Question

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

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

Answer 1

Answer: (B)

In diffraction
$dsin30^{\circ }=\lambda$
$\lambda =\frac{d}{2}$

Young’s fringe width [ $d^{{}^{\prime }}$ -the separation between two slits ]
$\beta =\frac{\lambda \times D}{d^{{}^{\prime }}}$
$10^{-2}=\frac{d}{2}\times \frac{50\times 10^{-2}}{d^{{}^{\prime }}}$
$10^{-2}=\frac{10^{-6}\times 50\times 10^{-2}}{2\times d^{{}^{\prime }}}$
$d^{{}^{\prime }}=25\mu m$