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In YDSE, having slits of equal width, le...

In YDSE, having slits of equal width, let `beta` be the fringe width and `I_(0)` be the maximum intensity. At a distance x from the central brigth fringe, the intensity will be

A

`l_(0)cos(x/beta)`

B

`l_(0)cos^(2)(x/beta)`

C

`l_(0)cos^(2)(pix/beta)`

D

`l_(0)/4 cos^(2)(pix/beta)`

Text Solution

Verified by Experts

The correct Answer is:
C
`Delta=x(d//D)`
Phase difference = `phi=2pi/lambda Delta`
Let a = amplitude at the screen due to each slit.
` :. L_(0)=k(2a)^(2)=4ka^(2)` , where k is a constant.
For phase difference `phi`, amplitude `= A = 2a cos` `cos(phi//2)`.
intensity, I=`KA^(2)=K(4a^(2))cos^(2)(phi/2) =l_(0)cos^(2)(piDelta lambda)`
`=l_(0)cos^(2)[pi/lambda x "xd"/D]=l_(0)cos^(2)[pi"x"/beta]`.
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