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In Young's double slit experiment, one o...

In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If `I_m` be the maximum intensity, the resultant intensity I when they interfere at phase difference `phi` is given by:

A

` (I_(m))/( 9) ( 4 + 5 cos phi)`

B

`(I_(m))/( 3) (1 + 2 cos ^(2)"" (phi)/( 2))`

C

`(I_(m))/( 5) (1 + 4 cos^(2) "" (phi)/(2))`

D

`(I_(m))/( 9) (1 + 8 cos ^(2) "" (phi)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
D
Intensity
`I = I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) cos phi `
`rArr I = I_(0) + 4 I_(0) + 2 sqrt(I_(0) xx 4 I_(0)) cos phi`
`rArr I = I_(0) + 4 I_(0) ( 1 + cos phi)`
`rArr I = I_(0) [ 1 + 8 cos^(2) phi // 2]`
`I_("max") = ( sqrt(I_(1)) + sqrt( I_(2)))^(2) = ( sqrt(I_(0) ) + sqrt(4 I_(0)))^(2) = 9I_(0)`
`rArr I_(0) = (I_("max"))/(9)`
`:. I = (I_("max"))/(9) [ 1 + 8 cos^(2) phi //2]`
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