Question

In a Young's double slit experiment, the slit separation 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 $1.0\text{mm}$. Which of the following statements(s) is/are correct?

• A. For $\alpha =\frac{{0.36}^{\circ }}{\pi }$, there will be destructive interference at point $O$
• B. For $\alpha =\frac{{0.36}^{\circ }}{\pi }$ , there will be destructive interference at point $P$
• C. For $\alpha =0^{\circ }$ there will be constructive interference at point $P$
• D. None of the above

Answer 1

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


$\text{Case (1)}$ For $\alpha =\frac{{0.36}^{\circ }}{\pi }$

$\alpha =\frac{0.36}{\pi }\times \frac{\pi }{180}=\frac{1}{500}\text{rad}$

The path difference between the waves at point $O$ is

$\Delta x=d\sin \alpha$

$\Rightarrow \Delta x=0.3\times {10}^{-3}\times \frac{1}{500}=6\times {10}^{-7}\text{m}$

[ for small $\theta ,\sin \theta \approx \theta$]

$\Rightarrow \Delta x=6\times {10}^{-7}=600\text{nm}=\lambda$

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

The path difference between the waves at point $P$ is

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

$\Rightarrow \Delta x=d(\frac{1}{500}+1\times {10}^{-3})$

$\Rightarrow \Delta x=\frac{d}{500}(1+\frac{1}{2})$

$\Rightarrow \Delta x=\frac{0.3\times {10}^{-3}}{500}\times 1.5$

Or, $\Delta x=1.5\lambda$

$\therefore$ For $\alpha =\frac{{0.36}^{\circ }}{\pi }$, there will be destructive interference at point $P$

$\text{Case (2)}$ For, $\alpha =0$

Path difference at $P$ is

$\Delta x=\frac{yd}{D}$

$\Rightarrow \Delta x=\frac{1\times {10}^{-3}\times 0.3\times {10}^{-3}}{1}$

$\Rightarrow \Delta x=\frac{0.3\times 2}{2}\times {10}^{-6}$

$\Rightarrow \Delta x=6\times {10}^{-7}\times \frac{1}{2}$

Or, $\Delta x=0.5\lambda$

Therefore, for $\alpha =0$ there will be destructive interference at $P$.

Hence, option $\left(\text{B}\right)$ is the only correct answer.