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Light of wavelength $5000 Å$ is incident normally on a slit. The first minimum of the diffraction pattern is observed to lie at distance of $5 \mathrm{~mm}$ from the central maximum on a screen placed at a distance of $2 \mathrm{~m}$ from the slit. The width of the slit.is
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The correct answer is:
$0.02 \mathrm{~cm}$
The slit width is given as:
$\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
$\therefore \quad \mathrm{d}=\frac{2 \times 5000 \times 10^{-10}}{5 \times 10^{-3}}=0.02 \mathrm{~cm}$
$\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$
$\therefore \quad \mathrm{d}=\frac{2 \times 5000 \times 10^{-10}}{5 \times 10^{-3}}=0.02 \mathrm{~cm}$
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