A lens has a power of +5 diaptre in air. Calculate its power if it is completely immersed in water? Given ${\mu }_g=\frac{3}{2}and{\mu }_w=\frac{4}{3}$ .

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

$P_a=\frac{1}{f_a} \text{ and } P_w=\frac{{\mu }_w}{f_w}$

Since lens has power $+5\mathrm{D}$ in air, so

$f_a=\frac{1}{P}=\frac{1}{5}=0.2 \mathrm{m}=20 \mathrm{cm}$

Using Lens Maker's formula, we get

$$\begin{gathered} \frac{1}{f_a}=\left({\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \text{ Similarly, }\frac{1}{f_w}=\left(\frac{{\mu }_g}{{\mu }_w}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ \Rightarrow P_w=\frac{{\mu }_w}{f_w}=\left({\mu }_g-{\mu }_w\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \end{gathered}$$

Dividing equation (2) by equation (1), we get,

$$\begin{gathered} \frac{P_w}{P_a}=\frac{\left({\mu }_g-{\mu }_w\right)}{\left({\mu }_g-1\right)}=\frac{1}{3} \\ \Rightarrow P_w=\frac{P_a}{3}=+\frac{5}{3}\mathrm{D} \end{gathered}$$