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

Options

A. `I_(0) cos ((x)/(beta))`
B. `I_(0) cos^(2) ((2 pi x)/(beta))`
C. `I_(0) cos^(2) ((pi x)/(beta))`
D. `(I_(0))/(4) cos^(2) ((pi x)/(beta))`

Answer

c

Explanation

At point P, `Delta phi = (2 pi)/(lambda) Delta x` and `Delta x = (d x)/(Delta)`
`Delta phi = (2pi)/(lambda) xx (d x)/(D) = (2 pi x)/(beta)`
`I_(P) = I_(1) + I_(2) + 2 sqrt(I_(1) I_(2)) ocs Delta phi, I_(1) = I_(2) = I`
`I_(P) = 4 I cos^(2) (Delta d)/(2)`
Given `4 I = I_(0)`
`implies I_(1) = I_(2) = (I_(0))/(4)`
`I_(P) = (I_(0))/(2) + (I_(0))/(2) cos Delta phi`
`I_(P) = I_(0) cos^(2) ((Delta phi)/(2)) = I_(0) cos^(2) ((pi x)/( beta))`