Question

A biconvex lens $(\mu =1.5)$ of focal length $0.2m$ acts as a divergent lens of power $1D$ when immersed in a liquid. Find the refractive index of the liquid.

Answer 1

New focal length

$f=\frac{1}{1D}=1m$

$f_1=0.2m$.

So we have,

$\frac{1}{f}=(n-1)(\frac{1}{R_1}+\frac{1}{R_2})$

for biconvex lens

$\frac{f}{f_1}=\frac{{\mu }_1-1}{{\mu }_2-1}$

$\frac{1}{0.2}=\frac{1.5-1}{{\mu }_2-1}$

$\therefore {\mu }_2=1.01$