Young's double slit experiment is performed inside water `(mu = 4/3)` with light of frequency `6 xx 10^(14)` Hz. If the slits are separated by 0.2 mm and the screen kept 1 m from the slits, find the fringe width. Using the same set-up, what will the fringe width be if the experiment is performed in air?

Explanation

Here the wavelength is much less than separation of the slits. Hence, we can use the formula for fringe width.
Calculations: The wavelength of given light in air can be obtained from the relation
`c=f lambda_("air")`
`lambda_("air")=(c)/(f)=(3 xx 10^(8))/(6 xx 10^(14))=5 xx 10^(-7) m`
The wavelength of light of same frequency (`6 xx 10^(14) Hz`) in water will be different and
`lambda_("water")=(lambda_("air"))/(mu_("water"))=(5 xx 10^(-7))/(4 // 3)=3.75 xx 10^(-7) m`.
As the experiment is performed in water, the fringe width is given by
`beta=(lambda_("water")D)/(d)`
Here D=1 m, `d = 0.2 mm = 0.2 xx 10^(-3)` m, therefore
`beta=(3.75 xx 10^(-7) xx 1)/(0.2 xx 10^(-3)) m=1.87 mm.`
If the experiment is performed in air, the fringe width is given by
`beta=(lambda_("air")D)/(d)=(5 xx 10^(-7) xx 1)/(0.2 xx 10^(-3)) m=2.5 mm`. (Answer)
Learn: The fringe width in water reduces by the factor of refractive index because the wavelength reduces by the same factor.