Answer to Question #118940 in Molecular Physics | Thermodynamics for Adan

Question

An incident angle in an unknown substance is 47.0 if this ray is refracted at an angle of 63.7 degree in ice what is the speed of light in the unknown substance

Answer 1

An angle of incident is the angle between the ray and the perpendicular to the surface (see https://en.wikipedia.org/wiki/Angle_of_incidence_(optics) ).

According to the Snell's law (see https://en.wikipedia.org/wiki/Snell%27s_law) for the incident angle $\theta_1$ and angle of refraction $\theta_2$

$\dfrac{\sin\theta_1}{\sin\theta_2} = \dfrac{n_2}{n_1}$ ; here $n_1$ and $n_2$ are refractive indices.

We know that $\theta_1 = 47.0^\circ, \theta_2 = 63.7^\circ, n_2 = 1.31$ (see https://en.wikipedia.org/wiki/Refractive_index), therefore

$n_1 = n_2\cdot \dfrac{\sin\theta_2}{\sin\theta_1} = 1.31\cdot\dfrac{\sin63.7^\circ}{\sin47.0^\circ} \approx 1.61.$

Refractive index is a ratio of the speed of light in vacuum and in the substance. Therefore, the speed of light in substance is

$v = \dfrac{c}{n_1} = \dfrac{3\cdot10^5\,\mathrm{km/s}}{1.61} \approx 1.9\cdot10^5\,\mathrm{km/s}.$