Two light beams of intensities in the ratio of 9: 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be:
1. 2: 3 
2. 16: 81 
3. 25: 169 
4. 25: 1 
Subtopic:  Superposition Principle |
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Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is \(16.\) The intensity of the waves are in the ratio:
1. \(16:9\)
2. \(25:9\)
3. \(4:1\)
4. \(5:3\)
Subtopic:  Superposition Principle |
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The interference pattern is obtained with two coherent light sources of intensity ratio 4 :1. And the ratio \(\frac{I_{\max }+I_{\min }}{I_{\max }-I_{\min }} \text { is } \frac{5}{x}\).  Then, the value of x will be equal to:
1. 3
2. 4
3. 2
4. 1
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Interference fringes are observed on a screen by illuminating two thin slits \(1\) mm apart with a light source (\(\lambda =632.8~\mathrm{nm}\)). The distance between the screen and the slits is \(100\) cm. If a bright fringe is observed on a screen at a distance of \(1.27\) mm from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to:
1. \(1.27~\mathrm{\mu m}\)
2. \(2~\mathrm{nm}\)
3. \(2.87~\mathrm{nm}\)
4. \(2.05~\mathrm{\mu m}\)

Subtopic:  Superposition Principle |
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In the phenomenon of interference of light, what happens to the energy?

1. It is conserved but redistributed.
2. It is the same at every point.
3. It is not conserved.
4. It is created at the bright fringes.
Subtopic:  Superposition Principle |
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An interference pattern can be observed due to the superposition of more than one of the following waves:
(A) \(y=a\sin(\omega t)\)
(B) \(y=a\sin(2\omega t)\)
(C) \(y=a\sin\left(\omega t-\phi\right)\)
(D) \(y=a\sin(3\omega t)\)
Identify the waves from the options given below:
1. (B) and (C) only 2. (B) and (D) only
3. (A) and (C) only 4. (A) and (B) only
Subtopic:  Superposition Principle |
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NEET - 2024
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The two light beams having intensities I and 9I interfere to produce a fringe pattern on a screen. The phase difference between the beams is \({\pi \over 2}\) at point P and \(\pi\) at point Q. Then the difference between the resultant intensities at P and Q will be: 
1. \(2~\text I\)
2. \(6~\text I\) 
3. \(5~\text I\)
4. \(7~\text I\)
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In a wave having wavelength \(\lambda\), the path difference corresponding to the phase difference of \(\phi\) is:
1. \(\dfrac{2\lambda}{\pi}\phi\) 2. \(\dfrac{2\pi}{\lambda}\phi\)
3. \(\dfrac{\lambda}{\pi}\phi\) 4. \(\dfrac{\lambda}{2\pi}\phi\)
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The equations of two light waves are given by:
\(y_1=6~\text{cos}(\omega t)\) & \(y_2=8~\text{cos}(\omega t+\phi).\)
What is the ratio of the maximum to the minimum intensities produced by the superposition of these waves?
1. \(49:1\)
2. \(1:49\)
3. \(1:7\)
4. \(7:1\)

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An interference pattern is obtained with two coherent light sources of intensity ratio \(n.\) In the interference pattern, the ratio of their intensities \(\left(\frac{I_{max}-I_{min}}{I_{max}+I_{min}}\right)\) will be:
1. \(\dfrac{\sqrt{n}}{n+1}\) 2. \(\dfrac{2\sqrt{n}}{n+1}\)
3. \(\dfrac{\sqrt{n}}{(n+1)^2}\) 4. \(\dfrac{2\sqrt{n}}{(n+1)^2}\)
Subtopic:  Superposition Principle |
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NEET - 2016
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