Total Internal Reflection

Using the Fresnel Equations, we can solve for the 'critical angle' where the reflection coefficient goes to one (in this case, all the light is reflected internally). We can do this with either the parallel or perpendicular component of the electric field. Let's use the parallel component:

$r_{\perp}=\frac{n_i\cos \left(\theta _i\right)-n_t\cos \left(\theta _t\right)}{n_t\cos \left(\theta _t\right)+n_i\cos \left(\theta _i\right)}=1$

this gives

$n_i\cos \left(\theta _i\right)-n_t\cos \left(\theta _t\right)=n_t\cos \left(\theta _t\right)+n_i\cos \left(\theta _i\right)$

or

$2n_t\cos \left(\theta _t\right)=0$

which means

$\theta _t=\pi /2$

and using Snell's law, solving for the critical angle now, which is also the incident angle

$\theta _c=\arcsin \left(\frac{n_t}{n_i}\sin \left(\theta _t\right)\right)=\arcsin \left(\frac{n_t}{n_i}\right)$

the plot below shows the critical angle where light is completely reflected when going from a low index material to a high index material i.e., n_i > n_t

Notice that you still get a polarization angle but now there appears another, critical angle where no light is reflected.

There are a few points to consider from the plot above. The first is that the amplitude transmission coefficients are greater than one. There is nothing wrong with this because these coefficients are not directly equal to the transmitted power, which we will calculate momentarily.

The second point is that, even in this case, there is a "polarization angle" where the amplitude reflection coefficient for the parallel component is again zero.

The last point, is there now appears a new, "critical angle" θ_c, where all the coefficients seem to disappear from the plot. At the critical angle the coefficients become complex and so are not plotted.