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In a Frounhofer diffraction experiment, a single slit of width $0.5 \mathrm{~mm}$ is illuminated by a monochromatic light of wavelength $600 \mathrm{~nm}$. The diffraction pattern is observed on a screen at a distance of $50 \mathrm{~cm}$ from the slit. What will be the linear separation of the first order minima?
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Verified Answer
The correct answer is:
$1.2 \mathrm{~mm}$
Hint:
$\mathrm{d}=0.5 \mathrm{~mm} .$
$\lambda=600 \mathrm{~nm} .$ Width of central maxima $=\frac{2 \lambda \mathrm{D}}{\mathrm{d}}$
$\mathrm{D}=50 \mathrm{~cm}$
$\frac{2 \lambda \mathrm{D}}{\mathrm{d}}=\frac{2 \times 600 \times 50 \times 10^{-2} \times 10^{-9}}{0.5 \times 10^{-3}}=1.2 \mathrm{~mm}$
$\mathrm{d}=0.5 \mathrm{~mm} .$
$\lambda=600 \mathrm{~nm} .$ Width of central maxima $=\frac{2 \lambda \mathrm{D}}{\mathrm{d}}$
$\mathrm{D}=50 \mathrm{~cm}$
$\frac{2 \lambda \mathrm{D}}{\mathrm{d}}=\frac{2 \times 600 \times 50 \times 10^{-2} \times 10^{-9}}{0.5 \times 10^{-3}}=1.2 \mathrm{~mm}$
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