Question

The power of a thin convex lens $(n_g=1.5)$ is $+5.0D,$ when this lens is dipped in a liquid of refractive index $n_l,$ it behaves like a concave lens whose focal length is $100$ cm, then what will be the value of $n_l?$

Answer 1

Given, $n_g=1.5=\frac{3}{2};P_1=+5D$
$\Rightarrow f_a=\frac{100}{P_a}=\frac{100}{5}=+20cm;f_l=-100cm;n_l=?$
Focal length of any lens in liquid
$f_l=\frac{{(}_a{n}_g-1)}{{(}_l{n}_g-1)}\times f_a$
or ${}_l{n}_g-1={(}_a{n}_g-1)\frac{f_a}{f_l}$
$=(1.5-1)\times \frac{20}{-100}=0.5\times (-\frac{1}{5})$
or ${}_l{n}_g-1=-0.1$ or ${}_l{n}_g=1-0.1=0.9$
or $\frac{n_g}{n_l}=0.9$
${\therefore }_a{n}_l=\frac{{}_a{n}_l}{0.9}=\frac{1.5}{0.9}=\frac{15}{9}=\frac{5}{3}$
or ${}_a{n}_g=\frac{5}{3}$