Question 215 of 295

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 $\mathrm{f}_2$ 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

A$1: 2$B$1: 3$C$3: 5$D$2: 3$

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Answer: (B) $1: 3$

For case (1)


$\begin{aligned} & \frac{1.5}{v_1}-\frac{1}{\propto}=\frac{(1.5-1)}{R_1} \\ & \frac{1}{f_1}-\frac{1.5}{v_1}=\frac{1-1.5}{\propto}\end{aligned}$
$\Rightarrow \frac{1}{f_1}=\frac{(1.5-1)}{R_1}$ ...(1)
$\Rightarrow \frac{1}{f_1}=\frac{1}{2 R_1}$
For case (2)
$\begin{aligned}
& \frac{1.5}{v_1}-\frac{1.2}{\infty}=\frac{(1.5-1.2)}{R_2} \\
& \frac{1.2}{f_2}-\frac{1.5}{v_1}=\frac{(1.2-1.5)}{\propto} \\
& \Rightarrow \frac{1.2}{f_2}=\frac{0.3}{R_2} \\
& \Rightarrow \frac{1}{f_2}=\frac{1}{4 R_2}
\end{aligned}$
So clearly, $\frac{f_1}{f_2}=\frac{2 \times 2}{4 \times 3}=\frac{1}{3}$