In a Young's double slit experiment sources of equal intensites are used. Distance between slits is d and wavelength of light used is `lambda(lambdaltltd)` . Angular separation of the nearst points on either side of central maximum where intensities become half of the maximum vlaue is -

Options

A. `(lambda)/(d)`
B. `(lambda)/(2d)`
C. `(lambda)/(4d)`
D. `(lambda)/(6d)`

Answer

B

Explanation

`I+I+2sqrt(I)+sqrt(I)cosphi`
`2I=2I+2Icosphi`
`cosphi=0`
`phi=(pi)/(2)`
`(2piDeltax)/(lambda)=(pi)/(2)`
`Deltax=(lambda)/(4)`
`(dx)/(D)=(lambda)/(4)`
`x=(Dlambda)/(4d)`
`theta=(2x)/(D)=(2)/(D)xx(Dlambda)/(4d)=(lambda)/(2d)`