Question:

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:

Updated On: Jul 8, 2024
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The Correct Option is B

Solution and Explanation

$\frac{I_{\max}}{I_{\min}} = 16 $ $ \Rightarrow \frac{A_{\max }}{A_{\min }} = 4 $ $ \Rightarrow \frac{A_{1} + A_{2}}{A_{1} - A_{2}} = \frac{4}{1} $ Using componendo & dividendo. $ \frac{A_{1} }{A_{2}} = \frac{5}{3} \Rightarrow \frac{I_{1}}{I_{2}} = \left(\frac{5}{3}\right)^{2} = \frac{25}{9} $
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Concepts Used:

Young’s Double Slit Experiment

  • Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
  • The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
  • Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.

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