Question:
In YDSE, intensity at two sources are in ratio of \(1: 9\). if source are incoherent then intensity at central point is \(I_1\), and if sources are coherent (and phase differs by \(60°\)) then intensity at central point is \(12\) then \(\frac{I_1}{I_2}\) is
In YDSE, intensity at two sources are in ratio of \(1: 9\). if source are incoherent then intensity at central point is \(I_1\), and if sources are coherent (and phase differs by \(60°\)) then intensity at central point is \(12\) then \(\frac{I_1}{I_2}\) is
Updated On: Jan 13, 2026
- \(\frac{10}{13}\)
- \(\frac{5}{13}\)
- \(\frac{8}{13}\)
- \(\frac{7}{11}\)
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The Correct Option is A
Solution and Explanation
The Correct Option is (A) :\(\frac{10}{13}\)
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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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