A biconvex lens of refractive index $$1.5$$ has a focal length of $$20$$ cm in air. Its focal length when immersed in a liquid of refractive index $$1.6$$ will be:

We need to find the focal length of a biconvex lens when immersed in a liquid, given its focal length in air.

For a biconvex lens with equal radii of curvature $$R$$:

$$\frac{1}{f} = (\mu_{rel} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu_{rel} - 1)\frac{2}{R}$$

where $$\mu_{rel}$$ is the refractive index of the lens relative to the surrounding medium.

In air ($$\mu_{medium} = 1$$), $$\mu_{rel} = 1.5/1 = 1.5$$:

$$\frac{1}{20} = (1.5 - 1)\frac{2}{R} = 0.5 \times \frac{2}{R} = \frac{1}{R}$$

$$R = 20 \text{ cm}$$

In liquid ($$\mu_{liquid} = 1.6$$), $$\mu_{rel} = 1.5/1.6 = 15/16$$:

$$\frac{1}{f'} = \left(\frac{15}{16} - 1\right)\frac{2}{20} = \left(-\frac{1}{16}\right)\frac{1}{10} = -\frac{1}{160}$$

$$f' = -160 \text{ cm}$$

The negative sign indicates that the lens behaves as a diverging lens in this liquid. This happens because the refractive index of the liquid (1.6) is greater than that of the lens (1.5), so the lens bends light in the opposite direction compared to air.

The correct answer is Option (2): $$-160$$ cm.