A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m away. The wavelength of light used is 'x' nm. The value of 'x' to the nearest integer is _ _ _ _ .
Solution
Fringe width = $6 mm = \frac {\lambda D}{d} = \frac {\lambda \times 10}{1\times 10^{-3}}$
$\Rightarrow \lambda = \frac {6 \times 10^{-3} \times 10^ {-3}}{10} = 6 \times 10^{-7} = 600 nm $
$\therefore x = 600$