Question

In a YDSE apparatus, separation between the slits d = 1mm,$\lambda =600nm$ and D = 1m. Assume that each slit produce same intensity on the screen. The minimum distance between two points on the screen having $75\%$ intensity of the maximum intensity is $n\times {10}^{-4}m$. Find n___

Answer 1

$75\%$ of $I_{max}=\frac{75}{100}\times 4I_0$
$=3I_0$
$3I_0=4I_{0}co{s}^2\frac{\Delta \varphi }{2}$
$\Delta \varphi =\left(\frac{\pi }{3}\right)$
Path difference $\Delta x=\frac{\Delta \varphi }{2\pi }\lambda =\frac{\lambda }{6}$
$\frac{\Delta x}{d}=\frac{y}{D}$
$\Delta x=\frac{yd}{D}$
$\frac{\lambda }{6}=\frac{yd}{D}$
$y=\frac{\lambda D}{6d}$
Minimum separation = 2y
$=\frac{2\lambda D}{6d}=\frac{\lambda D}{3d}$
$=\frac{(600\times {10}^{-9}1)}{3\times {10}^{-3}}$
$=200\times {10}^{-6}$
$=2\times {10}^{-4}m$