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 (`lamda=700` nm in vacuum). Find the fringe-width of the pattern formed on the screen.

Explanation

Given that: `d=0.28mm`
`=0.28xx10^-3m`
`D=48cm`
`lamda_a=700nm in vacuum`
Let `lamda_w` =wavelength of red light in water
Since,` m=v_a/v_w=lamda_a/lamda_2=4/3`
`rarr (3lamda_a)/4=(3xx700)/4=525mm`
so the fringe width of the pattern is given by
`B=lamda_wD`
`=525xx10^-9xx(0.48))/((0.28)xx10^-3)`
`=9x10^-4=0.90mm`