In a Young's double slit experiment the intensity at a point where the path difference is `(lambda)/(6)(lambda` being the wavelength of the light used) is I. If `I_(0)` denotes the maximum intensity then `(I)/(I_(0))=…`

Options

A. `(1)/(sqrt(2))`
B. `(sqrt(3))/(2)`
C. `(1)/(2)`
D. `(3)/(4)`

Answer

D

Explanation

Maximum intensity `I_(0)=I.+I+2 sqrt(I.I) cos0^(@)`
`=I.+2I. [ :. cos 0^(@)=1]`
`=4I.`
and intensity at any point,
`I=I.+I.+2sqrt(I.I) cos phi`
where `phi=(2pi)/(lambda)xx` path difference
`=(2pi)/(lambda)xx(lambda)/(6)=(pi)/(3)` rad
`:.I=2I.+2I"cos"(pi)/(3)`
`=2I.+I.=3I.`
`:.(I)/(I_(0))=(3I.)/(4I.)=(3)/(4)`