A Young’s double slit experimental set up is immersed in water of refractive index 1.33. It has slit separation 1mm and the distance between slits and screen is 1.33m. If the incident light has a wavelength of $6300\stackrel{\text{o}}{\text{A}}$ in vaccum, then the fringe width on the screen is

When Young’s double slit experiment set up is immersed in water then fringe width, $y=\frac{\lambda \text{D}}{\mu \text{d}}$

Given, wavelength of incident light on slits,

$\lambda =6300\stackrel{\text{o}}{\text{A}}=6300\times {10}^{-10}\text{m}$

distance between slit and screen, D = 1.33m, and refractive index of water, $\mu =1.33$

Hence, the fringe width of water,

$y=\frac{6300\times {10}^{-10}\times 1.33}{1.33\times 1\times {10}^{-3}}$

$y=6.3\times {10}^{-4}\text{m}=0.63\text{mm}$