Physics - Geometrical Optics Question with Solution | TestHub

Lens-maker's formula is given by

$\frac{1}{f}=\left({ }_a{\mathrm{\mu }}_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \dots \left(\mathrm{i}\right)$

If the lens is immersed in a liquid of refractive index${\mathrm{\mu }}_1$then

$\frac{1}{f_1}=\left({ }_l{\mathrm{\mu }}_{\mathrm{g}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \dots \left(\mathrm{i}\mathrm{i}\right)$

Here,${ }_l{\mathrm{\mu }}_{\mathrm{g}}$is refractive index of glass w.r.t liquid

Dividing Eq. (i) by Eq. (ii), we have

$\frac{f_1}{f}= \frac{\left({ }_a{\mathrm{\mu }}_{\mathrm{g}}-1\right)}{\left({ }_l{\mathrm{\mu }}_{\mathrm{g}}-1\right)}$

$\Rightarrow \frac{f_1}{f}=\left(\frac{1.5-1}{\frac{1.5}{1.25}-1}\right)$

$\Rightarrow \frac{f_1}{f}=\frac{0.5\times 1.25}{0.25}=2.5\Rightarrow f_1=\left(2.5\right)f$

Hence, focal length increases by a factor of 2.5.