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Commit 1ceca2f3 authored by Erik Strand's avatar Erik Strand
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Answer 9.2

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Use Fresnel’s equations and the Poynting vectors to find the reflectivity and transmissivity of a
dielectric interface, defined by the ratios of incoming and outgoing energy.
First note that
$$
\begin{align*}
\sqrt{\frac{\epsilon}{\mu}}
&= \sqrt{\frac{\epsilon_0 \epsilon_r}{\mu_0 \mu_r}} \\
&= \sqrt{\frac{\epsilon_0}{\mu_0}} \frac{n}{\mu_r} \\
&\approx \sqrt{\frac{\epsilon_0}{\mu_0}} n
\end{align*}
$$
since most dielectric materials have $$\mu_r \approx 1$$.
So for a plane wave in a homogeneous medium,
$$
\begin{align*}
P &= E \times H \\
&= E \times \left( \sqrt{\frac{\epsilon}{\mu}} \hat{k} \times E \right)
\end{align*}
$$
and
$$
\begin{align*}
\langle | P | \rangle
&= \frac{1}{2} \sqrt{\frac{\epsilon}{\mu}} E_\text{max}^2 \\
&\approx \frac{n}{2} \sqrt{\frac{\epsilon_0}{\mu_0}} E_\text{max}^2
\end{align*}
$$
There are two cases to consider. First, when the electric field is perpendicular to the plane of
incidence, the reflected and refracted electric fields will be (respectively)
$$
\begin{align*}
E_1 &= \frac{\sin(\theta_2 - \theta_0)}{\sin(\theta_2 + \theta_0)} E_0 \\
E_2 &= \frac{2 \cos \theta_0 \sin \theta_2}{\sin(\theta_0 + \theta_2)} E_0
\end{align*}
$$
So the reflectivity is
$$
\begin{align*}
R &= \frac{E_1^2}{E_0^2} \\
&= \frac{\sin^2(\theta_2 - \theta_0)}{\sin^2(\theta_2 + \theta_0)}
\end{align*}
$$
and the transmissivity
$$
\begin{align*}
T &= \frac{E_2^2}{E_0^2} \\
&= \frac{n_2}{n_1} \frac{4 \cos^2 \theta_0 \sin^2 \theta_2}{\sin^2(\theta_0 + \theta_2)}
\end{align*}
$$
When the electric field is in the plane of incidence,
$$
\begin{align*}
E_1 &= \frac{\tan(\theta_0 - \theta_2)}{\tan(\theta_0 + \theta_2)} E_0 \\
E_2 &= \frac{2 \cos \theta_0 \sin \theta_2}{\sin(\theta_0 + \theta_2) \cos(\theta_0 - \theta_2)} E_0
\end{align*}
$$
so
$$
\begin{align*}
R &= \frac{\tan^2(\theta_0 - \theta_2)}{\tan^2(\theta_0 + \theta_2)} \\
T &= \frac{n_2}{n_1} \frac{4 \cos^2 \theta_0 \sin^2 \theta_2}{\sin^2(\theta_0 + \theta_2)
\cos^2(\theta_0 - \theta_2)}
\end{align*}
$$
When the electric field has components out of and in the plane of incidence, the total reflectivity
and transmissivity can be found via adding the values above in quadrature.
### (b)
{:.question}
For a glass–air interface (n = 1.5) what is the reflectivity at normal incidence?
At normal incidence, there is no unique plane of incidence. So by symmetry it shouldn't matter which
reflection coefficients we use. Let's choose the perpendicular case. We would like to evaluate
$$
R = \frac{\sin^2(\theta_2 - \theta_0)}{\sin^2(\theta_2 + \theta_0)}
$$
for $$\theta = \theta_2 = 0$$. But this formula diverges, so instead we will compute its limit as
$$\theta_0$$ and $$\theta_2$$ approach zero.
As they approach zero we must still respect Snell's Law: $$n_1 / n_2 = \sin \theta_2 / \sin
\theta_0$$. Since we are interested in the limit, the small angle approximation is in this case not
an approximation at all. So we can write this as
$$
\frac{n_1}{n_2} = \frac{\theta_2}{\theta_0}
$$
or
$$
\theta_2 = \frac{\theta_0 n_1}{n_2}
$$
Thus the reflectivity is
$$
\begin{align*}
R
&= \lim_{\theta_0 \to 0} \frac{\sin^2 \left( \theta_0 \left( \frac{n_1}{n_2} - 1 \right) \right)}
{\sin^2 \left( \theta_0 \left( \frac{n_1}{n_2} + 1 \right) \right)} \\
&= \lim_{\theta_0 \to 0} \frac{\theta_0^2 \left( \frac{n_1}{n_2} - 1 \right)^2}
{\theta_0^2 \left( \frac{n_1}{n_2} + 1 \right)^2} \\
&= \frac{\left( \frac{n_1}{n_2} - 1 \right)^2}{\left( \frac{n_1}{n_2} + 1 \right)^2} \\
&= \frac{\left( n_1 - n_2 \right)^2}{\left( n_1 + n_2 \right)^2}
\end{align*}
$$
For an air to glass interface,
$$
\begin{align*}
R &= \frac{\left( 1 - 1.5 \right)^2}{\left( 1 + 1.5 \right)^2} \\
&= 0.04
\end{align*}
$$
### (c)
{:.question}
What is the Brewster angle?
$$
\theta_B = \tan^{-1} \left( \frac{n_2}{n_1} \right)
$$
So for air to glass,
$$
\theta_B = \tan^{-1} \left( \frac{1.5}{1} \right) = 0.98 \si{rad} = 56.3^\circ
$$
and for glass to air,
$$
\theta_B = \tan^{-1} \left( \frac{1}{1.5} \right) = 0.58 \si{rad} = 33.7^\circ
$$
### (d)
{:.question}
What is the critical angle?
$$
\begin{align*}
\theta_C &= \sin^{-1} \left( \frac{n_2}{n_1} \right) \\
&= \sin^{-1} \left( \frac{1}{1.5} \right) \\
&= 0.73 \si{rad} \\
&= 41.8^\circ
\end{align*}
$$
for glass to air. There isn't a critical angle for air to glass, since it requires $$n_2 \leq n_1$$.
## (9.3)
......
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