When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
[tex]\theta_c = \arcsin ( \frac{n_r}{n_i} )[/tex] (2)
where [tex]n_r[/tex] is the refractive index of the second medium and [tex]n_i[/tex] is the refractive index of the first medium.
We can find the ratio [tex]n_r / n_i[/tex] by using Snell's law:
[tex]n_i \sin \theta_i = n_r \sin \theta_r[/tex] (1)
where
[tex]\theta_i[/tex] is the angle of incidence
[tex]\theta_r[/tex] is the angle of refraction
By using the data of the problem and re-arranging (1), we find
[tex] \frac{n_r}{n_i} = \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566 [/tex]
and if we use eq.(2) we can now find the value of the critical angle:
[tex]\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}[/tex]
