Question

In Young's double slit experiment using monochromatic light of wavelengths $\lambda$, the intensity of light at a point on the screen with path difference $\lambda$ is M units. The intensity of light at a point where path difference is $\lambda /3$ is then

• A. $\frac{M}{2}$
• B. $\frac{M}{4}$
• C. $\frac{M}{8}$
• D. $\frac{M}{16}$

Answer 1

Answer: (B)

$\frac{M}{4}$

Resultant intensity
$I_R=I_1+I_2+2\sqrt{I_1{I}_2}cos\theta$
If path difference = $\lambda$ phase difference $2\pi$
$\therefore I_R=I+I+2\sqrt{I\times I}cos2\pi =4I=M$
If path difference $\frac{\lambda }{3}$, phase difference $\varphi =\frac{2\pi }{3}rad$
$I_R^{\prime }=I+I+2\sqrt{I\times I}cos\frac{2\pi }{3}=I=\frac{M}{4}$