A lens is made of flint glass (refractive index =1.5 ). When the...

Let $f$ be focal length of lens. \begin{aligned} & \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}{0.5 f} . . \end{aligned} Let $f^{\prime}$ be focal length of the lens when immersed in a liquid. \begin{aligned} & \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}{0.5 f} \text{ (using (i)) } \\ & \frac{1}{f^{\prime}}=\frac{1}{2.5 f} \text{ or, } f^{\prime}=2.5 f . \end{aligned} i.e. focal length of lens when immersed in a liquid is increased by a factor of 2.5 .