Question

A convex lens of focal length 10 cm and refractive index 1.5 is dipped in a liquid of refractive index 1.75. It will behave as

• A. a convex lens of focal length 10 cm
• B. a convex lens of focal length 35 cm
• C. a concave lens of focal length 10 cm
• D. a concave lens of focal length 35 cm

Answer 1

Answer: (D)

Given, focal length of convex lens, $f=10cm$
${\mu }_{\text{lens}}=1.5$ and ${\mu }_{\text{liquid}}=1.75$
As, lens maker’s formula
$\frac{1}{f}=\left({\mu }_{lens}-1\right)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
$\because$ For a equi convex lens $R_1=R$ and $R_2=-R$
$\frac{1}{f}=\left({\mu }_{lens}\right)\left[\frac{1}{R}-\frac{1}{\left(-R\right)}\right]$
$\Rightarrow \frac{1}{f}=\left({\mu }_{lens}-1\right)\frac{2}{R}$
$\Rightarrow \frac{1}{10}=\left(1.5-1\right)\frac{2}{R}$
$\Rightarrow R=10cm$
If lens is dropped in liquid, then
$\frac{1}{f^{\prime }}=\left[\frac{{\mu }_{lens}}{{\mu }_{liq}}-1\right]\frac{2}{R}$
$\Rightarrow \frac{1}{f^{\prime }}=\left[\frac{1.5}{1.75}-1\right]\frac{2}{10}$
$\Rightarrow f^{\prime }=-35.71cm$
Hence, the lens behaves like a concave(negative focal length) lens of focal length $35cm.$