Question
Prove that $n_{12}=\frac{1}{\sin i_C}$ where $I_c$ is the critical angle.
✓
Answer
The angle of incidence whose corresponding angle of refraction is $90^{\circ}$. It is called the critical angle $i_c$ for a given pair of mediums. From Snell's law $n_{21}=\frac{\sin i}{\sin r}$ we see that if the relative refractive index is less than one, then the maximum value of $\sin r$ is one. Therefore, there is an upper limit to the value of $\sin i$ up to which this rule can be applied.
This is $i=i_{ C }$, thus $\sin i_{ C }=n_{21}$
Snell's law of refraction cannot be applied to values of $i_c$ greater than $i$. Hence no refraction is possible. The refractive index of denser medium 2 will be relative to rarer medium 1.
$
n_{12}=\frac{1}{\sin i_C}
$
This is $i=i_{ C }$, thus $\sin i_{ C }=n_{21}$
Snell's law of refraction cannot be applied to values of $i_c$ greater than $i$. Hence no refraction is possible. The refractive index of denser medium 2 will be relative to rarer medium 1.
$
n_{12}=\frac{1}{\sin i_C}
$
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