Question

1. A ray of light (f = 5.09 X 10^14 Hz) is traveling through the glycerol toward air. The angle of incidence of the ray is measured to be 10 degrees. Part of the ray is reflected and part of the ray is refracted.
2. Find the angle of reflection.
3. Find the angle of refraction.
4. Find the speed of light in glycerol
5. Find the wavelength of this ray in glycerol
6. Find the critical angle for light traveling from water to air.

Answer 1

The angle of reflection is the same as the angle of incidence, that is, 10°.

The angle of refraction can be found from Snell's law:

$$n=\frac{\sin\theta_i}{\sin\theta_r},\\\space\\ \theta_r=\arcsin\frac{\sin\theta_i}{n}=\arcsin\frac{\sin10°}{1.47}=6.78°.$$

The speed of light in glycerol:

$$c_n=\frac cn=\frac{3·10^8}{1.47}=2.04·10^8\text{ m/s}.$$

The wavelength of this ray in glycerol:

$$\lambda=\frac{c_n}{f}=4.01·10^{-9}\text{ m}.$$

The critical angle for light traveling from water to air:

$$\theta_c=\arcsin\frac{1}{n_w}=\arcsin\frac{1}{1.33}=48.6°.$$