From d5322149dc5f3c8e9dc4f70de38eabebb8d659c4 Mon Sep 17 00:00:00 2001
From: Erik Strand <erik.strand@cba.mit.edu>
Date: Wed, 6 Mar 2019 16:26:43 -0500
Subject: [PATCH] Answer 6.6
---
_psets/4.md | 55 +++++++++++++++++++++++++++++++++++++++++++++++++++++
1 file changed, 55 insertions(+)
diff --git a/_psets/4.md b/_psets/4.md
index 09d314b..6120ef7 100644
--- a/_psets/4.md
+++ b/_psets/4.md
@@ -278,8 +278,63 @@ Assume that sunlight has a power energy density of $$1 \si{kW/m^2}$$ (this is a
typical average value in the continental USA is $$\approx 200 \si{W/m^2}$$). Estimate the electric
field strength associated with this radiation.
+The energy delivered per square meter in one second by sunlight is the field energy stored in a 1
+meter by 1 meter by 1 light second volume. So the average energy density of sunlight must be
+
+$$
+\frac{10^3 \si{J}}{1 \si{m} \cdot 1 \si{m} \cdot \num{3e8} \si{m}} = \num{0.33e-5} \si{J/m^3}
+$$
+
+The energy density of light in a vacuum is
+
+$$
+\begin{align*}
+U &= \frac{1}{2} (E \cdot D + B \cdot H) \\
+&= \frac{1}{2} \left( \epsilon_0 |E|^2 + \mu_0 |H|^2 \right) \\
+&= \frac{1}{2} \left( \epsilon_0 |E|^2 + \epsilon_0 |E|^2 \right) \\
+&= \epsilon_0 |E|^2
+\end{align*}
+$$
+
+since $$\vert E \vert / \vert H \vert = \sqrt{\mu_0/\epsilon_0}$$.
+
+Thus the root mean square magnitude of the electric field must be
+
+$$
+\frac{\num{0.33e-5} \si{J/m^3}}{\epsilon_0} = \num{3.8e5} \si{V^2/m^2}
+$$
+
+Since the electric field is a sinusoidal wave, the root mean square amplitude is half the maximum
+amplitude. So the peak field strength is $$\num{7.5e5} \si{V/m}$$.
+
+
### (b)
{:.question}
If 1 W of power is focused in a laser beam to a square millimeter, what is the field strength? What
about if it is focused to the diffraction limit of $$\approx 1 \si{\mu m^2}$$?
+
+Summarizing the previous section, we found that the root mean square field strength necessary to
+deliver $$x$$ watts of power to an area $$A$$ is
+
+$$
+\frac{x \si{W}}{c \si{m/s}} \cdot \frac{1}{A \si{m^2}} \cdot \frac{1}{\epsilon_0 \si{F/m}}
+$$
+
+So for one watt to be delivered to a square millimeter, the RMS field strength is
+
+$$
+\frac{1 \si{W}}{\num{3e8} \si{m/s}} \cdot \frac{1}{10^{-6} \si{m^2}} \cdot \frac{1}{\num{8.85e-12} \si{F/m}}
+= \num{3.8e8} \si{V^2/m^2}
+$$
+
+which means the peak field strength is $$\num{7.5e8} \si{V/m}$$.
+
+If it's focused down to a square micrometer, the RMS field strength is
+
+$$
+\frac{1 \si{W}}{\num{3e8} \si{m/s}} \cdot \frac{1}{10^{-12} \si{m^2}} \cdot \frac{1}{\num{8.85e-12} \si{F/m}}
+= \num{3.8e14} \si{V^2/m^2}
+$$
+
+and the peak field strength is $$\num{7.5e14} \si{V/m}$$.
--
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