A convex lens of refractive index 1.5 and focal length $f=18 \mathrm{~cm}$ is immersed in water. The

Question

A convex lens of refractive index 1.5 and focal length $f=18 \mathrm{~cm}$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha 	imes f$. The value of $\alpha$ is $\_\_\_\_$ .
(refractive index of water $=4 / 3$ )

Professor PC Thomas Classes · Answer

Using the Lens Maker's formula, the focal length in air is $\frac{1}{f} = (\mu_g - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)C = 0.5C$. When immersed in water, the focal length $f_w$ is $\frac{1}{f_w} = (\frac{\mu_g}{\mu_w} - 1)C = (\frac{1.5}{4/3} - 1)C = (\frac{4.5}{4} - 1)C = \frac{0.5}{4}C$. Taking the ratio $\frac{f_w}{f} = \frac{0.5C}{0.5C / 4} = 4$, we find $f_w = 4f$. The difference in their focal lengths is $f_w - f = 4f - f = 3f$. Thus, $\alpha = 3$.