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)(x/beta)`
C. `I_(0)cos^(2)((pix)/beta)`
D. `(I_(0)/4)cos^(2)((pix)/beta)`

Answer

C

Explanation

`Delta=xd/D`, where `Delta` is path difference between two waves.
`therefore` phase difference = `phi=(2pi)/lamdaDelta`.
Let a = amplitude at the screen due to each slit.
`thereforeI_(0)=k(2a)^(2)=4ka^(2)`, where k is a constant.
For phase difference `phi`,
amplitude = A = `2acos(phi//2)`.
[Since, `a^(2)=a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cosphi,"here "a_(1)=a_(2)`]
Intensity I,
`I=kA^(2)=k(4a^(2))cos^(2)(phi//2)=I_(0)cos^(2)((pix)/betaDelta)`
= `I_(0)cos^(2)(pi/lamda*(xd)/D)=I_(0)cos^(2)((pix)/beta)`