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…

Question

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

Aicharya · Accepted Answer

Both lenses are plano-convex, so the second surface is plane ( $R=\infty$), which simplifies the lens maker's formula:
$\frac{1}{f}=\left(\frac{n}{n_n}-1\right)\left(\frac{1}{R}-0\right)$
where:
- In is the refractive index of the lens material,
- $n_n$ is the refractive index of the surrounding medium,
- $R$ is the radius of eurvature of the curved surface.
Step E Calculate $f_1$ (Lens in Air)
For Lens 1 in Air:
$\begin{gathered}
\frac{1}{f_1}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{2}\right) \\
\frac{1}{f_1}=(1.5-1) \times \frac{1}{2} \\
\frac{1}{f_1}=0.5 \times \frac{1}{2}=\frac{0.5}{2}=\frac{1}{4} \\
f_1=4 \mathrm{~cm}
\end{gathered}$
Step 2: Calculate $f z$ (Lens in Liquid)
For Lens $\mathrm{2 i n}$ Liquid:
$\begin{aligned}
& \frac{1}{T_2}=\left(\frac{1.5}{1.2}-1\right)\left(\frac{1}{3}\right) \\
& \frac{1}{T_2}=\left(\frac{1.5-1.2}{1.2}\right) \times \frac{1}{3} \\
& \frac{1}{T_2}=\left(\frac{0.3}{1.2}\right) \times \frac{1}{3} \\
& \frac{1}{T_2}=\frac{0.3}{3.6} \\
& f_2=\frac{3.6}{1.3}=12 \mathrm{~cm}
\end{aligned}$
Step 3: Find the Ratio $\frac{4}{12}$
$\frac{f_1}{f_2}=\frac{4}{12}=\frac{1}{3}$
Final Answer:
$\frac{1}{3}$

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of $f_1$ in air.…

Solution