Question

Light of wavelength $6000\stackrel{\circ }{A}$ is incident normally on a slit of width $24\times {10}^{-5}\text{cm}$. What is the angular position of second minimum from central maximum?

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$


The position of dark fringes in single slit experiment is given by the formula
$w\sin \theta =m\lambda$ where $m$ is the no. of dark fringe.
For second minima/dark fringe, $m=2$
Then, $\sin {\theta }_2=\frac{2\lambda }{w}=\frac{2\times 6000\times {10}^{-10}}{24\times {10}^{-7}}=\frac{1}{2}$
$\therefore {\theta }_2={30}^{\circ }$ is the angular position of second minimum.