In Young's double slit experiment, the slits are horizontal. The intensity at a point P as shown in figure is `(3)/(4) I_(0)`,where`I_(0)` is the maximum intensity.
Then the value of `theta` is,
(Given the distance between the two slits `S_(1)` and `S_(2)` is `2 lambda`)

Options

A. `cos^(-1)((1)/(12))`
B. `sin^(-1)((1)/(12))`
C. `tan^(-1)((1)/(12))`
D. `sin^(-1)((3)/(5))`

Answer

A

Explanation

In Young.s double slit experiment, intensity at a point is given by
`I=I_(0)cos^(2)((phi)/(2))`
But `(I)/(I_(0))=(3)/(4)` (given) or `cos^(2)((phi)/(2))=(3)/(4)`
or `cos.(phi)/(2)=(sqrt(3))/(2)` or `phi=60^(@)=(pi)/(3)`
Phase difference, `phi=(2pi)/(lambda)xx` path difference

From the figure, path difference is
`dcostheta=2lambdacostheta( :.d=2lambda)`
`:.(pi)/(3)=(2pi)/(lambda)2lambdacostheta`
`:.costheta=(1)/(12)`
`:.theta=cos^(-1)((1)/(12))`