Question #1d195

May 3, 2015

The critical angle of a light ray, passing from a dense medium(glass) to a less dense medium(air), is angle angle at which the refracted ray is exactly ${90}^{0}$.

Now, all you do is to use Snell's law:

${n}_{g} \sin i = {n}_{a} \sin r$

where ${n}_{g}$ is the refractive index of glass
${n}_{a}$ is the refractive index of air
$i$ is the angle of incidence
$r$ is the angle of refraction

For critical angle $i$ becomes $c$, and $r$ becomes ${90}^{0}$
And we know that ${n}_{a} \cong 1$

$\implies {n}_{g} \sin c = 1$

$\implies \sin c = \frac{1}{n} _ g$

$\implies c = {\sin}^{-} 1 \left(\frac{1}{n} _ g\right)$

This could also be written as $c = {\sin}^{-} 1 \left({v}_{g} / {v}_{v a c}\right)$

Where ${v}_{g}$ is the speed of light in glass
${v}_{v a c}$ is the speed of light in a vacuum or air.