Explanation

(a) As we know
$\beta = \frac{D}{d}\lambda$
$\lambda \propto \frac{1}{\mu}$
From Eqs. (i) and (ii),
$\beta \propto \lambda \propto \frac{1}{\mu}$
$\beta \propto \frac{1}{\mu}$
The refractive index of water is greater than air, therefore fringe width will decrease.