Question:

The ratio of intensity at the centre of a bright fringe to the intensity at a point distant one fourth of the distance between two successive bright fringes will be

Updated On: Jul 7, 2022
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The Correct Option is C

Solution and Explanation

Intensity at the centre of bright fringe, $ I_0 = I + I + 2 \sqrt{II} cos 0^\circ$ $ I_0 = 2 I + 2 I$ $ I_0 = 4 I$ Intensity at a point distant $\frac{\beta}{4}$ (with a phase diffrerence = $\frac{2 \pi}{4} = \frac{\pi}{2})$ is $ I' = I + I + \sqrt{II} cos \frac{\pi}{2}$ $ I' = 2 I \big[ \because \, cos \frac{\pi}{2} = 0 \big]$ $\therefore\frac{I_0}{I'} = \frac{4 I}{2 I} = 2$
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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.

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