A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of $f_1$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of $f_2$ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5 , the ratio of $f_1$ and $f_2$ will be

<p>$$ \frac{1}{f} = \left(\frac{n}{n'} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $$</p> <p>For a thin plano-convex lens, one surface is flat, meaning $R_2 = \infty$ and $\frac{1}{R_2} = 0.$ The formula reduces to:</p> <p>$\frac{1}{f} = \left(\frac{n}{n'} - 1\right)\frac{1}{R_1}$</p> <p>Thus, the focal length is:</p> <p>$f = \frac{R_1}{\left(\frac{n}{n'} - 1\right)}$</p> <h3>For the First Lens (in Air)</h3> <p><p>Refractive index of lens, $n = 1.5$</p></p> <p><p>Surrounding medium (air), $n' = 1$</p></p> <p><p>Radius of curvature, $R_1 = 2 \text{ cm}$</p></p> <p>Substitute into the formula:</p> <p>$$ f_1 = \frac{2}{\left(\frac{1.5}{1} - 1\right)} = \frac{2}{(1.5 - 1)} = \frac{2}{0.5} = 4 \text{ cm} $$</p> <h3>For the Second Lens (in Liquid)</h3> <p><p>Refractive index of lens, $n = 1.5$</p></p> <p><p>Surrounding medium (liquid), $n' = 1.2$</p></p> <p><p>Radius of curvature, $R_1 = 3 \text{ cm}$</p></p> <p>Substitute into the formula:</p> <p>$f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1\right)}$</p> <p>First, calculate the term inside the parentheses:</p> <p>$\frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25$</p> <p>Now, compute $f_2$:</p> <p>$f_2 = \frac{3}{0.25} = 12 \text{ cm}$</p> <h3>Ratio of the Focal Lengths</h3> <p>$\frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3}$</p> <p>Thus, the ratio of $f_1$ to $f_2$ is:</p> <p>$1 : 3$</p>