Explanation

(d) If a lens of focal length $f$ is divided into two equal parts as shown in figure (i) and each has a focal length $f'$ then
$\frac{1}{f} = \frac{1}{f'} + \frac{1}{f'}\ ie,\ f' = 2f$
$ie$, each part will have focal length 2$f$
Now if these parts are put in contact as in figure (2), then resultant focal length of the combination will be
$\frac{1}{F} = \frac{1}{2f} + \frac{1}{2f}ie,\ F = f\ (initial\ value)$
For this combination,
$\frac{1}{F} = \left( \ _{a}\mu_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ \ \ \ \ \ \ \ \ \ldots\ldots\ldots\ldots..(i)$
Now, if this combination is immersed in liquid, then
$\frac{1}{F'} = \left( \ _{1}\mu_{g} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)\ \ \ \ \ \ \ldots..(ii)$
$\frac{F'}{f} = \frac{(\ _{a}\mu_{g} - 1)}{(\ _{1}\mu_{g} - 1)} = \frac{(1.5) - 1)\ }{\left( \frac{\frac{3}{2}}{\frac{4}{3}} - 1 \right)}$
$or\ \frac{F'}{f} = \frac{0.5}{\left( \frac{9}{8} - 1 \right)} = 0.5 \times 8$
∴ F’$= 0.5 \times 8 \times 10 = 40cm$