Example 12,2: Refraction

Question: A light-ray of wavelength $\lambda_1 = 589$nm traveling through air is incident on a smooth, flat slab of crown glass (refractive index 1.52) at an angle of $\theta_1= 30.0^\circ$ to the normal. What is the angle of refraction? What is the wavelength $\lambda_2$ of the light inside the glass? What is the frequency $f$ of the light inside the glass?
 
Answer: Snell's law can be written

$$\sin\theta_2 = \frac{n_1}{n_2} \,\sin\theta_1.$$

In this case, $\theta_1=30^\circ$, $n_1\simeq 1.00$ (here, we neglect the slight deviation of the refractive index of air from that of a vacuum), and $n_2= 1.52$. Thus,

$$\sin\theta_2 =\frac{(1.00)}{(1.52)}\,(0.5)=0.329,$$

giving

$$\theta_2 = 19.2^\circ$$

as the angle of refraction (measured with respect to the normal).

The wavelength $\lambda_2$ of the light inside the glass is given by

$$\lambda_2 = \frac{n_1}{n_2}\,\lambda_1 = \frac{(1.00)}{(1.52)}\,(589) = 387.5\,{\rm nm}.$$

The frequency $f$ of the light inside the glass is exactly the same as the frequency outside the glass, and is given by

$$f = \frac{c}{n_1\,\lambda_1} = \frac{(3\times 10^8)}{(1.00)\,(589\times 10^{-9})} =5.09\times 10^{14} \,{\rm Hz}.$$