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+2"cos"^(2)(phi)/(2))`
B. `(I_(m))/(5)(1+4"cos"^(2)(phi)/(2))`
C. `(I_(m))/(9)(1+8"cos"^(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)=4 " or " I_(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_(1)+4I_(1)+2sqrt(I_(1)(4I_(1)))cos phi`
`" " =5I_(1)+4I_(1)cos phi=I_(1)+4I_(1)+4I_(1)cos phi=I_(1)+4I_(1)(1+cos phi)`
` " "=I_(1)(1+8"cos"^(2)(phi)/(2)) " "( :' 1+cos phi=2"cos"^(2)(phi)/(2))`
Putting the value of `I_(1)` from eqn. (i),we get
` I=(I_(m))/(9)(1+8"cos"^(2)(phi)/(2))`