Question

The angular width of the central maximum in a single slit diffraction pattern is $60^\circ $ . The width of the slit is $1 \, μm$ . 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 $50 \, cm$ from the slits. If the observed fringe width is $1 \, cm$ , what is slit separation distance? (i.e. the distance between the centres of each slit.)

• A. $100 \, μm$
• B. $25 \, μm$
• C. $50 \, μm$
• D. $75 \, μm$

Answer 1

Answer: (B)

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

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