6.5 Some consequences of the fresnel equations

The fresnel equations

$$r_{\perp }\equiv {\left(\frac{E_{0r}}{E_{0i}}\right)}_{\perp }=\frac{n_i\cos {\theta }_i-n_t\cos {\theta }_t}{n_i\cos {\theta }_i+n_t\cos {\theta }_t}$$ $$t_{\perp }\equiv {\left(\frac{E_{0t}}{E_{0i}}\right)}_{\perp }=\frac{2n_i\cos {\theta }_i}{n_i\cos {\theta }_i+n_t\cos {\theta }_t}$$

$$r_{\parallel }\equiv {\left(\frac{E_{0r}}{E_{0i}}\right)}_{\parallel }=\frac{n_t\cos {\theta }_i-n_i\cos {\theta }_t}{n_t\cos {\theta }_i+n_i\cos {\theta }_t}$$ $$t_{\parallel }\equiv {\left(\frac{E_{0t}}{E_{0i}}\right)}_{\parallel }=\frac{2n_i\cos {\theta }_i}{n_t\cos {\theta }_i+n_i\cos {\theta }_t}$$

Why polaroid sunglasses work

Lets put these equations to work and figure out something practical. Consider light reflecting off a surface, such as the road in front of you when you aredriving a car. The light hitting the road surface can have any polarization but that will be some addition of light that has $\vec{E}$ $\perp$ and $\parallel$ to the plane of incidence. From $$r_{\parallel }\equiv {\left(\frac{E_{0r}}{E_{0i}}\right)}_{\parallel }=\frac{n_t\cos {\theta }_i-n_i\cos {\theta }_t}{n_t\cos {\theta }_i+n_i\cos {\theta }_t}$$ and $$r_{\perp }\equiv {\left(\frac{E_{0r}}{E_{0i}}\right)}_{\perp }=\frac{n_i\cos {\theta }_i-n_t\cos {\theta }_t}{n_i\cos {\theta }_i+n_t\cos {\theta }_t}$$ we see there is an angle where there is no $\parallel$ and only $\perp$ light, namely $$n_t\cos {\theta }_i-n_i\cos {\theta }_t=0$$ or $$n_t\cos {\theta }_i=n_i\cos {\theta }_t\text{.}$$ Now we can use Snell's law rewritten as $$\frac{\sin {\theta }_i}{n_t}=\frac{\sin {\theta }_t}{n_i}$$ and multiply both sides to get $$n_t\cos {\theta }_i=n_i\cos {\theta }_t$$ $$n_t\cos {\theta }_i\frac{\sin {\theta }_i}{n_t}=n_i\cos {\theta }_t\frac{\sin {\theta }_t}{n_i}$$ $$\cos {\theta }_i\sin {\theta }_i=\cos {\theta }_t\sin {\theta }_t$$ which can only be true if $${\theta }_t=\pi /2-{\theta }_i\text{.}$$ In this case Snell's law can be written $${\displaystyle \begin{array}{c}\frac{n_t}{n_i}=\frac{\sin {\theta }_i}{\sin {\theta }_t}\\ =\frac{\sin {\theta }_i}{\sin (\pi /2-{\theta }_i)}\\ =\tan {\theta }_i\\ \end{array}}$$ The angle that gives this effect is known as Brewster's angle $${\theta }_i\equiv {\theta }_{Brewster}={\tan }^{-1}\frac{n_t}{n_i}$$ At this angle light is completely polarized, it only has $\vec{E}$ $\perp$ to the plane of incidence (or parallel to the surface). Thus the glare you get from reflected light tends to be polarized in this way. In 1929 Edwin Landinvented a method for making celluoid filters to filter out light with given polarizations. He then manufactured sunglasses with these polarizers lined upto filter out the $E_{\perp }$ light and thereby reduce glare.

Other cases

Look at the Fresnel equations again and examine what happens when ${\theta }_i$ approaches 90 degrees. The reflection approaches 1 (ignore the signs) Thus at Glancing incidence you get lots of reflection. In fact X-ray telescopes usethis to focus the x-rays onto their detector.

Another effect, if $n_i>n_t$ then there is an angle of incidence beyond which light is only reflected. That is the angle where ${\theta }_t=90$ degrees. $$n_t\sin {\theta }_t=n_i\sin {\theta }_i$$ $${\theta }_t\to \pi /2$$ $$\sin {\theta }_t\to 1$$ $$\sin {\theta }_i\to \frac{n_t}{n_i}$$ The critical angle at which this occurs is $${\theta }_c={\sin }^{-1}\frac{n_t}{n_i}$$