Physics - Geometrical Optics Question with Solution | TestHub

Let $f$ be focal length of lens. $$\begin{array}{rl}& \frac{1}{f}=\left({}^a{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ & \text{ or, }\frac{1}{f}=0.5\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\text{ or, }\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{0.5f}..\end{array}$$ Let $f^{\prime }$ be focal length of the lens when immersed in a liquid. $$\begin{array}{rl}& \frac{1}{f^{\prime }}=\left({}^L{\mu }_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{1.5}{1.25}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\\ & =\left(\frac{0.25}{1.25}\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{0.25}{1.25}\times \frac{1}{0.5f}\text{ (using (i)) }\\ & \frac{1}{f^{\prime }}=\frac{1}{2.5f}\text{ or, }{f}^{\prime }=2.5f.\end{array}$$ i.e. focal length of lens when immersed in a liquid is increased by a factor of 2.5 .