Question

The angular width of the central maximum in a single slit diffraction pattern is ${60}^o$. The width of the slit is $1$ $\mu$m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance $50$cm from the slits. If the observed fringe width is $1$cm, what is slit separation distance? (i.e., distance between the centres of each slit.)

• A. $75\mu$m
• B. $100\mu$m
• C. $25\mu$m
• D. $50\mu$m

Answer 1

Answer: (C)

$25\mu$m

Angular width of the central maxima ${\theta }_c={60}^o$
Half angle $\theta =\frac{{60}^o}{2}={30}^o$
Given : $b={10}^{-6}m$
Using $\sin \theta =\frac{\lambda }{b}$
Or $\sin {30}^o=\frac{\lambda }{{10}^{-6}}$
$⟹\lambda =0.5\times {10}^{-6}m$
Given : $D=50cm=0.5m$
Fringe width $\beta =1cm=0.01m$
Using $\beta =\frac{\lambda D}{d}$
$\therefore$ $0.01=\frac{0.5\times {10}^{-6}\times 0.5}{0.01}=25\mu m$