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Commit 7f38f5fe authored by Erik Strand's avatar Erik Strand
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Answer 9.3 and 9.4

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...@@ -242,11 +242,103 @@ between layers with indices $$n_1$$ and $$n_3$$. ...@@ -242,11 +242,103 @@ between layers with indices $$n_1$$ and $$n_3$$.
What is the reflectivity? Think about matching the boundary conditions, or about the multiple What is the reflectivity? Think about matching the boundary conditions, or about the multiple
reflections. reflections.
Using Snell's Law and the small angle "approximation" as in the previous problem, we can find the
the coefficients of reflection and transmission (this time for field strength instead of power, for
reasons that will become clear). For this problem it's important to keep track of the direction in
which light crosses (or reflects from) the dielectric barriers, so I'll use subscripts throughout.
$$
\begin{align*}
R_{12}
&= \lim_{\theta_0 \to 0} \frac{\sin \left( \theta_0 \left( \frac{n_1}{n_2} - 1 \right) \right)}
{\sin \left( \theta_0 \left( \frac{n_1}{n_2} + 1 \right) \right)} \\
&= \lim_{\theta_0 \to 0} \frac{\theta_0 \left( \frac{n_1}{n_2} - 1 \right)}
{\theta_0 \left( \frac{n_1}{n_2} + 1 \right)} \\
&= \frac{\left( \frac{n_1}{n_2} - 1 \right)}{\left( \frac{n_1}{n_2} + 1 \right)} \\
&= \frac{n_1 - n_2}{n_1 + n_2} \\
T_{12}
&= \lim_{\theta_0 \to 0} \frac{2 \sin \left( \theta_0 \frac{n_1}{n_2} \right) \cos \theta_0}
{\sin \left( \theta_0 \left( \frac{n_1}{n_2} + 1 \right) \right)} \\
&= \lim_{\theta_0 \to 0} \frac{2 \theta_0 \frac{n_1}{n_2}}
{\theta_0 \left( \frac{n_1}{n_2} + 1 \right)} \\
&= \frac{2 \frac{n_1}{n_2}}{\left( \frac{n_1}{n_2} + 1 \right)} \\
&= \frac{2 n_1}{n_1 + n_2}
\end{align*}
$$
Note that $$R_{12} = -R_{21}$$. We'll use this later.
Now the reflections can be modeled as an infinite sum, taking into account the phase shift that
results from travel through the middle medium.
$$
\begin{align*}
E^-
&= E^+ R_{12} + E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d}
+ E^+ T_{12} R_{23}^2 T_{21} R_{21} e^{4 i k_2 d} \\
&= E^+ R_{12} + E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d}
\sum_{n = 0}^\infty \left( R_{21} R_{23} e^{2 i k_2 d} \right)^n \\
&= E^+ R_{12} + \frac{E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d}}
{1 - R_{21} R_{23} e^{2 i k_2 d}} \\
&= \frac{E^+ R_{12} \left( 1 - R_{21} R_{23} e^{2 i k_2 d} \right)
+ E^+ T_{12} R_{23} T_{21} e^{2 i k_2 d}} {1 - R_{21} R_{23} e^{2 i k_2 d}} \\
&= E^+ \frac{R_{12} \left( 1 + R_{12} R_{23} e^{2 i k_2 d} \right)
+ T_{12} R_{23} T_{21} e^{2 i k_2 d}} {1 + R_{12} R_{23} e^{2 i k_2 d}} \\
&= E^+ \frac{R_{12} + \left( R_{12}^2 + T_{12} T_{21} \right) R_{23} e^{2 i k_2 d}}
{1 + R_{12} R_{23} e^{2 i k_2 d}} \\
&= E^+ \frac{R_{12} + R_{23} e^{2 i k_2 d}} {1 + R_{12} R_{23} e^{2 i k_2 d}}
\end{align*}
$$
The last line follows since
$$
\begin{align*}
R_{12}^2 + T_{12} T_{21}
&= \frac{(n_1 - n_2)^2}{(n_1 + n_2)^2} + \frac{2 n_1}{n_1 + n_2} \frac{2 n_2}{n_1 + n_2} \\
&= \frac{(n_1 - n_2)^2 + 4 n_1 n_2}{(n_1 + n_2)^2} \\
&= \frac{n_1^2 - 2 n_1 n_2 + n_2^2 + 4 n_1 n_2}{(n_1 + n_2)^2} \\
&= \frac{(n_1 + n_2)^2}{(n_1 + n_2)^2} \\
&= 1
\end{align*}
$$
So the total reflectivity in terms of power is
$$
\begin{align*}
R &= \frac{(E^-)^2}{(E^+)^2} \\
&= \frac{(R_{12} + R_{23} e^{2 i k_2 d})^2}{(1 + R_{12} R_{23} e^{2 i k_2 d})^2}
\end{align*}
$$
### (b) ### (b)
{:.question} {:.question}
Can you find values for $$n_2$$ and $$d$$ such that the reflection vanishes? Can you find values for $$n_2$$ and $$d$$ such that the reflection vanishes?
