1.

In Young's double slit experiment, the intensity at a point is 1/4 of maximum intensity, angular position of this point is ?

Answer»

`sin^(-1) (lambda)/(d)`
`sin^(-1)(lambda)/(2D)`
`sin^(-1)(lambda)/(3d)`
`sin^(-1)(lambda)/(4d)`

Solution :`I = (I_(0))/(4) = I_(1) = I_(2)`
`phi` is phase difference,
then `I = I_(1) + I_(2) + 2 SQRT(I_(1)I_(2)) cos phi`
`(I_(0))/(4) = (I_(0))/(4) + (I_(0))/(4) + 2.(I_(0))/(4) cos phi`
`thereforecos phi = -1/2, phi = (2pi)/(3)`
Now `Deltax = (lambda)/(2pi) phi =(lambda)/(2pi) .(2 pi)/(3) = (lambda)/(3)`
If `theta` is angular SEPARATION the d `sin theta = Deltax = (lambda)/(3)`
`therefore sin theta =(lambda)/(3d)`
`theta = sin^(-1)((lambda)/(3d))`


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