Question

In Young’s double-slit experiment using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where path difference is $\lambda$, is K units. What is the intensity of light at a point where path difference is $\frac{\lambda }{3}$?

Answer 1

Given,
When path difference $\Delta x=\lambda$,
$I=K$
Phase -difference:
$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$
$\Delta \varphi =\frac{2\pi }{\lambda }\times \lambda =2\pi$
Resultant intensity when both sources are monochromatic $(I_1=I_2=I_0):$
$I=4I_0{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$
$K=4I_0{\cos }^2\left(\frac{2\pi }{2}\right)$
$I_0=\frac{K}{4}$

When path difference $\Delta x=\frac{\lambda }{3}$,
Phase -difference:
$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$
$\Delta \varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}$
Resultant intensity:
$I=4I_0{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$
$I=4\left(\frac{K}{4}\right){\cos }^2\left(\frac{\pi }{3}\right)=\frac{K}{4}$
$I=\frac{K}{4}$
Final Answer : $\frac{K}{4}$.