In Young's double-slit experiment `d//D = 10^(-4)` (d = distance between slits, D = distance of screen from the slits) At point P on the screen, resulting intensity is equal to the intensity due to the individual slit `I_(0)`. Then, the distance of point P from the central maximum is `(lambda = 6000 Å)`

Options

A. (a) `2mm`
B. (b) `1mm`
C. (c) `0.5mm`
D. (d) `4mm`

Answer

A

Explanation

`I=4I_0cos^2(phi/2)`
`I_0=4I_0cos^2(phi/2)`
`cos(phi/2)=1/2`
or `phi/2=pi/3`
or `phi=(2pi)/(3)=((2pi)/(lambda))*Deltax`
or `1/3=(1/lambda)y*d/D` `(Deltax=(yd)/(D))`
`impliesy=(lambda)/((3xxd/D))=(6xx10^-7)/(3xx10^-4)`
`=2xx10^-3m=2mm`