Question

In Young's double slit interference experiment, the distance between two sources is $0.1/\pi mm$. The distance of the screen from the source is $25cm$. Wavelength of light used is $5000\mathring{A}$. Then the angular position of the first dark fringe is

• A. $0.10^{\circ }$
• B. $0.15^{\circ }$
• C. $0.30^{\circ }$
• D. $0.45^{\circ }$

Answer 1

Answer: (D)

The angular position
$\theta =\frac{\beta }{D}=\frac{\lambda }{d}\left(\because \beta =\frac{\lambda D}{d}\right)$
The first dark fringe will be at half the fringe width from the mid point of central maximum. Thus the angular position of first dark fringe will be-
$\alpha =\frac{\theta }{2}=\frac{1}{2}\left[\frac{\lambda }{d}\right]$
$\alpha =\frac{1}{2}\left[\frac{5000\times \pi }{1\times 10^{-3}}\times 10^{-10}\right]\frac{180}{\pi }$
$\alpha =0.45^{\circ }.$