_2}\cos\phi \).
2. At A: \( I_A=I+4I+0=5I \). At B: \( I_B=5I-4I=I \).
3. Difference \( =5I-I=4I \).
Teacher's Note:
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92. .
(a) 4
(b) 5
(c) 3
(d) 2
Show Answer & Explanation
Answer: (a) 4
Explanation:
1. \( n\lambda_r=(n+1)\lambda_b \), so \( 600n=480(n+1) \).
2. \( 120n=480 \), so \( n=4 \).
Teacher's Note:
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102. .
(a) 6000 \( \mathring{A} \)
(b) 4000 \( \mathring{A} \)
(c) 1200 \( \mathring{A} \)
(d) 2400 \( \mathring{A} \)
Show Answer & Explanation
Answer: (a) 6000 \( \mathring{A} \)
Explanation:
1. \( \beta=\lambda D/d \), so \( \lambda=\beta d/D=6\times10^{-4}\times10^{-3}/1 \).
2. \( \lambda=6\times10^{-7} \) m = 6000 Å.
Teacher's Note:
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97. .
(a) 0.12 mm
(b) 0.3 mm
(c) 3.0 mm
(d) 4.0 mm
Show Answer & Explanation
Answer: (b) 0.3 mm
Explanation:
1. \( \beta=\dfrac{\lambda D}{d}=\dfrac{6\times10^{-7}\times2}{4\times10^{-3}} \).
2. \( \beta=3\times10^{-4} \) m = 0.3 mm.
Teacher's Note:
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(a) 5
(b) 4
(c) 6
(d) 7
Show Answer & Explanation
Answer: (a) 5
Explanation:
1. Add the phasors: 10 along 0 degrees, 4 along 90 degrees, 7 along 180 degrees.
2. Horizontal: \( 10-7=3 \); vertical: 4.
3. Resultant \( =\sqrt{3^2+4^2}=5 \) mm.
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(a) 4
(b) 1
(c) 2
(d) 3
Show Answer & Explanation
Answer: (a) 4
Explanation:
1. \( n_1(650)=n_2(520) \), so \( n_1:n_2=520:650=4:5 \).
2. The smallest integers are \( n_1=4 \) (650 nm) and \( n_2=5 \) (520 nm).
Teacher's Note:
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120. .
(a) 100
(b) 200
(c) 300
(d) 400
Show Answer & Explanation
Answer: (a) 100
Explanation:
1. Number of fringes shifted \( =\dfrac{(\mu-1)t}{\lambda} \).
2. \( =\dfrac{0.5\times10^{-4}}{5\times10^{-7}}=100 \).
Teacher's Note:
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75. .
(a) 5 mm
(b) 2.5 mm
(c) 1.25 mm
(d) 1.0 mm
Show Answer & Explanation
Answer: (b) 2.5 mm
Explanation:
1. Five minima intervals span 0.5 mm, so each is 0.1 mm: \( \lambda D/a=0.1 \) mm.
2. \( a=\dfrac{\lambda D}{0.1\ \text{mm}}=\dfrac{5\times10^{-7}\times0.5}{10^{-4}}=2.5\times10^{-3} \) m.
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127. .
(a) 2.5 cm
(b) 2 cm
(c) 25 cm
(d) 2 mm
Show Answer & Explanation
Answer: (a) 2.5 cm
Explanation:
1. For the first minimum \( a\sin\theta=\lambda \).
2. \( \lambda=5\times\sin30^\circ=2.5 \) cm.
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102. .
(a) 0.3125 cm
(b) 3.125 cm
(c) 0.03125 cm
(d) 312 cm
Show Answer & Explanation
Answer: (a) 0.3125 cm
Explanation:
1. Width of the central maximum \( =\dfrac{2\lambda D}{a} \).
2. \( =\dfrac{2\times6.25\times10^{-7}\times0.5}{2\times10^{-4}}=3.125\times10^{-3} \) m = 0.3125 cm.
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(a) 0.3 mm
(b) 0.2 mm
(c) 0.15 mm
(d) 0.1 mm
Show Answer & Explanation
Answer: (a) 0.3 mm
Explanation:
1. \( a\sin\theta\approx a\dfrac yD=\lambda \).
2. \( a=\dfrac{\lambda D}{y}=\dfrac{6\times10^{-7}\times2}{4\times10^{-3}}=3\times10^{-4} \) m = 0.3 mm.
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123. .
(a) 5000 \( \mathring{A} \)
(b) 2500 \( \mathring{A} \)
(c) 7500 \( \mathring{A} \)
(d) 8500 \( \mathring{A} \)
Show Answer & Explanation
Answer: (a) 5000 \( \mathring{A} \)
Explanation:
1. For the nth minimum \( a\,y_n/f=n\lambda \).
2. \( \lambda=\dfrac{a\,y}{3f}=\dfrac{3\times10^{-4}\times5\times10^{-3}}{3\times1}=5\times10^{-7} \) m = 5000 Å.
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