diff --git a/_psets/7.md b/_psets/7.md
index bd383a139499663f3526d0368eccb3ad9b768580..d116708025cdbf560d6871585d362c6e43963e96 100644
--- a/_psets/7.md
+++ b/_psets/7.md
@@ -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
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)
{:.question}
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)
@@ -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
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)