Question

A Young's double slit apparatus has slits separated by 0.28 mm and a screen 48 cm away from the slits. The whole apparatus is immersed in water and the slits are illuminated by the red light ($\lambda =700$ nm in vacuum). Find the fringe-width of the pattern formed on the screen.

Answer 1

Given that, $d=0.28mm=0.28\times {10}^{-3}m,D=48cm=0.48m,{\lambda }_a=700nm$ in vacum
Let ${\lambda }_w=$ wavelength of red light in water
Since, the fringe width of the pattern is given by.
$\beta =\frac{{\lambda }_{w}D}{d}=\frac{525\times {10}^{-9}\times 0.48}{0.28\times {10}^{-3}}=9\times {10}^{-4}m=0.90mm$