Question

A double convex lens made of material of refractive index $1.5$ & having radius of curvature of each surface of $10\text{cm}$ is immersed in a liquid of refractive index $3.0$. The lens will behave as:

• A. converging lens of $f=10\text{cm}$
• B. diverging lens of $f=10\text{cm}$
• C. converging lens of $f=5\text{cm}$
• D. diverging lens of $f=5\text{cm}$

Answer 1

The correct option is B diverging lens of $f=10\text{cm}$

Given,


${\mu }_1=3,{\mu }_2=1.5$
$R_1=+10\text{cm}$ , $R_2=-10\text{cm}$

Using Lens maker's formula $$\frac{1}{f}=(\frac{{\mu }_1}{{\mu }_2}-1)(\frac{1}{R_1}-\frac{1}{R_2})$$Subtituting the values,

$\Rightarrow \frac{1}{f}=(\frac{1.5}{3}-1)(\frac{1}{10}-\frac{1}{-10})$

$\Rightarrow \frac{1}{f}=(\frac{1}{2}-1)\left(\frac{2}{20}\right)$

$\Rightarrow \frac{1}{f}=\frac{-1}{2}\times \frac{2}{20}=\frac{-1}{10}$

$\therefore$ focal length $f=-10\text{cm}$

Since, the focal length of the lens is negative so it is diverging in nature.

$$\begin{array}{c}\text{Why this Q:}\\ \text{To understand the concept of lens maker's formula}\end{array}$$