Question

A thin equi convex lens is made of glass of refractive index 1.5 and its focal length in air is 0.2 m. If it acts as a concave lens of 0.5 m focal length when dipped in a liquid, the velocity of light in the liquid is

• A. $1.2\times {10}^{8}m{s}^{-1}$
• B. $1.6\times {10}^{8}m{s}^{-1}$
• C. $1.8\times {10}^{8}m{s}^{-1}$
• D. $2.4\times {10}^{8}m{s}^{-1}$

Answer 1

The correct option is B $1.6\times {10}^{8}m{s}^{-1}$

Given,

The refractive index of glass ${\mu }_g=1.5$

As the lens is equi-convex hence, the focal length is given as $f=0.2m$ (positive)

Now the focal length is given as

$\frac{1}{f}=\left(\frac{{\mu }_g}{{\mu }_a}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{0.2}=\left(1.5-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ……. (1)

Now when it is dipped in liquid it acts as a concave lens, hence

The focal length is given as $f=-0.5m$

Therefore the focal length is given as

$\frac{1}{f}=\left(\frac{{\mu }_g}{{\mu }_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$\frac{1}{-0.5}=\left(\frac{1.5}{{\mu }_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ …… (2)

Taking the ratio of equation (1) and (2)

$\frac{\frac{1}{0.2}}{\frac{1}{-0.5}}=\frac{\left(1.5-1\right)}{\left(\frac{1.5}{{\mu }_l}-1\right)}$

$-5=\frac{1}{\left(\frac{1.5}{{\mu }_l}-1\right)}$

${\mu }_l=\frac{15}{8}$

Now as the absolute refractive index is given as

$\mu =\frac{{\mu }_a}{{\mu }_l}=\frac{v}{c}$

Where $v$ is the velocity of light in the given medium and $c$ is the velocity of light in vacuum.

Hence,

$v=\frac{{\mu }_a\times c}{{\mu }_l}$

$v=\frac{8}{15}\times 3\times {10}^{8}m/s$

$v=1.6\times {10}^{8}m/s$