Question

A thin coverging lens is made up of glass of refractive index $1.5$. It acts like a concave lens of focal length $50\text{cm}$, when immersed in a liquid of refractive index $\frac{15}{8}$, focal length of converging lens in air (in metre) is

• A. $0.15$
• B. $0.20$
• C. $2.2$
• D. $0.40$

Answer 1

The correct option is B $0.20$

Focal length of lens $\frac{1}{f}=(\frac{{\mu }_1}{{\mu }_2}-1)(\frac{1}{R_t}-\frac{1}{R_2})$
Here, $\mu =1.5$
and when lens is in air ${\mu }_s=1\text{(for air)}So,\frac{1}{f}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\frac{1}{f}=0.5(\frac{1}{R_1}-\frac{1}{R_2})\dots \left(1\right)$

When lens is in liquid medium ${\mu }_s=\frac{15}{8}$ and lens become concave lens of focal length $50\text{cm}$.
focal length of concave lens$=-50\text{cm}$
So, $\frac{1}{f^{\prime }}=(\frac{1.5}{\frac{15}{8}}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
$\Rightarrow \frac{-1}{50}=\frac{-2}{10}(\frac{1}{R_1}-\frac{1}{R_2})$
$\Rightarrow \frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{10}$
Using this value in equation (1)
$\frac{1}{f}=0.5\times \frac{1}{10}=\frac{1}{20}\Rightarrow f=20\text{cm}$