Question

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

• A. $si{n}^{-1}$ 0.01
• B. $si{n}^{-1}$ 0.0001
• C. $si{n}^{-1}$ 0.001
• D. $si{n}^{-1}$ 0.1

Answer 1

The correct option is C $si{n}^{-1}$ 0.001

In young's double slit experiment half angular width is given by$,$

$\sin \theta =\frac{\lambda }{d}$

$=\frac{589\times {10}^9}{0.589\times {10}^{-3}}={10}^{-3}$

$\theta ={\sin }^{-1}\left(0.001\right)$

Hence,

option $\left(c\right)$ is correct answer.