Explanation

(d) Given, $f_{a}ₐ = 0.15\ m,\ \mu_{g} = \frac{3}{2}and\ \mu_{w}\ ᷿ = \frac{4}{3}$
According to Lens maker’s formula
$\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ where\ \mu = \frac{\mu_{L}}{\mu_{M}}$
$\frac{1}{f_{a}ₐ} = \left( \frac{\mu_{g}}{\mu_{a}ₐ} - 1 \right)\ \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ $
$= \left( \frac{(3/2)}{1} - 1 \right)C\ where\ C = \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ $
$Or\ \frac{1}{f_{a}ₐ} = \frac{C}{2}\ \ \ \ \ \ldots(i)$
$Also,\ \ \ \frac{1}{f_{w}\ ᷿} = \left( \frac{\mu_{g}}{\mu_{w}} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ $
$= \left( \frac{(3/2)}{(4/3)} - 1 \right)C$
$Or\ \ \ \frac{1}{f_{w}\ ᷿} = \frac{C}{8}\ \ \ \ \ \ \ \ \ \ \ (ii)$
From Eqs. (i) and (ii), we get
$\frac{f_{w}\ ᷿}{f_{a}ₐ} = \frac{C}{2} \times \frac{8}{C} = 4\ $
$Or\ \ \ \ \ f᷿_{w} = 4f_{a}ₐ = 4 \times 0.15 = 0.6\ m$