Question

An equi-convex lens of focal length 10 cm in air and refractive index $({\mu }_g=1.5)$ is placed in a liquid whose refractive index varies with time as $\mu \left(t\right)=1.0+\frac{1}{10}t$.If the lens was placed in the liquid at t = 0, lens will act as concave lens of focal length 20 cm at t = 5n. Find n.

Answer 1

Given focal length of equiconvex lens = $10\text{cm}$.
Refractive index varies with time as
$\mu \left(t\right)=1.0+\frac{1}{10}t....\left(1\right)$
$\frac{f_e}{f_a}=\frac{{(}_a{\mu }_g-1{)}_a{\mu }_l}{{}_a{\mu }_g{-}_a{\mu }_l}$
$\frac{-20}{10}=\frac{(1.5-1{)}_a{\mu }_l}{1.5{-}_a{\mu }_l}$
$3={1.5}_a{\mu }_l$
${}_a{\mu }_l=2$
Now substituting the value of µ in equation (1), we get
$2=1+\frac{1}{10}t\Rightarrow 20=10+t$
$\Rightarrow t=10\text{sec}$.
So comparing with 5n, we get n = 2.