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(a) Derive an expression for path difference in Young's double slit experiment and obtain the conditions for constructive and destructive interference at a point on the screen.
(b) The intensity at the central maxima in Young's double slit experiment is $I_{o}$ . Find out the intensity at a point where the path difference is $\frac{\lambda }{6},\frac{\lambda }{4}$ and $\frac{\lambda }{3}$ .

Answers (1)

$d<<D$

For constructive interference

$S_{2}P-S_{1}P=n\lambda$

$\left [ D^{2} +\left ( x+\frac{d}{2} \right )^{2}\right ]-\left [ D^{2} +\left ( x-\frac{d}{2} \right )^{2}\right ]$

$=S_{2}P^{2}-S_{1}P^{2}$

$\Rightarrow S_{2}P^{2}-S_{1}P^{2}=xd$

$(S_{2}P-S_{1}P)(S_{2}P+S_{1}P)=xd$

$S_{2}P-S_{1}P=\frac{xd}{D}\; \; \; \; \; \; \; \; (d<<D)$

$\therefore \frac{xd}{D}=n\lambda$

or $x_{n}=\frac{n\lambda D}{d}\; \; \; \; (n=o,\pm 1,\pm 2)$

for destructive interference

$\frac{xd}{D}=\left ( n+\frac{1}{2} \right )\lambda$ $(n=o,\pm 1,\pm 2)$

(b) i)

Path difference $=\frac{\lambda }{6},\phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}$

$=\frac{\pi }{3}=60^{o}$

$I=I_{o}\cos ^{2}\frac{\phi }{2}$

$I=\frac{3I_{o}}{4}$

Similarly

ii) for path difference $=\frac{\lambda }{4}$

$\phi =90$

$\therefore I=\frac{I_{o}}{2}$

iii) for path difference $=\lambda /3,\phi =120^{o}$

$\therefore I=\frac{I_{o}}{4}$

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