Question 145 of 191

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 $50 cm$ 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$100\mu m$B$25\mu m$C$50\mu m$D$75\mu m$

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Answer: (B) $25\mu m$

In diffraction, $d\sin 30^{\circ}=\lambda$, $\lambda =\dfrac{d}{2}$



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

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