1.

In Young' s double slit experiment, one of the slit is wider than other, s that the 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 :

Answer»

`(I_(m))/(9)( 1+ 8 cos^(2) "(theta)/(2))`
`(I_(m))/(9)( 4 + 5 cos phi)`
`(I_(m))/(3)(1 + 2 cos ^(2) " (phi)/(2))`
`(I_(m))/(5)(1 + 4 cos ^(2) "(phi)/(2))`

SOLUTION :`I_(m) = I_(0) + 4I_(0) + 2 sqrt(I_(0) xx 4I_(0)) cos phi `
` = I_(0) + 4I_(0) + 4I_(0)cos phi = I_(0)(5 + 4 cos phi)`
from EQUATION (1) put `phi = 0^(@), I_(m) = 9I_(0)`
`I_(0) = (I_(m))/(9)` from equation (2)
`I = (I_(m))/( 1 + cos^(2)phi/2)`
`because cos phi = 2 cos^(2) phi/2 - 1`


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