In Young's double slit experiment, one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If `I_m` be the maximum intensity, the resultant intensity I when they interfere at phase difference `phi` is given by:

Options

A. `(I_m)/(9)(4+5cosphi)`
B. `(I_m)/(3)(1+2cos^(2)((phi)/(2))`
C. `(I_m)/(5)(1+4cos^2((phi)/(2))`
D. `(I_m)/(9)(1+8cos^2((phi)/(2))`

Answer

D

Explanation

Let `a_1=a`,`I_1=a_1^2=a^2`
`a_2=2a`,`I_2=a_2^2=4a^2` therefore `I_2=4I_1`
`I_r=I_1+I_2+2sqrt(I_1I_2)cosphi`
`I_r=I_1+4I_1+2sqrt(4I_1^2)cosphi`
`impliesI_r=5I_1+4I_1cosphi`...(1)
Now, `I_(max)=(a_1+a_2)^2=(a+2a)^2=9a^2`
`I_(max)=9I_1impliesI_1=(I_(max))/(9)`
Substituting in equation (1)
`I_(r)=(5I_(max))/(9)+(4I_(max))/(9)cosphi`
`I_(r)=(I_(max))/(9)[5+4cosphi]`
`I_r=(I_max)/(9)[5+8cos^2(phi)/(2)-4]`
`I_r=(I_max)/(9)[1+8cos^2((phi)/(2))]`