__Atmospheric Extinction and Refraction__

** **

Ah, our atmosphere. It protects us, nourishes us, and
gives us life. However, it also acts as a distorting filter when we view the
sky. Another tutorial addresses the atmosphere’s turbulence effects that reduce
resolution of telescopes. We’ll now cover two more areas that affect our view
of the heavens. It should be pointed out that major observatories have
established unique algorithms and procedures to deal with extinction and
refraction for their specific locations. The formulas below are more useful for
amateur use.

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## EXTINCTION

Atmospheric
extinction is the reduction in brightness of stellar objects as their photons
pass through our atmosphere. The effects of extinction depend on transparency,
elevation of the observer, and the zenith angle, the angle from the zenith to
one’s line of sight. Therefore, looking vertically, the zenith angle is 0^{0},
and is 90^{0} at the horizon.

It’s
obvious that as the zenith angle increases, light from stellar objects must
pass through more atmosphere, decreasing brightness. Therefore, a star viewed
near the zenith appears much brighter than when it nears the horizon. We’ll
examine a formula (reference 1.) that can quantify this effect.

If we assume that one’s air mass at the zenith is unity,
the number of air masses X at zenith angle z is

_{}_{}

Where e has the value of 2.71828…

For example, if z = 45^{0 },

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Therefore,
at a zenith angle of 45^{0}, stellar
light passes through about 41.4% more air mass than at the zenith.

Note that the number of air masses at the horizon is 40. That is, stellar light passes through 39x more atmosphere than at the zenith.

There
are three factors that can be quantitively considered to assess the effect of
extinction. Molecular absorption, mainly due to atmospheric ozone and water, is
a minor one, about 0.02 magnitudes per air mass. More importantly, Rayleigh
scattering by air molecules accounts for up to 0.14 magnitude increases per air
mass. Finally, aerosol scattering (dust, water and manmade pollutants) adds
about 0.12 magnitudes per air mass. The average total effect at sea level is
the sum of these factors, in the order of 0.28 magnitudes per air mass at
Standard Temperature and Pressure, (STP = 760 mm Hg, 0^{0} C). Note
that stellar objects are, therefore, 0.28 magnitudes brighter at the top of our
atmosphere. At elevations of 0.5 km, 1.0 km, and 2.0 km, the extinction effects
are about 0.24, 0.21, and 0.16 magnitudes per air mass, respectively. Therefore,
mountain observatories have smaller extinction (and also refraction) effects. Extinction
in winter is smaller than in summer due to less atmospheric water. Finally,
Rayleigh scattering affects blue light more than red so as zenith angle
increases there’s a corresponding reddening of a stellar object.

Extinction
becomes significant when altitudes are lower than about 45^{0}. At sea level, zenith extinction is 0.28 magnitudes, and
at an altitude of 45^{0} it’s 0.40 magnitudes, an increase of only 0.12
magnitudes. However, at an altitude of 12.5^{0}, extinction is 1.28 magnitudes, an increase of 1.00
magnitudes greater than the zenith’s. The
effect becomes much more dramatic at even lower altitudes. At the horizon, the
magnitude effect is 11.2!

The
Sun’s magnitude at noon is about –26.7, but at the horizon, it would only be
in the order of –15.8, a loss of brilliance almost 23,000x from that of noon. In addition,
reddening would be very prominent.

## REFRACTION

As stellar light passes through the atmosphere, it is
refracted just as through a lens. Blue light is refracted more than red light. Like
extinction, the amount of atmospheric refraction depends on the amount of air
mass that light has to traverse.

There are many formulas and computer programs available to
correct for refraction. Duffett-Smith has several formulas for accurate
refraction predictions. A simple approximation (reference 2.) that is useful
for zenith angles up to 60^{0} is

_{}_{}

where

r is the increased perceived altitude due to refraction

z is the zenith angle (90^{0} – altitude)

k depends on the wavelength of the light and the elevation of the observer above sea level.

Note that the actual altitude of a refracted object is
(perceived altitude – r). This refraction effect is relatively minor until the
zenith angle approaches 90^{0 }.

At STP, k averages about 60.3”. However, at an elevation
of 2 km, k is about 48.8”, a 19% reduction in refraction from sea level. For
400 (blue), 500, 600 and 700 (red) nm wavelengths, the corresponding STP k
values are 60.4”, 57.8”, 57.4”, and 57.2”, respectively. So as z increases, a
tiny spectrum becomes visible in good telescopic seeing, with blue at the top
and red at the bottom. For example, at an altitude of 30^{0}, the size
of the spectrum would be about 5.5 arc seconds.

We’re familiar with the distortion of a setting Sun or
Moon. Recall that these bodies average about 30 arc minutes in size. The
following image illustrates the flattening effect. It also includes the effects
of atmospheric turbulence.

To
explain this oval shape, we must examine how rapidly atmospheric refraction
increases as we near the horizon. At altitudes of 2^{0}, 1^{0}
and 0^{0} the respective angles of refraction, r,^{ }are 18.4’,
24.75’, and 35.35’. The differential refraction of an extended body like the Sun
or Moon, where the angle of refraction is greater at the bottom of the body
than at its top results in its flattened shape. When the Sun appears to be at
the horizon, it has actually set.

### REFERENCES

- http://cfa-www.harvard.edu/icq/ICQExtinct.html
- http://ganymede.nmsu.edu/holtz/a535/ay535notes/node6.html

Mike Luciuk