Question

State the law of refraction of light that defines the refractive index of a medium with respect to the other. Express it mathematically. How is the refractive index of any medium 'A' with respect to medium 'B' related to the speed of propagation of light in two media A and B? State the name of this constant when one medium is vacuum or air.
The refractive indices of glass and water with respect to vacuum are 3/2 and 4/3 respectively. If the speed of light in glass is $2\times {10}^8$, find the speed of light in:
(i) vacuum
(ii) water.

Answer 1

Snell's law: When a ray of light of a particular colour passing through a medium 1 is incident on another medium 2, the ratio of sine of the angle of incidence (i) and sine of the angle of refraction (r) is a constant.
This constant is called the refractive index of medium 2 with respect to medium 1, denoted as $n_{21}$.

Mathematically,
$n_{21}=\frac{\sin i}{\sin r}$
In this case, if medium B is vacuum or air, then $n_{ab}$ is called the absolute refractive index of medium A, and is simply written as $n_A$.

Absolute refractive index of glass, $n_g$ = 3/2

Absolute refractive index of water, $n_w$ = 4/3

Speed of light in glass, $v_g=2\times {10}^8$m/s.

(i) Now,
$n_g=\frac{speedoflightinvacuum,c}{v_w}\Rightarrow c=n_g\times v_w=\frac{3}{2}\times 2\times {10}^8=3\times {10}^{8}m/s$


(ii) let the speed of light in water be $v_w$,
$n_w=\frac{c}{v_w}\Rightarrow v_w=\frac{c}{n_w}=\frac{3\times {10}^8}{4/3}=\frac{9}{4}\times {10}^8=2.25\times {10}^{8}m/s$