To make $$R = 0$$ we need $$R_{12} = -R_{23} e^{2 i k_2 d}$$. Since $$R_{12}$$ is a positive real
number, the only way this can work is if $$e^{2 i k_2 d} = -1$$ and $$R_{12} = R_{23}$$. (Consider
that $$e^{2 i k_2 d}$$ lies on the complex unit circle. So to be real it has to be 1 or -1, and 1
would leave us with a negative number.) Thus $$2 k_2 d = \pi$$, or
$$
d = \frac{\pi}{2 k_2} = \frac{\lambda}{4}
$$
since $$k_2 = 2 \pi / \lambda$$. And $$R_{12} = R_{23}$$ implies
$$
\begin{align*}
\frac{n_1 - n_2}{n_1 + n_2} &= \frac{n_2 - n_3}{n_2 + n_3} \\
(n_1 - n_2) (n_2 + n_3) &= (n_1 + n_2) (n_2 - n_3) \\
n_1 n_2 + n_1 n_3 - n_2^2 - n_2 n_3 &= n_1 n_2 - n_1 n_3 + n_2^2 - n_2 n_3 \\
n_2^2 &= n_1 n_3 \\
n_2 &= \sqrt{n_1 n_3} \\
\end{align*}
$$
## (9.4) ## (9.4)
...@@ -255,6 +347,73 @@ Consider a ray starting with a height $$r_0$$ and some slope, a distance $$d_1$$ ...@@ -255,6 +347,73 @@ Consider a ray starting with a height $$r_0$$ and some slope, a distance $$d_1$$
lens with focal length $$f$$. Use ray matrices to find the image plane where all rays starting at lens with focal length $$f$$. Use ray matrices to find the image plane where all rays starting at
this point rejoin, and discuss the magnification of the height $$r_0$$. this point rejoin, and discuss the magnification of the height $$r_0$$.
$$
\begin{align*}
M &=
\begin{bmatrix}
1 & d_2 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
-1/f & 1
\end{bmatrix}
\begin{bmatrix}
1 & d_1 \\
0 & 1
\end{bmatrix}
\\
&=
\begin{bmatrix}
1 & d_2 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & d_1 \\
-1/f & -d_1/f + 1
\end{bmatrix}
\\
&=
\begin{bmatrix}
1 - d_2/f & d_1 + d_2 - d_1 d_2/f \\
-1/f & 1 - d_1/f
\end{bmatrix}
\end{align*}
$$
So
$$
M
\begin{bmatrix}
r_0 \\ s_0
\end{bmatrix}
=
\begin{bmatrix}
r_0 (1 - d_2/f) + s_0 (d_1 + d_2 - d_1 d_2/f) \\
-r_0/f + s_0 (1 - d_1/f)
\end{bmatrix}
$$
The position (i.e. 0th element) doesn't depend on $$s_0$$ if
$$
d_1 + d_2 - \frac{d_1 d_2}{f} = 0
$$
i.e.
$$
\frac{1}{d_1} + \frac{1}{d_2} = \frac{1}{f}
$$
Note that this doesn't depend on $$r_0$$, so there really is a focal plane.
At the $$d_2$$ satisfying the relation above, each light ray's position is $$r_0 (1 - d_2/f)$$. So
the whole plane is stretched by a factor of $$1 - d_2 / f$$.
## (9.5) ## (9.5)
......
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