In a Young's double slit experiment, D equals the distance of screen and d is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is same as that due to a single slit is equal to

Options

A. `(Dlambda)/d`
B. `(Dlambda)/(2d)`
C. `(Dlambda)/(3d)`
D. `(2Dlambda)/d`

Answer

C

Explanation

Intensity, `I=4I_(0)"cos"^(2)phi/2`
`because I=I_(0)`
`:. Phi=(2pi)/3=(2pi)/lambda. Deltax=(2pi)/lambda ((yd)/D)`
`:. Y=(lambdaD)/(3d)`