Question

A lens is made of flint glass (refractive index $=1.5$ ). When the lens is immersed in a liquid of refractive index $1.25$, the focal length

• A. increases by a factor of $1.25$
• B. increases by a factor of $2.5$
• C. increases by a factor of $1.2$
• D. decreases by a factor of $1.2$

Answer 1

Answer: (B)

Lens-maker's formula is given by
$\frac{1}{f}=\left({ }_a \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$....(i)
where ${ }_a \mu_g$ is refractive index of glass w.r.t. air, $R_1$ and $R_2$ are radii of curvature of two surfaces of lens and $f$ is focal length of the lens.
If the lens is immersed in a liquid of refractive index $\mu_p$, then
$\frac{1}{f_l}=\left({ }_l \mu_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$....(ii)
Here, ${}_{l}\mu_{g}$ is refractive index of glass w.r.t. liquid.
Dividing Eq. (i) by Eq. (ii), we have
$ \frac{f_l}{f}=\frac{\left({ }_a \mu_g-1\right)}{\left({ }_l \mu_g-1\right)} \Rightarrow \frac{f_l}{f}=\left(\frac{1.5-1}{\frac{1.5}{1.25}-1}\right) $
$ \Rightarrow \frac{f_l}{f}=\frac{0.5 \times 1.25}{0.25}=2.5$
Hence, focal length increases by a factor of $2.5$.