Extinction, airmass and optical depth

Builds on: Sections 4.2, 4.3 and 1.2
Reading time: 35 min
Exercises: 4
Olympiad problems: 2

The Sun at noon is far too bright to look at, but as it sets it turns orange and can sometimes be watched without discomfort. Its light has not changed. Near the horizon the light crosses a far longer path through the air than at noon, and the air takes away more of it on the way, the blue part most of all. Every star suffers the same loss, more as it sinks lower, and an astronomer who measures magnitudes has to undo it. In this section we describe how a layer of air or dust dims light, find how the dimming grows with the length of the path, and see how measurements at a few altitudes give the magnitude a star would have above the air.

4.4.1Optical depth

Before you readMake a guess first

A layer of fog lets through some fraction of the light that enters it. A second, identical layer is placed behind the first. What fraction of the light gets through both?

Light crossing a layer of gas or dust loses some of its photons. Some are absorbed and turned into heat, and some are scattered, sent off in other directions so that they never reach us. In each thin slice of the layer the same fraction of the light that enters is lost, so the light falls off exponentially with the thickness crossed, just as the guess above showed for whole layers. The number that measures how much light a layer removes is its optical depth $τ$ .

ResultLight through an absorbing layer(4.6)

\begin{aligned} I & = I_{0} e^{- τ} \\ A & = 2.5 \log_{10} (e) τ = 1.086 τ \end{aligned}

$I_{0}, I$	the flux before and after the layer
$τ$	the optical depth of the layer, a pure number
$A$	the extinction, the dimming in magnitudes

The second line follows from (4.2), since

A = - 2.5 \log_{10} (I / I_{0}) = 2.5 τ \log_{10} e

. A layer of optical depth 1 lets through

e^{- 1} = 37 %

of the light and dims it by 1.086 magnitudes. A layer with

τ

much less than 1 is called optically thin, and one with

τ

much more than 1 optically thick.

Small molecules of air scatter blue light much more strongly than red, roughly as $1 / λ^{4}$ , which is why the daytime sky is blue and the setting Sun red. Dust scatters all colours more evenly, and ozone and water vapour absorb at particular wavelengths. The optical depth therefore depends on the wavelength, and every band of magnitudes has its own.

Quick checkA thin layer

A layer of dust has optical depth 0.50 in the V band. By how many magnitudes does it dim a star behind it?

magnitudes

4.4.2Airmass

The light of a star at the zenith crosses the least air it can. We call that amount of air one airmass. If we treat the atmosphere as a flat layer of the same thickness everywhere, light from a star at zenith distance $z$ crosses it on a slanted path $1 / \cos z = \sec z$ times as long, and so crosses $\sec z$ airmasses. The optical depth along the path is $\sec z$ times the optical depth $τ_{z}$ straight up, and the extinction is $\sec z$ times the extinction at the zenith.

ResultExtinction and airmass(4.7)

\begin{aligned} X & = \sec z \\ m & = m_{0} + k X \end{aligned}

$X$	the airmass, the amount of air crossed in units of the amount straight up
$z$	the zenith distance of the star, which is 90° minus its altitude (Section 1.2)
$m, m_{0}$	the magnitude measured on the ground and the magnitude above the air
$k$	the extinction coefficient, the extinction at the zenith in magnitudes per airmass, which is $1.086 τ_{z}$

The flat-layer picture works well up to a zenith distance of about 70°, where the airmass is 2.9. Nearer the horizon the curve of the Earth and the bending of light by refraction make the true airmass smaller than

\sec z

, about 38 at the horizon itself instead of infinity. In the V band,

k

is about 0.1 to 0.2 at a good mountain site and 0.25 or more near sea level.

Figure 4.2Extinction grows with the airmassinteractive

AltitudeExtinction coefficient

Airmass
Extinction
Measured magnitude
Light that reaches the ground

The figure shows the layer of air above an observer O, between the ground and the dotted line at the top of the air, with the path straight up to the zenith Z, one airmass long, and the heavy path of light from a star S at zenith distance

z

. The graph shows the star's measured magnitude against airmass for a star of magnitude 2.00 above the air, measured at altitudes of 90°, 60°, 45°, 30° and 20°, with the star at the chosen altitude marked S. The points lie on a straight line of slope

k

, and the dashed line carries it back to airmass 0, where it meets A, the star's magnitude above the air, here 2.00. The sliders set the altitude and the extinction coefficient. With it loses only 13 per cent of its light, while more than half of its light is lost.

