When light travels from a medium with greater refractive index [tex]n_1[/tex] to a medium with smaller refractive index [tex]n_2[/tex], there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
[tex]\theta_c = \arcsin ( \frac{n_2}{n_1} )[/tex]
In this problem, we know that the critical angle is[tex]\theta_c = 53.7^{\circ}[/tex], so we can find the ratio between the refractive indices of the two mediums:
[tex] \frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81 [/tex]
and since the second medium is air (n=1.00), the refractive index of the first medium is
[tex]n_1= \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23 [/tex]
In the second part of the problem, we have light entering from air ([tex]n_i = 1.00[/tex]) at angle of incidence of [tex]\theta_i = 45.0 ^{\circ}[/tex], into the second medium with [tex]n_r = 1.23[/tex]. By using Snell's law, we can find the angle of refraction of the light inside the medium:
[tex]n_i \sin \theta_i = n_r \sin \theta_r[/tex]
[tex]\sin \theta_r = \frac{n_i}{n_r} \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574[/tex]
[tex]\theta_r = \arcsin(0.574)=35.1^{\circ}[/tex]
