Question

In a Young's double slit experiment, the slit separation $d$ is $0.3\text{mm}$ and the screen distance $D$ is $1\text{m}$. A parallel beam of light of wavelength $600\text{nm}$ is incident on the slits at angle $\alpha$ as shown in figure. On the screen, the point $O$ is equidistant from the slits and distance $PO$ is $11.0\text{mm}$. Which of the following statement(s) is /are correct?

• A. For $\alpha =0^{\circ }$, there will be constructive interference at point $P$.
• B. For $\alpha =\frac{{0.36}^{\circ }}{\pi }$, there will be constructive interference at point $P$.
• C. For $\alpha =\frac{{0.36}^{\circ }}{\pi }$, there will be destructive interference at point $O$.
• D. Fringe spacing depends on $\alpha$.

Answer 1

The correct option is C For $\alpha =\frac{{0.36}^{\circ }}{\pi }$, there will be destructive interference at point $O$.

In order to check the conditions of Constructive Interference and Destructive Interference, we need to find the path difference.


Clearly, path difference at $P$

$\Delta x=d\sin \alpha +d\sin \theta$

If $\theta$ and $\alpha$ are very small

$\sin \theta \approx \tan \theta =\frac{y}{D}$ and $\sin \alpha \approx \alpha$

$\therefore \Delta x=d\alpha +\frac{dy}{D}$

$\text{Case (1)}$ For $\alpha =0^{\circ }$

Path difference at $P$

$\therefore \Delta x=\frac{dy}{D}=\frac{0.3\times 11}{1000}=33\times {10}^{-4}\text{mm}$

Also, in terms of $\lambda$,

$\Delta x=\frac{33\times {10}^{-4}}{600\times {10}^{-6}}\lambda =\frac{11}{2}\lambda$

as $\Delta x=(2n-1)\frac{\lambda }{2}$

Therefore, there will be destructive interference at $P$ for $\alpha =0^{\circ }$

$\text{Case (2)}$ For $\alpha =\frac{{0.36}^{\circ }}{\pi }$

Path difference at $P$

$\Delta x=0.3\times \frac{0.36}{\pi }\times \frac{\pi }{180}+\frac{0.3\times 11}{1000}$

$\therefore \Delta x=39\times {10}^{-4}\text{mm}$

Also, in terms of $\lambda$,

$\Delta x=\frac{39\times {10}^{-4}}{600\times {10}^{-6}}\lambda =\frac{13}{2}\lambda$

as $\Delta x=(2n-1)\frac{\lambda }{2}$

Therefore, there will be destructive interference at $P$ for $\alpha =\frac{{0.36}^{\circ }}{\pi }$

$\text{Case (3)}$ For $\alpha =\frac{{0.36}^{\circ }}{\pi }$

Path difference at $O$

$\Delta x=3\times \frac{0.36}{\pi }\times \frac{\pi }{180}+0=6\times {10}^{-3}\text{mm}$

Also, in terms of $\lambda$,

$\Delta x=\frac{6\times {10}^{-3}}{600\times {10}^{-6}}\lambda =10\lambda$

as $\Delta x=n\lambda$

Therefore, there will be Constructive interference at $O$ for $\alpha =\frac{{0.36}^{\circ }}{\pi }$

Also, fringe width does not depend on $\alpha$.

Hence, option $\left(\text{C}\right)$ is correct.