1.

In Young's double slit experiment, the intensity at a point P on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is lambda and d is the distance between the slits, the angular separation between point P and the centre of the screen is ......

Answer»

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

Solution :`I=I_(0)cos^(2){(K(r_(1)-r_(2)))/(2)}`
`(I_(0))/(2)=I_(0) cos^(2){(2pi)/(lambda)XX(d sin THETA)/(2)}`
`(1)/(2)=cos^(2){(PI d sin theta)/(lambda)}`
`:.(1)/(sqrt(2))= cos {(pi d sin theta)/(lambda)}`
`:.(pi)/(4)=(pi d sin theta)/(lambda)`
`:. (lambda)/(4)= d sin theta :. theta sin^(-1)((lambda)/(4d))`


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