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+5 cos phi)`
B. `I_(m)/3(1+2 (cos^(2)) phi/2)`
C. `I_(m)/5(1+4 (cos^(2)) phi/2)`
D. `I_(m)/9(1+8 (cos^(2)) phi/2)`

Answer

D

Explanation

`a_(1)=2a_(2)`
`I_(1)=4I^('),I_(2)^(')=I^(')`
`I_(m)=(sqrt(I_(1))+sqrt(I_(2)))^(2)= (sqrt(4I^('))+sqrt(I^(')))^(2)=9I^(')`
`I=I_(1)+I_(2)+2sqrt(I_(1)I_(2)) cos phi`
`=4I^(')+I^(')+2sqrt(4I^(').I^(2)) cos phi`
`=4I^(')+I^(')+4I^(') cos phi`
`=I^(')(5+4 cos phi)`
`=I_(m)/9[5+4(2 cos^(2). (phi)/(2)-4)]`
`=I_(m)/9[5+8 cos^(2) ((phi)/2)-4]`
`=I_(m)/9[1+8 cos^(2)(phi/2)]`