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+5cosphi)`
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

The slits are of different widths and the amplitudes of the waves are such that `A_(1)=2A_(2)`
and `because` intensity `IpropA^(2)" "therefore I_(2)=I_(0) and I_(1)=4I_(0)`
`therefore ` The resultant intensity.
`I=I_(1)=I_(2)+2sqrt(I_(1)I_(2))cosdelta`
`=4I_(0)+I_(0)+2sqrt(4I_(0)xxI_(0))cosdelta`
the intensity is maximum, when `cosdelta=1`
`therefore I_(m)=5I_(0)+4I_(0)=9I_(0)" "therefore I_(0)=(I_(m))/(9)` . . (1)
when the phase difference is `phi`, then
`I=4I_(0)+I_(0)+2sqrt(4I_(0)xxI_(0))cosphi`
`=I_(0)+4I_(0)(1+cosphi)`
`=I_(0)+4I_(0)*2"cos"^(2)(phi)/(2)`
`=I_(0)(1+8"cos"^(2)(phi)/(2))`
`therefore I=(I_(m))/(9)(1+8"cos"^(2)(phi)/(2))`