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Maximum intensity in YDSE is I_0. Find the intensity at a point on the screen where (a) the phase difference between the two interfering beams is pi//3 . (b) the path difference between them is lamda//4 |
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Answer» Solution :(a) We have `I = I_(max) cos^2 (PHI/2) `. Here , `I_(max)` is `I_0` (i.e., intensity due to independent sources is `I_0//4` ) (b) Phase difference corresponding to the given path difference `Delta X = lamda/4` is `phi = ((2PI)/(lamda)) (lamda/4) ( because phi = (2pi)/(lamda)Delta x) = pi/2 " or " phi/2 =pi/4` `I = I_0 cos^2 (pi/4) = (I_0)/(2)` |
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