4.4.3The magnitude above the air

Since the measured magnitude rises in a straight line with the airmass, two or more measurements of one star at different altitudes give both the extinction coefficient and the magnitude above the air. The slope of the line is $k$ , and carrying the line back to $X = 0$ , where there would be no air at all, gives $m_{0}$ . This is called the Bouguer method, after the French scientist Pierre Bouguer who used it in the 1720s, and it appears in data-analysis problems as a graph of magnitude against airmass (Section D5).

Example 4.7One star at two altitudes

A star is measured at magnitude 3.39 at an altitude of 60° and at magnitude 3.69 later the same night at an altitude of 25°. Find the extinction coefficient and the star's magnitude above the air.

Solution

The airmasses are $X_{1} = 1 / \sin 60 ° = 1.155$ and $X_{2} = 1 / \sin 25 ° = 2.366$ . By (4.7) the slope of the line is

k = \frac{3.69 - 3.39}{2.366 - 1.155} = 0.25 mag per airmass .

Carrying the line back to $X = 0$ ,

m_{0} = 3.39 - 0.25 \times 1.155 = 3.10 .

AnswerThe extinction coefficient is 0.25 magnitudes per airmass, and the star has magnitude 3.10 above the air.

Example 4.8Light lost at the zenithfinish it

At a site where the extinction coefficient in the V band is 0.20, what percentage of a star's light reaches the ground when the star is at the zenith?

The start

At the zenith $X = 1$ , so the star is dimmed by $k X = 0.20$ magnitudes.

Your stepNow turn this dimming into a fraction of the light.

per cent

Common errorsSection 4.4

The airmass is the secant of the zenith distance, not of the altitude. A star at an altitude of 30° has zenith distance 60° and airmass 2.
An optical depth is not an extinction in magnitudes. The two differ by the factor $2.5 \log_{10} e = 1.086$ , because optical depth uses powers of $e$ and magnitudes use powers of 10.
The rule $X = \sec z$ is good only up to a zenith distance of about 70°. Close to the horizon it gives far too much air.

ExercisesSection 4.44 questions

4.4.1
What percentage of the light gets through a cloud of optical depth 2.0?
per cent
4.4.2
What is the airmass towards a star at an altitude of 30°?
4.4.3
The extinction coefficient is 0.30 magnitudes per airmass. By how much is a star at an altitude of 45° dimmed by the air?
magnitudes
4.4.4
In which band of the UBV system is the extinction coefficient of the air largest?

Olympiad problemsPast IOAA problems, restated2 problems

2013The Sun through London fog10 points

Adapted from IOAA 2013 (Volos), theory problem S13. We have restated it in our own words and kept the original data, adding the magnitudes of the Sun and the full Moon. The solution was last checked on 20 October 2026.

On a day of very thick fog in London, the Sun looked only as bright as the full Moon usually does on a clear night. Assume that the fog dims light exponentially, and find the optical depth of the fog. Take the magnitude of the Sun as −26.74 and that of the full Moon as −12.74.

I solved it without the solution

AnswerThe fog had an optical depth of about 12.9.

2015A radio calibrator seen through the air

Adapted from IOAA 2015 (Magelang), theory problem S6. We have restated it in our own words and kept the original data. The solution was last checked on 20 October 2026.

Before each observation a radio telescope is pointed at a calibration source whose flux density above the atmosphere is known to be 21.86 Jy. On one day the telescope measured only 14.27 Jy from it, with the source at an altitude of 35°. Estimate the optical depth of the atmosphere at the zenith, $τ_{z}$ . A jansky (Jy) is a unit of flux density used in radio astronomy.

I solved it without the solution

AnswerThe optical depth at the zenith was about 0.24.