Question

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is $I_0$. The intensity of light at a point where the path difference becomes $\frac{\lambda }{3}$ is

• A. $I_0$
• B. $\frac{I_0}{4}$
• C. $\frac{I_0}{3}$
• D. $\frac{I_0}{2}$

Answer 1

Answer: (A)

$I_0$

Intensity is given by

$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$

$I=I_1=I_2=I_0$

$K=I_0+I_0+2I_0\cos 2\pi =4I_0$

$\theta =\frac{2\pi \Delta x}{\lambda }$

$\Delta x=\lambda /3$

therefore we have

$K^{{}^{\prime }}=I_0+I_0+2I_0\cos 2\pi /3$

$K^{{}^{\prime }}=I_0+I_0+2I_0\left(\frac{1}{2}\right)=I_0$