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. (a) `I_0`
B. (b) `I_0/4`
C. (c) `3/4I_0`
D. (d) `I_0/2`

Answer

D

Explanation

Path difference
`=S_2P-S_1P`
`=sqrt(D^2+d^2)-D`
`=D(1+1/2d^2/D^2)-D`
`=D[1+(d^2)/(2D^2)-1]=(d^2)/(2D)`
`Deltax=(2pi)/(lambda)*pi/4=pi/2`
So, intensity at the desired point is
`I=I_0cos^2phi/2=I_0cos^2pi/4=I_0/2`