Question

A double-slit apparatus is immersed in a transparent liquid of refractive index 1.33. The slit separation is 1mm and the distance between the plane of slits and screen is 1.33m. The slits are illuminated by a parallel beam of light whose wavelength in air is $6300\stackrel{\circ }{A}$. Find the fringe width.

• A. 0.47 m
• B. 0.63 m
• C. 0.84 m
• D. 1.11 m

Answer 1

The correct option is B

0.63 m

The wavelength in liquid is

${\lambda }_1=\frac{{\lambda }_a{\mu }_a}{{\mu }_1}=\frac{6300\stackrel{\circ }{A}\times 1}{1.33}=\frac{6.3\times {10}^{-7}}{1.33}$

Therefore, fringe width in liquid is

${\beta }_1=\frac{{\lambda }_{1}D}{d}=\frac{6.3\times {10}^{-7}\times 1.33}{1.33\times 1\times {10}^{-3}}=6.3\times {10}^{-4}m=0.63m$