Question

In Young's double slit experiment distance, between two sources is $0.1mm$. The distance of the screen from the source is $20cm$. Wavelength of light used is $5460\stackrel{0}{A}$. Then, angular position of the first dark fringe is:

• A. ${0.08}^{\circ }$
• B. ${0.16}^{\circ }$
• C. ${0.20}^{\circ }$
• D. ${0.31}^{\circ }$

Answer 1

Answer: (B)

${0.16}^{\circ }$

The position of the dark fringe in the interference is given by:

$\theta =(2n-1)\frac{\lambda }{2d}$

Angular position of first dark fringe $(n=1)$,

${\theta }_1=(2\times 1-1)\frac{\lambda }{2d}=\frac{\lambda }{2d}$

$\Rightarrow \frac{5460\times {10}^{-10}}{2\times 0.1\times {10}^{-3}}=2730\times {10}^{-6}rad$

$=2730\times {10}^{-6}\times \frac{180}{\pi }={0.16}^{\circ }$