The angular width of the central maximum in a single slit diffrac… - Vidyadip

MCQ

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

A
$50$
B
$75 $
C
$100$
✓
$25$

✓

Answer

Correct option: D.

$25$

d
Angular width of central maxima $=\frac{2 \lambda}{\mathrm{d}}$

or, $\lambda=\frac{\mathrm{d}}{2} ;$ Fringe width, $\beta=\frac{\lambda \times \mathrm{D}}{\mathrm{d}^{\prime}}$

$10^{-2}=\frac{d}{2} \times \frac{50 \times 10^{-2}}{d^{\prime}}=\frac{10^{-6} \times 50 \times 10^{-2}}{2 \times d^{\prime}}$

Therefore, slit separation distance, $\mathrm{d}^{\prime}=25 \,\mu \mathrm{m}$

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