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

Here `(I)/(I_(0))=(3)/(4)`(given)
`implies "cos"^(2)((phi)/(2))=(3)/(4) " "( :' I=I_(0)"cos"^(2)((phi)/(2)))`
`"or cos"(phi)/(2)=(sqrt(3))/(2)" or " phi=60^(@)=(pi)/(3)`
Phase difference `phi=(2pi)/(lambda)xx` path difference
From the figure, pathe difference is
d `cos theta = 2lambda cos theta " "( :' d=2lambda)`
` :. (pi)/(3)=(2pi)/(lambda)2lambda cos theta`
` :. cos theta =(1)/(12)`
` :. theta = "cos"^(-1)((1)/(12))`