The intensity at the maximum in a Young's double slit experiment is `I_(0^.)` Distance between two slits is `d=5 lamda,` where `lamda` is the wavelength of light used in the experment. What will be that intensity in front of one of the slit on the screen placed at a distance at a distance D=10 d ?

Options

A. `I_(0)`
B. `(I_(0))/(4)`
C. `(3)/(4)I_(0)`
D. `(I_(0))/(2)`

Answer

D

Explanation

Path difference `(Deltax)= S_(2)P-S_(1)P= sqrt(D^(2)+d^(2))-D`

`=D(1+ (d^(2))/(D^(2)))^((1)/(2))-D`
`= D[1+ (1)/(2)(d^(2))/(D^(2))-1]= (d^(2))/(2D)`
`Deltax= (d^(2))/(2 xx 10d)= (d)/(20)= (5lambda)/(20)= (lambda)/(4)`
`Deltaphi= (2pi)/(lambda) xx (lambda)/(4)= (pi)/(2)`
Intensity(I)= `I_(0)cos^(2)((phi)/(2))= I_(0)cos^(2)((pi)/(4)) = (I_(0))/(2)`