In a Young's double slit experiment, the fringe width is found to be `0.4mm`. If the whole apparatus is immersed in water of refractive index `4//3` without disturbing the geometrical arrangement, the new fringe width will be

Options

A. 0.30 mm
B. 0.40 mm
C. 0.53 mm
D. 450 `mu`m

Answer

A

Explanation

In Young's double slit experiment if distance between coherent source is d and distance between screen and source is D, then fringe width `beta` is given by
`beta=(Dlambda)/(d) " " `…(i)
where, `lambda` is wavelength of monochromatic light source.
Aslo, from Huygens' principle, `mu=(lambda)/(lambda_(w))=` refractive index
In water, `" " beta'=(Dlambda_(w))/(d) " " `...(ii)
Dividing Eq. (i) by Eq. (ii), we get
`(beta)/(beta')=(lambda)/(lambda_(w))rArr beta'=(lambda_(w))/(lambda)beta=(beta)/(mu)`
Given `beta=0.4 mm, mu=(4)/(3)`, we have, `beta'=(0.4)/((4//3))=0.3mm`