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+5cos phi)`
B. `(I_(m))/(3)(1+2"cos"^(2)(phi)/(2))`
C. `(I_(m))/(5)(1+4"cos"^(2)(pi)/(2))`
D. `(I_(m))/(2)(1+8"cos"^(2)(phi)/(2))`

Answer

D

Explanation

Given, `a_(1)=2a_(2)`
`rArr I_(1)=4I_(2)=4I_(0)`
`therefore I_("max")=(sqrt(I_(1))=sqrt(I_(2)))^(2)=(3sqrt(I_(2)))^(2)=9I_(2)=9I_(0)`
Now, `I=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cos phi=4I_(0)+I_(0)+2sqrt(4I_(0)*I_(0))cos phi`
`=5I_(0)+4I_(0)cos phi =(I_("max"))/(9)(5+4cosphi)`
`=(I_("max"))/(9)[1+(4(1+cos phi)]`
`=(I_("max"))/(9)(1+8cos^(2)phi//2)`