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

Answer

C

Explanation

Here, `A_(2)=2A_(1)`
`:.` Intensity `prop ("Amplitude)^(2)`
`:.(I_(2))/(I_(1))=((A_(2))/(A_(1)))^(2)=((2A_(1))/(A_(1)))^(2)=4impliesI_(2)=4I_(1)`
Maximum intensity, `I_(m)=(sqrt(I_(1))+sqrt(I_(2)))^(2)`
`I_(m)=(sqrt(I_(1))+sqrt(4I_(1)))^(2)=(3sqrt(I_(1)))^(2)=9I_(1)"or"I_(1)=(I_(m))/(9)` ...(i)
Resultant intensity, `I=I_(1)+I_(2)+2sqrt(I_(1)I_(2))cosphi`
`I=I_(1)+4I_(1)+2sqrt(I_(1)(4I_(1)))cosphi`
`=5I_(1)+4I_(1)cosphi=I_(1)+4I_(1)+4I_(1)cosphi=I_(1)+4I_(1)(1+cosphi)`
`=I_(1)+8I_(1)cos^(2).(phi)/(2)=I_(1)(1+8cos^(2).(phi)/(2))`
Putting the value of `I_(1)` from eqn. (i) we get
`I=(I_(m))/(9)(1+8cos^(2).(phi)/(2))`