Question

how to derive the relation between crictical angle and refractive index ?

Answer 1

We know that TIR (Total Internal Reflection) is responsible for the critical angle, Derivation is given as follows,
When light passes from medium 2 to medium 1,
$n_{21} = \frac{sin i}{sin r} = \frac{sin c}{sin 90} = \frac{sinC[tex] n_{12} = \frac{1}{sinC}$}{1} [/tex]

refractive index of denser medium w.r.t to rarer medium is $n_{12} = \frac{1}{ n_{12} }$
∴