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In Young's double slit experiment with sodium vapour lamp of wavelength $589 \mathrm{~nm}$ and slit $0.589 \mathrm{~mm}$ apart, the half angular width of the central maxima is
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The correct answer is:
$\sin ^{-1}(0.001)$
Given, $\lambda=589 \mathrm{~nm}=589 \times 10^{-9} \mathrm{~m}$
$$
d=0589 \mathrm{~mm}=0.589 \times 10^{-3} \mathrm{~m}
$$
In Young's double slit experiment, half angular width of central maxima is
$$
\sin \theta=\frac{\lambda}{d}=\frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}=10^{-3} \Rightarrow \theta=\sin ^{-1}(0.001)
$$
$$
d=0589 \mathrm{~mm}=0.589 \times 10^{-3} \mathrm{~m}
$$
In Young's double slit experiment, half angular width of central maxima is
$$
\sin \theta=\frac{\lambda}{d}=\frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}=10^{-3} \Rightarrow \theta=\sin ^{-1}(0.001)
$$
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