Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be :

Young's double slit experiment in a liquid of refractive index $$\mu = 1.44$$.

Slit separation $$d = 1.5$$ mm, wavelength in air $$\lambda_0 = 690$$ nm, screen distance $$D = 0.72$$ m.

Wavelength in liquid: $$\lambda = \frac{\lambda_0}{\mu} = \frac{690}{1.44} = 479.17$$ nm.

Fringe width: $$\beta = \frac{\lambda D}{d} = \frac{479.17 \times 10^{-9} \times 0.72}{1.5 \times 10^{-3}}$$

$$= \frac{345 \times 10^{-9}}{1.5 \times 10^{-3}} = 230 \times 10^{-6} = 0.23 \times 10^{-3}$$ m $$= 0.23$$ mm.

The correct answer is Option A: 0.23 mm.