kruby/notes-archive
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

Electrodynamics Fix

Browse Source
This commit is contained in:
Indigo5684
2025-09-30 13:19:35 -05:00
parent cf704d8f94
commit ef30e73651
1 changed files with 6 additions and 6 deletions
Download Patch File Download Diff File Expand all files Collapse all files

12
docs/physics/electrodynamics/10-electromagnetic-waves.md
View File

@@ -230,15 +230,15 @@ Then, continuity of $\mathbf{H}^\parallel$ implies that $H_i + H_r = H_t$, and c
From this, we can solve for $H_r$ and $H_t$ to see that From this, we can solve for $H_r$ and $H_t$ to see that
$$\begin{align} $$\begin{align}
H_r &= \frac{Z_1 \cos \theta_i - Z_2 \cos \theta_t}{\Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_i \\ H_r &= \frac{Z_1 \cos \theta_i - Z_2 \cos \theta_t}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_i \\
H_t &= \frac{2Z_1 \cos \theta_i}{\Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_t H_t &= \frac{2Z_1 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_t
\end{align}$$ \end{align}$$
Alongside $E_i = Z_1 H_i$ and $H_t = Z_2 H_t$, we can see that Alongside $E_i = Z_1 H_i$ and $H_t = Z_2 H_t$, we can see that
$$\begin{align} $$\begin{align}
E_r &= \frac{Z_2 \cos \theta_t - Z_1 \cos \theta_i}{\Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_i \\ E_r &= \frac{Z_2 \cos \theta_t - Z_1 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_i \\
E_t &= \frac{2Z_2 \cos \theta_i}{\Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_t E_t &= \frac{2Z_2 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t} H_t
\end{align}$$ \end{align}$$
The relative signs of $E_r$ and $H_r$ was chosen to agree with their relationship for normal incidence. The relative signs of $E_r$ and $H_r$ was chosen to agree with their relationship for normal incidence.
@@ -246,8 +246,8 @@ The relative signs of $E_r$ and $H_r$ was chosen to agree with their relationshi
Again, we can define $R$ and $T$ as Again, we can define $R$ and $T$ as
$$\begin{align} $$\begin{align}
R &= (\frac{Z_2 \cos \theta_t - Z_1 \cos \theta_i}{\Z_1 \cos \theta_i + Z_2 \cos \theta_t})^2 \\ R &= (\frac{Z_2 \cos \theta_t - Z_1 \cos \theta_i}{Z_1 \cos \theta_i + Z_2 \cos \theta_t})^2 \\
T &= \frac{4Z_1 Z_2 \cos \theta_i \cos \theta_t}{(\Z_1 \cos \theta_i + Z_2 \cos \theta_t)^2} T &= \frac{4Z_1 Z_2 \cos \theta_i \cos \theta_t}{(Z_1 \cos \theta_i + Z_2 \cos \theta_t)^2}
\end{align}$$ \end{align}$$
Interestingly, if $Z_2 \cos \theta_t = Z_1 \cos \theta_i$ for a wave with any polarization, the reflection's transverse magnetic component vanishes. If $\mu_1 \approx \mu_2$, we see that this condition becomes $n_1 \cos \theta_t = n_2 \cos \theta_i$, which using Snell's law, becomes Interestingly, if $Z_2 \cos \theta_t = Z_1 \cos \theta_i$ for a wave with any polarization, the reflection's transverse magnetic component vanishes. If $\mu_1 \approx \mu_2$, we see that this condition becomes $n_1 \cos \theta_t = n_2 \cos \theta_i$, which using Snell's law, becomes