KCET · Physics · Wave Optics
In Young's double slit experiment with sodium vapour lamp of wavelength \(589 \mathrm{~nm}\) and the slits \(0.589 \mathrm{~mm}\) apart, the half angular width of the central maximum is
- A \(\sin ^{-1}(.001)\)
- B \(\sin ^{-1}(.00001)\)
- C \(\sin ^{-1}(.0001)\)
- D \(\sin ^{-1}(.) 01\)
Answer & Solution
Correct Answer
(C) \(\sin ^{-1}(.0001)\)
Step-by-step Solution
Detailed explanation
In Young's double slit experiment half angular width is given by
\[
\begin{aligned}
\sin \theta &=\frac{\lambda}{d}=\frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}=10^{-3} \\
\Rightarrow \quad \theta &=\sin ^{-1}(0.001)
\end{aligned}
\]
\[
\begin{aligned}
\sin \theta &=\frac{\lambda}{d}=\frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}=10^{-3} \\
\Rightarrow \quad \theta &=\sin ^{-1}(0.001)
\end{aligned}
\]
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