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

Answer 1

The Correct Option is A

$d \sin \theta=\lambda$

$\lambda=\frac{d}{2} \,\,\,\,\,\left[d=1 \times 10^{-6} m \right]$
$\Rightarrow \lambda=5000\, ?$
Fringe width, $B=\frac{\lambda D}{d'}\left(d'\right.$ is slit separation)
$10^{-2}=\frac{5000 \times 10^{-10} \times 0.5}{d'}$
$\Rightarrow d'=25 \times 10^{-6} m =25\, \mu \,m$