# Dust Absorption and Scattering

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle #1\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}}

\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \def\lya{Ly\alpha}

\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.

\end{document} <\latex>