Transmission of atmosphere

Earth's atmosphere doesn't transmit 100% of light. Various things contribute to the absorption of light:

• scattering

• aerosols

• ozone

• $H_2O$

• $O_2$

• $CO_2$

• $N_2O$

• $CH_4$

All are functions of wavelength, time to some extent, and position in sky.

Airmass and zenith distance dependence

Definition of airmass: path length that light takes through atmosphere relative to length at zenith: $X\equiv 1$ vertically (at $z=0$). Given the zenith distance, $z$, which can be computed from:

$$\sec z = (\sin \phi \sin \delta + \cos \phi \cos \delta \cos h )^{-1}$$

where $\phi$ is the latitude, $\delta$ is the declination, and $h$ is the hour angle ( $=\textrm{local sidereal time} - \alpha$, where $\alpha$ is the right ascension), we have

$$X \sim \sec z$$

which is exactly true in the case of a plane parallel atmosphere. Since the earth's atmosphere is not a plane, the plane parallel approximation breaks down for larger airmasses. For $X\mathrel{\vcenter{\offinterlineskip \hbox{$>$} \kern 0.3ex \hbox{$\sim$}}}2$, a more precise formula needed:

$$X= \sec z - 0.0018167 (\sec z - 1) - 0.002875 (\sec z -1)^2 -0.0008083 (\sec z -1)^3$$

Consider thin sheet of atmosphere, with incident flux $F$, and outcoming flux $F+dF$. Let the thin sheet have opacity $\kappa = N\sigma$, where $N$ is the number density of absorbers/scatterers, and $\sigma$ is the cross-section/absorber-scatterer.

$$dF = -\kappa F dx$$

$$F = F_0 e^{-\int \kappa dx } \equiv F_0 e^{-\tau}$$

where $\tau$ is the optical depth of atmosphere.

$$\tau(X) \sim \tau(z=0) X$$

$$F = F_0 e^{-\tau_0 X}$$

$$m = m_0 + 1.086 \tau_0 X$$

We can define the extinction coefficient $k_\lambda$:

$$m = m_0 + k_\lambda X$$

$$k_\lambda \equiv -1.086\tau(z=0)$$

so the extinction can be specified by a set of extinction coefficients. Of course, use of the scaling of $\tau$ or $k$ with airmass assumes photometric weather!!

Sources of extinction

In the optical part of the spectrum, extinction is a roughly smooth function of wavelength and arises from a combination of ozone, Rayleigh scattering, and aerosols, as shown in this plot. The optical extinction can vary from night to night or season to season, as shown in this plot. Because of this variation, you must determine the amount of extinction on each night separately if you want accuracy better than a few percent. Generally, the shape of the extinction curve as a function of wavelength probably varies less than the amplitude at any given wavelength. Because of this, one commonly uses mean extinction coefficients when doing spectroscopy where one often only cares about relative fluxes.

In the infrared, the extinction does not vary so smoothly with wavelength because of the effect of molecular absorption. In fact, significant absorption bands define the so-called infrared windows (JHKLM), as shown in the near IR in this plot. At longer wavelengths, the broad absoprtion band behavior continues, as shown in this plot. In this figure, $transmission = f(b_\lambda l)$ where $l$ is path length (units of airmass):

$b_\lambda l$	f
-3	1
-2	0.97
-1	0.83
0	0.5
1	0.111
2	0.000

The L band is at 3.5$\mu$, M band at 5$\mu$.