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

  $1: 2$
• Correct answer
  B

  $1: 3$
• C

  $3: 5$
• D

  $2: 3$

Solution

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, $\frac{f_1}{f_2}=\frac{2 \times 2}{4 \times 3}=\frac{1}{3}$

Hence, the correct answer is option: 'B'.