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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

(i)We know that, intensity
`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
phase difference, `phi=pi/3 or phi/2=pi/6`
:. Intensity, `I=I_(0)cos^(2)(pi/6)=3/4 I_(0)`
(ii) Phase difference corresponding to the given path difference, `Deltax=lambda/4` is,
`phi=((2pi)/lambda)(lambda/4) (:. phi=(2pi)/lambda Deltax)`
` or phi/2=pi/4`
`:. I = I_(0) cos^(2)(pi/2)=I_(0)/2`
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