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Through a narrow slit of width $2 \mathrm{~mm}$, diffraction pattern is formed on a screen kept at a distance $2 \mathrm{~m}$ from the slit. The wavelength of the light used is $6330 Å$ and falls normal to the slit and screen. Then, the distance between the two minima on either side of the central maximum is
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The correct answer is:
$1.27 \mathrm{~mm}$
Width of central maxima $=\frac{2 \lambda D}{a}$
$$
=\frac{2 \times 6330 \times 10^{-10} \times 2}{2 \times 10^{-3}}=1.27 \mathrm{~mm}
$$
$$
=\frac{2 \times 6330 \times 10^{-10} \times 2}{2 \times 10^{-3}}=1.27 \mathrm{~mm}
$$
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