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 light, and all the light is reflected. The value of this angle is given by
[tex]\theta_c = \arcsin ( \frac{n_2}{n_1} )[/tex]
where n2 and n1 are the refractive indices of the second and first medium, respectively.
In the first part of the problem, light moves from glass to air ([tex]n_a=1.00 [/tex]) and the critical angle is [tex]\theta_c = 30.8^{\circ}[/tex]. This means that we can find the refractive index of glass by re-arranging the previous formula:
[tex]n_g=n_1 = \frac{n_2}{\sin \theta_c}= \frac{1.00}{\sin 30.8^{\circ}}=1.95 [/tex]
Now the glass is put into water, whose refractive index is [tex]n_w = 1.33[/tex]. If light moves from glass to water, the new critical angle will be
[tex]\theta_c = \arcsin ( \frac{n_2}{n_1} )=\arcsin( \frac{n_w}{n_g} )=\arcsin( \frac{1.33}{1.95} )=\arcsin(0.68)=43.0^{\circ}[/tex]
