In Young's double slit experiment, one of the slit is wider than other, so that the 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 interface at phase difference `phi` is given by .......

Options

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

Answer

D

Explanation

`I prop A^(2)`
`:.(I_(1))/(I_(2)=(A_(1)^(2))/(A_(2)^(2))=((2A_(2))^(2))/(A_(2)^(2))`
`:.(I_(1))/(I_(2))=4 " " :. I_(1)=4I_(2)`
and maximum intensity for constructive interference,
`I_("max") =I_(m) (A_(1)+ A_(2))^(2) `
(Taking value of proportionally constant =1)
`:. I_(m)=(2A_(2)+A_(2))^(2)` [ as given `A_(1)=2A_(2)]`
`:. I_(m)=(3A_(2))^(2)`
`:. I_(m)=9A_(2)^(2) " "...(1)`
but `I_(2) prop A_(2)^(2)`
`:.I_(2)=A_(2)^(2)` [ Taking proportionally constant]
`:. I_(m)=9I_(2)` [ From equation (1)]
`:. I_(2)=(I_(m))/(9) " "...(2)`
Now intensity of light at a point of both waves,
`I=I_(1)+I_(2)+2sqrt(I_(1)I_(2)) cos phi`
`=4I_(2)+I_(2)+2 sqrt((4I_(2))+(I_(2))) cos phi`
`=5I_(2)+4I_(2) cos phi =I_(2)[1+4+4cos phi]`
`=I_(2)[1+4(1+cos phi)]=(I_(m))/(9)[1+4xx"cos"^(2)(phi)/(2)]`
`[ :. I_(2)=(I_(m))/(9),1+cos phi=2"cos"^(2) (phi)/(2)]`
`=(I_(m))/(9)[1+8"cos"^(2)(phi)/(2)]`