Guest

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?

Grade:12th pass

1 Answers

Saurabh Kumar
askIITians Faculty 2406 Points
6 years ago
Phase difference = k x path difference
Then calculating Intensity, as is is a function of phase difference.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free