# A lens has a power of +5 diopters in air. What will be power if completely immersed in water? $_a \mu_w =\large\frac{4}{3}$ and $_a \mu_g =\large\frac{3}{2}$

$(a)\;1.25\;diopter \\ (b)\;2 diopter \\ (c)\;2.25\;diopter \\ (d)\;1\;diopter$

Let $f_a$ and $f_m$ be focal length of lens in air and water respectively , then
$P_a= \large\frac{1}{f_a}$
or $+5 =\large\frac{1}{f_a}$
$f_a =0.2 \quad m=20\;cm$
Now $\large\frac{1}{f_a}=(_a \mu _g-1) \bigg[ \large\frac{1}{R_1}-\frac{1}{R_2} \bigg]$
and $\large\frac{1}{f_w}=(_w \mu _g-1) \bigg[ \large\frac{1}{R_1}-\frac{1}{R_2} \bigg]$
Divide equations we get,
$\large\frac{f_w}{f_a} =\bigg[ \large\frac{a \mu _g-1}{w \mu _g -1}\bigg]$
Again $_w \mu _g=\large\frac{_a \mu _g}{_a \mu _w}$
$\qquad= \large\frac{\Large\frac{3}{2}}{\Large\frac{4}{3}}$
$\qquad =\large\frac{9}{8}$
$\therefore \large\frac{f_w}{f_a} = \large\frac{\Large\frac{3}{2}-1}{\Large\frac{9}{8} -1}$