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 $25\, cm$. 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} $
$\alpha=0.45^{\circ} .$