Question

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

Answer 1

$ \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(4 f _{ air }\right)}$
So, $f _{ H _2 O }=4 f _{\text {air }}=72\, cm$
So change in focal length $=72-18=54 \,cm$