QJEEM25Phbbfb: JEE Main 2025 Physics - Class 12 JEEM25Phbbfb Solution: A plano-convex lens having radius of curvature of first surface …

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

For a thin plano-convex lens, one surface is flat, meaning $R_2 = \infty$ and $\frac{1}{R_2} = 0.$ The formula reduces to:

$\frac{1}{f} = \left(\frac{n}{n'} - 1\right)\frac{1}{R_1}$

Thus, the focal length is:

$f = \frac{R_1}{\left(\frac{n}{n'} - 1\right)}$

For the First Lens (in Air)

Refractive index of lens, $n = 1.5$

Surrounding medium (air), $n' = 1$

Radius of curvature, $R_1 = 2 \text{ cm}$

Substitute into the formula:

$f_1 = \frac{2}{\left(\frac{1.5}{1} - 1\right)} = \frac{2}{(1.5 - 1)} = \frac{2}{0.5} = 4 \text{ cm}$

For the Second Lens (in Liquid)

Refractive index of lens, $n = 1.5$

Surrounding medium (liquid), $n' = 1.2$

Radius of curvature, $R_1 = 3 \text{ cm}$

Substitute into the formula:

$f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1\right)}$

First, calculate the term inside the parentheses:

$\frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25$

Now, compute $f_2$ :

$f_2 = \frac{3}{0.25} = 12 \text{ cm}$

Ratio of the Focal Lengths

$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$

Thus, the ratio of $f_1$ to $f_2$ is:

$1 : 3$