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

• A

  $100\mathrm{ }\mathrm{\mu m}$
• Correct answer
  B

  $25\mathrm{ }\mathrm{\mu m}$
• C

  $50\mathrm{ }\mathrm{\mu m}$
• D

  $75\mathrm{ }\mathrm{\mu m}$

Solution

In diffraction, $d\sin 30°=\mathrm{\lambda }$, $\mathrm{\lambda }=\frac{d}{2}$



Young’s fringe width [ $d^{\text{'}}$- the separation between two slits] 

$\mathrm{\beta }=\frac{\mathrm{\lambda }\times D}{d^{\mathrm{\text{'}}}}$$\Rightarrow {10}^{-2}=\frac{d}{2}\times \frac{50\times {10}^{-2}}{d^{\mathrm{\text{'}}}}$
$\Rightarrow {10}^{-2}=\frac{{10}^{-6}\times 50\times {10}^{-2}}{2\times d^{\mathrm{\text{'}}}}$$\Rightarrow d^{\mathit{\text{'}}}=25 \mathrm{\mu m}$.

Hence, the correct answer is option: 'B'.