Maximum intensity in Young's double slit experiment is I. Then what is the intensity at a point on the screen where the path difference between them is lambda/4?

Question

Saurabh Kumar · Answer

Phase difference = k x path difference Then calculating Intensity, as is is a function of phase difference.

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Maximum intensity in Young's double slit experiment is I. Then what is the intensity at a point on the screen where the path difference between them is lambda/4?

Maximum intensity in Young's double slit experiment is I. Then what is the intensity at a point on the screen where the path difference between them is lambda/4?

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Maximum intensity in Young's double slit experiment is I. Then what is the intensity at a point on the screen where the path difference between them is lambda/4?

Maximum intensity in Young's double slit experiment is I. Then what is the intensity at a point on the screen where the path difference between them is lambda/4?

1 Answers

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