Question

A double slit is illuminated by light of wave length 6000 $A^o$. The slits are 0.1 cm apart and the screen is placed 1 m away. Then the angular position of ${10}^{th}$ maxima is

Answer 1

$\begin{array}{c}\beta =\frac{\lambda D}{d}\\ =\frac{6000\times {10}^{-10}\times 1\times 100}{0.1}\\ \beta =6\times {10}^{-4}\\ \therefore 10\beta =6\times {10}^{-4}\times 10=6mm\\ \therefore 10\theta =\frac{10\beta }{D}=6^0\end{array}$

$\therefore$ Angular position $=6^0$.