In a Young's double slit experiment, the slit separation d is 0.3 m m and the screen distance D is 1 m . A…

1. For $\mathrm{\alpha }=\frac{0.36}{\mathrm{\pi }}$ degree, there will be destructive interference at point $\mathrm{O}.$
2. Fringe spacing depends on $\mathrm{\alpha }$
3. For $\mathrm{\alpha }=\frac{0.36}{\mathrm{\pi }}$ degree, there will be destructive interference at point $\mathrm{P}$
4. For $\mathrm{\alpha }=0,$ there will be constructive interference at point $\mathrm{P}.$

Solution

Path difference at $“P”$ , $∆x=d\sin \alpha +d\sin \theta$
$\begin{array}{cc}\theta , \alpha : \mathrm{s}\mathrm{m}\mathrm{a}\mathrm{l}\mathrm{l}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e},& \sin \theta \cong \tan \theta =\frac{y}{D}\end{array}$ and $\sin \alpha \approx \alpha$
$\Rightarrow ∆x=d\alpha +\frac{dy}{D}$
$\left(\mathrm{A}\right) ∆x=3 \mathrm{m}\mathrm{m}\times \frac{0.36}{\pi }\times \frac{\pi }{180}+0=600 \mathrm{n}\mathrm{m}$
$600 \mathrm{n}\mathrm{m}=n\lambda$
$n=1$
Constructive interference $\Rightarrow \left(\mathrm{A}\right)$ is incorrect
$\left(\mathrm{B}\right)$ Fringe width does not depend on $\alpha .\Rightarrow \left(\mathrm{B}\right)$ is incorrect
$\left(\mathrm{C}\right) ∆x=0.3 \mathrm{m}\mathrm{m}\times \frac{0.36}{\pi }\times \frac{\pi }{180}+\frac{0.3 \mathrm{m}\mathrm{m}\times 11\mathrm{ }\mathrm{m}\mathrm{m}}{1000}=39\times {10}^{-4}\mathrm{m}\mathrm{m}$
Now, $∆x=39\times {10}^{-4}=\left(2n-1\right)\times \frac{600\times {10}^{-9}\times {10}^3}{2}$
$\Rightarrow n=7$
Destructive interference happened $\Rightarrow$ correct
$\left(\mathrm{D}\right) \begin{array}{cc}\alpha =0& \therefore ∆x=dy/D=\frac{0.3\times 11}{1000}=33\times {10}^{-4}\mathrm{m}\mathrm{m}\end{array}$
$∆x$ in terms of $\lambda =\frac{33\times {10}^{-4}}{600\times {10}^{-6}}\lambda =\frac{11\lambda }{2}$
as $∆x=\left(2n-1\right)\frac{\lambda }{2}$
There will be destructive interference. $\Rightarrow \left(\mathrm{D}\right)$ is incorrect.

Asked in: JEE Advanced 2019 (Paper 2)

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