Question

Define critical angle. Prove that the refractive index of the denser medium is reciprocal of the sine of the critical angle .

Answer 1

The critical angle is the angle of incidence where the angle of refraction ${90}^0.$ The light must travel from an optically more dense medium to an optically less dense medium$.$

Prof$:-$

Instead of always having to measure the critical angles of different materials$,$ it is possible to calculate the critical angle at the surface between two media using snell's Law$.$ To recap$,$ sneel's Law states$:$

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Where $n_1$ is the refractive index of material $1,$ $n_2$ is the refractive index of material $2,$ ${\theta }_1$ is the angle of incidence and ${\theta }_2$ is the angle of refraction $.$ For total internal reflection we know that the angle of incidence is the critical angle$,$ so$,$

${\theta }_1={\theta }_c.$

However$,$ we also know that the angle of refraction at the critical angle is ${90}^0.$ So we have

${\theta }_2={90}^0$

WE can then write snell's Law as$:$

$n_1\sin {\theta }_c=n_2\sin {90}^0$

Solving for ${\theta }_c$ gives$:$

$n_1\sin {\theta }_c=n_2\sin {90}^0$

$\sin {\theta }_c=\frac{n_2}{n_1}\left(1\right)$

${\theta }_c={\sin }^{-1}\left(\frac{n_2}{n_1}\right)$

Hence$,$ we proved$.$