Question

State the relationship between the critical angle and the absolute refractive index of a medium.

Answer 1

Hint: The relationship between the critical angle and the refractive index be
$\Rightarrow \sin C = \dfrac{1}{{{\mu _b}^a}}$
From here we can say that sine of critical angle is inversely proportional to the absolute refractive index of the medium.
$\Rightarrow \sin C \propto \dfrac {1}{{\mu_b}^a}$

Solution step by step
The angle of incidence for which the angle of refraction is 90$^0$ is called critical angle and the ratio of velocity of light from vacuum to another medium is known as absolute refractive index.
To find the relationship between the absolute refractive index and critical angle let take a light ray and the angle of incidence be the critical angle.
So the critical angle of refraction is 90$^0$.
For absolute refractive index one medium is air (rare) and the other medium is denser.
$μ_a$ = refractive index one medium i.e. air
$μ_b$ = refractive index second medium which is denser
Applying Snell’s law
$\Rightarrow \dfrac{{\sin i}}{{\sin r}} = \dfrac{{{\mu _a}}}{{{\mu _b}}}$
at $ \sin r = 90^0$, $ \sin i = \sin C$
$\Rightarrow \dfrac {\sin C}{\sin 90^0} = \dfrac{{{\mu _a}}}{{{\mu _b}}}$
$\therefore \sin 90^0 =1$
$\Rightarrow \sin C = \dfrac{{{\mu _a}}}{{{\mu _b}}}$
Where sinc is equal to the sine of critical angle
Or it can be also written as
$\Rightarrow \sin C$ = $\dfrac{1}{{{\mu _b}^a}}$
Hence we can say that critical angle is inversely proportional to the absolute refractive index of the medium.

Note:
Refractive index is the ratio of velocity of light travel from one medium to another but in case of the absolute refractive index, the ratio of velocity of light from vacuum to another medium is known as absolute refractive index. Here both are different and in case of the critical angle, the angle of refraction is always perpendicular or 90$^0$ from the separating media.