Question

In Young's double-slit experiment, the intensity of light at a point on the screen where path difference is $\lambda$ is $I$. If intensity at a point is $I/4$, then possible path difference at this point are

• A. $\lambda /2,\lambda /3$
• B. $\lambda /3,2\lambda /3$
• C. $\lambda /3,\lambda /4$
• D. $2\lambda /3,\lambda /4$

Answer 1

The correct option is B $\lambda /3,2\lambda /3$

As Intensity at any point, $I=I_{center}co{s}^2\frac{\delta }{2}$

where, $\delta =\frac{2\pi }{\lambda }\Delta x$

Now for $\Delta x=\lambda$, $⟹\delta =2\pi$

Thus $I=I_{center}co{s}^2\pi$

Hence, $I=I_{center}$

Now, for intensity decreases by 4 times, $\frac{I}{4}=Ico{s}^2\frac{\delta }{2}$

$⟹cos\frac{\delta }{2}=+\frac{1}{2},-\frac{1}{2}$

For $\cos \frac{\delta }{2}=+\frac{1}{2}⟹\delta =\frac{2\pi }{3}$

Using, $\delta =\frac{2\pi }{\lambda }\Delta x$ gives $\Delta x=\frac{\lambda }{3}$

For $\cos \frac{\delta }{2}=+\frac{-1}{2}⟹\delta =\frac{4\pi }{3}$ gives $\Delta x=\frac{2\lambda }{3}$