Question

A parallel beam of light 500 nm is incident at an angle ${30}^o$ with the normal to the slit plane in a young's double-slit experiment. The intensity due to each slit is $I_o$. Point O is equidistant from $s_{1}and{s}_2$. The distance between slits is 1 mm.

• A. The intensity at O is 4Io
• B. The intensity at O is zero.
• C. The intensity at a point on the screen 4mm from O is 41o
• D. The intensity at a point on the screen 4 mm from O is zero

Answer 1

Answer: (A)

The intensity at O is 4Io

Given,

wavelength $\lambda =500nm=500\times {10}^{-9}m$

Angle of incident $\theta ={30}^{\circ }$

$d=1mm$

Path difference is

$△x=n\lambda =dsin\theta n=\frac{dsin\theta }{\lambda }$

Substitute all value in above equation

$n=\frac{1\times {10}^{-3}\times sin30}{500\times {10}^{-9}}={10}^3=integer$

Here n is an integer therefore it will make maximum intensity at point P.

$I_{max}={(\sqrt{I_1}+\sqrt{I_2})}^2={(\sqrt{I_0}+\sqrt{I_0})}^2=4I_0$

Option A is correct option.