Q143 / 180 · KCET 2007 · Physics

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
    $\sin ^{-1}(.00001)$
  • Correct answer
    C
    $\sin ^{-1}(.0001)$
  • D
    $\sin ^{-1}(.) 01$

Solution

In Young's double slit experiment half angular width is given by $$ \begin{aligned} \sin \theta &=\frac{\lambda}{d}=\frac{589 \times 10^{-9}}{0.589 \times 10^{-3}}=10^{-3} \\ \Rightarrow \quad \theta &=\sin ^{-1}(0.001) \end{aligned} $$

Hence, the correct answer is option: 'C'.

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