A thin convex lens of refractive index 1.5 has focal length $f_1$. When immersed in a liquid of refractive index $\frac{5}{3}$ length is found to be $f_2$. The relationship between $f_1$ and $f_2$ is:

$ f_2=-5f_1$

The correct answer is Option (1) → $ f_2=-5f_1$

Lens maker’s formula: $\frac{1}{f} = \left(\frac{n_\text{lens}}{n_\text{medium}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

In air ($n_\text{medium} = 1$):

$\frac{1}{f_1} = (1.5 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0.5 \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

In liquid ($n_\text{medium} = \frac{5}{3}$):

$\frac{1}{f_2} = \left(\frac{1.5}{5/3} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

$\frac{1.5}{5/3} = 1.5 \cdot \frac{3}{5} = 0.9$

So, $\frac{1}{f_2} = (0.9 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = -0.1 \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

From air case: $\frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{0.5 f_1} = \frac{2}{f_1}$

Thus, $\frac{1}{f_2} = -0.1 \cdot \frac{2}{f_1} = -\frac{0.2}{f_1}$

⟹ $f_2 = -5 f_1$

Answer: $f_2 = -5 f_1$