The radius of curvature of each surface of a convex lens having refractive index 1.8 is $20 \mathrm{~cm}$. 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 _________.

<p>Let&#39;s find the focal length of the lens in air and in the liquid. We will use the lens maker&#39;s formula:</p> <p>$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$</p> <p>where $f$ is the focal length, $\mu$ is the refractive index of the lens material, and $R_1$ and $R_2$ are the radii of curvature of the lens surfaces.</p> <p>Since the lens is convex, both surfaces have the same radius of curvature (positive), so $R_1 = R_2 = 20 \mathrm{~cm}$.</p> <p>First, let&#39;s find the focal length of the lens in air:</p> <p>$$\frac{1}{f_\text{air}} = (1.8 - 1)\left(\frac{1}{20} - \frac{1}{20}\right) = 0.8\left(\frac{1}{20}\right)$$</p> <p>$f_\text{air} = \frac{1}{0.8\left(\frac{1}{20}\right)} = 25 \mathrm{~cm}$</p> <p>Now, let&#39;s find the focal length of the lens in the liquid. The relative refractive index of the lens with respect to the liquid is:</p> <p>$\mu_\text{rel} = \frac{1.8}{1.5} = 1.2$</p> <p>$\frac{1}{f_\text{liquid}} = (1.2 - 1)\left(\frac{1}{20}\right)$</p> <p>$f_\text{liquid} = \frac{1}{0.2\left(\frac{1}{20}\right)} = 100 \mathrm{~cm}$</p> <p>The power of a lens is given by:</p> <p>$P = \frac{1}{f}$</p> <p>Now, we can find the ratio of the power of the lens in air to its power in the liquid:</p> <p>$$\frac{P\text{air}}{P</em>\text{liquid}} = \frac{f\text{liquid}}{f</em>\text{air}} = \frac{100}{25} = 4$$</p> <p>So, the ratio of the power of the lens in air to its power in the liquid is $x:1$, where $x = 4$.</p>