Physics - Geometrical Optics Question with Solution | TestHub

From Lens Maker's formula,

For air: $\frac{1}{f_a}=\left(\frac{{\mu }_g}{{\mu }_a}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

For water: $\frac{1}{f_w}=\left(\frac{{\mu }_g}{{\mu }_w}-1\right) \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\therefore \frac{\frac{1}{f_a}}{\frac{1}{f_w}}=\frac{\left(\frac{{\mu }_g}{{\mu }_a}-1\right)}{\left(\frac{{\mu }_g}{{\mu }_w}-1\right)}$

$\Rightarrow \frac{f_w}{f_a}=\frac{\left(1.5-1\right)}{\left({\displaystyle \frac{1.5\times 3}{4}}-1\right)}\Rightarrow f_w=\frac{0.5}{0.125}\times 18=72 \mathrm{cm}$

Thus, change in focal length of the lens is $72-18=54 \mathrm{cm}$