Physics - Geometrical Optics Question with Solution | TestHub

Using the lens maker's formula for the lens in air:

$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

When the lens is immersed in a liquid of refractive index ${\mu }_0$, the relative refractive index of the lens becomes $\frac{\mu }{{\mu }_0}$. The focal length $f_0$ in the liquid is given by:

$\frac{1}{f_0}=\left(\frac{\mu }{{\mu }_0}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{f_0}=\left(\frac{\mu -{\mu }_0}{{\mu }_0}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

Dividing the first equation by the second equation:

$\frac{f_0}{f}=\frac{\mu -1}{\frac{\mu -{\mu }_0}{{\mu }_0}}$

$\frac{f_0}{f}=\frac{{\mu }_0(\mu -1)}{\mu -{\mu }_0}$

$f_0=\frac{{\mu }_0(\mu -1)f}{\mu -{\mu }_0}$

Answer: $\frac{{\mu }_0(\mu -1)f}{(\mu -{\mu }_0)}$