Explanation

(a) Given that, the refractive index of the lens wrt air,
$\ _{a}\mu_{w} = 1.60$
And the refractive index of water wrt air $\ _{a}\mu_{w} = 1.33$
The focal length of the lens in air, $f = 20cm$
We know that for a lens
$\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
When the lens is in the air
$\frac{1}{20} = \left( \ \ _{a}\mu_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$or\frac{1}{20} = (1.60 - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$or\frac{1}{20} = 0.60 \times \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ \ \ \ \ \ \ \ldots(i)$
When the lens is in the water
$\frac{1}{f^{'}} = \left( \ \ _{w}\mu_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$or\frac{1}{f^{'}} = \left( \frac{\ _{a}\mu_{g}}{\ _{a}\mu_{w}} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$or\frac{1}{f^{'}} = \left( \frac{\ _{a}\mu_{g} -_{a}\mu_{w}}{\ _{a}\mu_{w}} \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ $
$\therefore\frac{1}{f^{'}} = \left( \frac{1.60 - 1.33}{1.33} \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ $
$or\frac{1}{f^{'}} = \frac{27}{133}\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ \ \ \ \ \ldots\ldots\ldots.(ii)$
On dividing Eq. (i) by Eq. (ii), we get
$\frac{f^{'}}{20} = \frac{0.60 \times 1.33}{27}$
$or\ \ f^{'} = 20 \times 2.95\ cm \approx 60cm$