The radius of curvature of each surface of a convex lens having refractive index 1.8 is 20 cm. The lens is now immersed in a liquid of refractive index 1.5. The ratio of power of lens in air to its power in the liquid will be $$x$$ : 1. The value of $$x$$ is _______

The power of a lens is given by the lensmaker's equation:

$$P = \left(\frac{n_l}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For a convex lens with both surfaces having radius of curvature 20 cm: $$R_1 = +20$$ cm, $$R_2 = -20$$ cm.

Power in air ($$n_m = 1$$):

$$P_{air} = (1.8 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) = 0.8 \times \frac{2}{20} = 0.8 \times 0.1 = 0.08 \text{ cm}^{-1}$$

Power in liquid ($$n_m = 1.5$$):

$$P_{liq} = \left(\frac{1.8}{1.5} - 1\right)\left(\frac{1}{20} + \frac{1}{20}\right) = (1.2 - 1) \times 0.1 = 0.2 \times 0.1 = 0.02 \text{ cm}^{-1}$$

Ratio:

$$\frac{P_{air}}{P_{liq}} = \frac{0.08}{0.02} = 4$$

Therefore, $$x = 4$$.