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In Young's double slit experiment with sodium vapour lamp of wavelength 589 nm and the slits 0.589 mm apart, the half angular width of the central maximum is
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The correct answer is:
$\sin ^{-1}(0.001)$
In Young's double slit experiment half angular width is given by
$\sin \theta=\frac{\lambda}{d}$
$=\frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}=10^{-3}$
$\Rightarrow \quad \theta=\sin ^{-1}(0.001)$
$\sin \theta=\frac{\lambda}{d}$
$=\frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}=10^{-3}$
$\Rightarrow \quad \theta=\sin ^{-1}(0.001)$
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