$(a)\;\frac{1}{3}\;D \\ (b)\; \frac{2}{3}\;D \\ (c)\;\frac{2}{3}\;D \\ (d)\;\frac{5}{3}\;D $

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Let $f_a$ and $f_w$ be the focal length of lens in air water respectively , then

$P_a=\large\frac{1}{f_a} $ and $ P_w= \large\frac{\mu_w}{f_w}$

$f_a=0.2 \;m=20 \;cm$

Using lens maker's formula :

$P_a= \large\frac{1}{f_a} $

$\quad= ( \mu_g-1) \bigg[ \large\frac{1}{R_1} -\frac{1}{R_2}\bigg]$

$\large\frac{1}{f_w} =\bigg(\large\frac{\mu_g}{\mu_w} -1 \bigg) \bigg[ \large\frac{1}{R_1} -\frac{1}{R_2} \bigg]$

=> $ P_w =\large\frac{\mu _w}{f_w} $$=(\mu_g -\mu_w)\bigg[ \large\frac{1}{R_1} -\frac{1}{R_2}\bigg]$

=>$\large\frac{P_w}{P_a} =\large\frac{(\mu_g -\mu_w)}{(\mu_g-1)}$

or $P_w=\large\frac{P_a}{3} =\frac{5}{3} $$D$

Hence D is the correct answer.

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