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 $50cm$ from the slits. If the observed fringe width is $1cm$, 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

Answer: (A)

$d\sin \theta =\lambda$

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