# Dust Absorption and Scattering

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\section{ Interaction of Radiation with Grains}

In general, we will be talking about a plane wave of wavelength $\lambda$
which is incident upon a particle (grain) of radius a. The cross-section
for absorption, scattering, and emission will all be proportional to the
physical cross-section of the grain, with some {\it absorption efficiency}
Q:
$$\sigma_i=Q_i\pi a^2$$
Kirchkoff's Law requires that $Q_{emit}=Q_{abs}$. If $a\gg\lambda$, then
\def\qscat{Q_{scat}}
\def\qabs{Q_{abs}}
we have the geometric optics limit, and $\qscat+\qabs\sim$. In fact,
Babinet's Principle says:
$$\boxed{\qscat+\qabs=2}$$
The proof of this goes as follows: suppose we have an infinite plane wave
focused by an infinite lens onto a point on a wall. The power pattern
should be a delta function at that point on the wall. Now suppose you
place a screen with an aperture of diameter $a$ between the plane wave
and the lens. We should now see an interference pattern on the wall. Call
the power incident on a point on the wall $P_1$. Now put in a new aperture
which is the exact compliment of our previous one (it is a scattering
body of diameter $a$), we should see a new power at our point on the wall:
$P_2$.
However, the sum of waves incident on the wall under these two apertures
should be the same as the wave from the sum of the apertures, which was
a delta function. Thus:
$$P_1=P_2$$
The first aperture (the slit) represented the scattering power from
diffraction alone $(\qscat=1)$, and the second represented the absorbed
power. The sum of these two powers, for everywhere but the true focus
point, is actually twice the incident power, and since $P_1=P_2$, we
get:
$$\qscat+\qabs=2$$
Note that if $\qscat>1$, then the object in ``shiny*: more light gets*
diffracted and less gets absorbed. Thus, although $\qscat$ changes,
the sum remains the same.

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