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 to a factor of $1.25$
• B. Increases to a factor of $2.5$
• C. Increases to a factor of $1.2$
• D. Decreases to a factor of $1.2$

Answer 1

Answer: (B)

Increases to a factor of $2.5$

From lens maker's formula,

$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

When lens is in air,

$\frac{1}{f_{air}}=(1.5-1)(\frac{1}{R_1}-\frac{1}{R_2})⟶\left(1\right)$

When lens is in air of R.I.$\mu$

$\frac{1}{f}=(\frac{1.5}{1.25}-1)(\frac{1}{R_1}-\frac{1}{R_2})⟶\left(II\right)$

(I)/(II) = $\frac{f}{fair}=\frac{1/2}{\frac{1}{5}}=\frac{5}{2}$

Hence focal length of lens increases by a factor of 2.5.