QJEEM23Phbefd: JEE Main 2023 Physics - Class 12 JEEM23Phbefd Solution: The radius of curvature of each surface of a convex lens having …

Let's find the focal length of the lens in air and in the liquid. We will use the lens maker's formula:

$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$

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.

Since the lens is convex, both surfaces have the same radius of curvature (positive), so $R_1 = R_2 = 20 \mathrm{~cm}$ .

First, let's find the focal length of the lens in air:

$\frac{1}{f_\text{air}} = (1.8 - 1)\left(\frac{1}{20} - \frac{1}{20}\right) = 0.8\left(\frac{1}{20}\right)$

$f_\text{air} = \frac{1}{0.8\left(\frac{1}{20}\right)} = 25 \mathrm{~cm}$

Now, let's find the focal length of the lens in the liquid. The relative refractive index of the lens with respect to the liquid is:

$\mu_\text{rel} = \frac{1.8}{1.5} = 1.2$

$\frac{1}{f_\text{liquid}} = (1.2 - 1)\left(\frac{1}{20}\right)$

$f_\text{liquid} = \frac{1}{0.2\left(\frac{1}{20}\right)} = 100 \mathrm{~cm}$

The power of a lens is given by:

$P = \frac{1}{f}$

Now, we can find the ratio of the power of the lens in air to its power in the liquid:

$\frac{P\text{air}}{P\text{liquid}} = \frac{f\text{liquid}}{f\text{air}} = \frac{100}{25} = 4$

So, the ratio of the power of the lens in air to its power in the liquid is $x:1$ , where $x = 4$ .