Explanation

(a) Here, $\mu_{g} = 1.5 = \frac{3}{2},\ \mu_{w}\ ᷿ = \frac{4}{3}$
$R_{1} = R, \Rightarrow R_{2} = - R$
From, lens maker’s formula
$\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)where\ \mu = \frac{\mu_{L}}{\mu_{M}}$
$\therefore\frac{1}{f_{a}ₐ} = \left( \frac{\mu_{g}}{\mu_{a}ₐ} - 1 \right)\left( \frac{1}{R} - \frac{1}{( - R)} \right)$
$Or\ \frac{1}{f_{a}ₐ} = \left( \frac{3/2}{1} - 1 \right)\frac{2}{R}$
$\Rightarrow \frac{1}{f_{a}} = \frac{1}{R}$
Also
$\frac{1}{f_{w}\ ᷿} = \left( \frac{\mu_{g}}{\mu_{w}ₐ} - 1 \right)\left( \frac{1}{R} - \frac{1}{( - R)} \right)$
$\ \frac{1}{f_{2}} = \left( \frac{3/2}{4/3} - 1 \right)\frac{2}{R}$
$Or\ \ \frac{1}{f_{w}} = \frac{1}{4R}$
Or $f_{w}\ ᷿ = 4R$