Question

A convex lens of refractive index $1.5$ and focal length $18cm$ in air is immersed in water. The change in focal length of the lens will be_______ $cm$.
$($ Given refractive index of water $=\frac{4}{3})$

Answer 1

Answer: 54

$\frac{I}{f_{H_{2}O}}=\left(\frac{{\mu }_g}{{\mu }_{H_{2}O}}-1\right)\left(\frac{2}{R}\right)$
$=\frac{1}{8}\left(\frac{2}{R}\right)$
$=\frac{1}{\left(4f_{air}\right)}$
So, $f_{H_{2}O}=4f_{\text{air }}=72cm$
So change in focal length $=72-18=54cm$