Question

In a Young's double slit experiment, the slit separation d is $0.3$ mm and the screen distance D is $1m$. A parallel beam of light of wavelength $600nm$ 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 $11.0$mm. Which of the following statement (s) is/are correct?

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

Answer 1

The correct option is B

For $\alpha =\frac{0.36}{\pi }$ degree, there will be destructive interference at point P.

$\Delta x=d\sin \alpha +d\sin \theta$
For small angles $\theta ,\alpha$ ; $\sin \theta \simeq \tan \theta =\frac{y}{b},\sin \alpha =\alpha .$
$\Delta x=d\alpha +\frac{dy}{D}$

$\left(1\right)\alpha =0$ $\therefore \Delta x=dy/D=\frac{0.3\times 11}{1000}=33\times {10}^{-4}mm$
$\Delta x$ in terms of $\lambda \Rightarrow \frac{33\times {10}^{-4}}{600\times {10}^{-6}}\lambda =\frac{11\lambda }{2}$
as $\Delta x=(2n-1)\frac{\lambda }{2}$
There will be destructive interference.

$\left(2\right)\Delta x=0.3mm\times \frac{0.36}{\pi }\times \frac{\pi }{180}+\frac{0.3mm\times 11mm}{1000}=39\times {10}^{-4}mm$
$39\times {10}^{-4}=(2n-1)\times \frac{600\times {10}^{-9}\times {10}^3}{2}$
$n=7$
There will be destructive interference.

$\left(3\right)\Delta x=3mm\times \frac{0.36}{\pi }\times \frac{\pi }{180}+0=600nm$
$600nm=n\lambda$
$n=1$
constructive interference.

(4) Fringe width does not depend on $\alpha$.