Question:
The intensity at each slit is equal for a YDSE and it is maximum \(I_{max}\) at \(7π\) central maxima. If I is intensity for phase difference \(\frac{7π}{2}\) between two waves on screen. Then \(\frac{I}{I_{max}}\) is?
The intensity at each slit is equal for a YDSE and it is maximum \(I_{max}\) at \(7π\) central maxima. If I is intensity for phase difference \(\frac{7π}{2}\) between two waves on screen. Then \(\frac{I}{I_{max}}\) is?
Updated On: Jan 22, 2025
- \(\frac{1}{2}\)
- \(\frac{1}{4}\)
- \(\frac{3}{8}\)
- \(\frac{1}{\sqrt2}\)
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The Correct Option is A
Solution and Explanation
The correct option is (A): \(\frac{1}{2}\)
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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.
Read More: Young’s Double Slit Experiment