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Maximum intensity in YDSE is I0. Find th...

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 `lambda/4`.

Text Solution

Verified by Experts

(a) We know that
`I=I_(max) cos^2 (phi/2)`
Here, `I_(max)` is `I_0` (i.e. intensity due to independent sources is `I_0//4`). Therefore, at
`phi = pi/3`
or `phi/2 = pi/6 `
` I= I_0 cos^2 (pi/6) = 3/4 I_0`
(b) Phase difference corresponding to the given path difference `Deltax = lambda/4` is
` phi = ((2pi)/lambda)(lambda/4)= pi/2` or phi/2= pi/4`
`I= I_0 cos^2 (pi/4) = I_0/2` .
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