Physics Question on Ray optics and optical instruments

Question

A thin lens made of glass of refractive index $\mu = 1.5$ has a focal length equal to 12 cm in air. lt is now immersed in water $\bigg(\mu = \frac{4}{3}\bigg).$ Its new focal length is

Answer 1

Solution

Focal length in air is given by,
$\frac{1}{f_a}=(_a\mu_g-1)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)$
The focal length of lens immersed in water is given by
$\frac{1}{f_l}=(_l \mu_g-1)\bigg(\frac{1}{R_1}-\frac{1}{R_2}\bigg)$
where $R_1, R_2$ are radii of curvatures of the two surfaces oflens
and $_1 \mu_g$ is refractive index of glass with respect to liquid.

Also $_1 \mu_g = \frac{_a \mu_g}{_a \mu_l }$

Given, $_a\, n _g = 1.5, f_a = 12\, cm, _a n_l = \frac{4}{3}$
$\therefore \frac{f_1}{f_a}=\frac{(_a\mu_g-1)}{(_l\mu_g-1)}$
$\frac{f_1}{12}=\frac{(1.5-1)}{\bigg(\frac{1.5}{4/3}-1\bigg)}=\frac{0.5 \times 4 }{0.5}$
$\Rightarrow f_1 = 4 \times 12$
$= 48\, cm$