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5. Astronomy Through the Atmosphere

Ground based observations are affected by:

o Absorption

o Refraction

o Scattering

o Scintillation

In this section we will briefly consider some of the effects of

these four phenomena.

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1nm 10nm 100nm 1m 10m 100m 1mm 1cm 10cm 1m 10m 100m

Wavelength

Fraction

ofenergytransmitted

0

1

Completely transparent

Completely

opaque

Ionisation of

air molecules

(starting with

ozone)

Optical window Radio window

Ion

ospherereflective

Molecular transitions (H2O)

X-rays UV Visible Infrared sub mm Radio

softhard EUV (FIR) mm & microwave

Absorption

The Earths atmosphere is opaque to E-M radiation, apart from two windows:in the optical and radio regions of the E-M spectrum.

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Absorption

Between the optical and radio windows (i.e. in the infra-red) there are

numerous absorption bands due to molecular transitions (mainly of water)

Altitude(km) 4

2

Temperature (C)0 20

Inversion

layer

It is possible to get abovetheclouds containing this

water vapour because of the

temperature structure

of the atmosphere. Aboveabout 2km there is a thin

inversion layer, where the

temperature increaseswith

height. Clouds form at the baseof the inversion layer,

leaving generally clear, dry air

above.

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Absorption

Between the optical and radio windows (i.e. in the infra-red) there are

numerous absorption bands due to molecular transitions (mainly of water)

The worlds bestobservatories (e.g.

La Palma, Hawaii,

La Silla, Paranal) are

all at altitudeswhich place them

above inversion

layers.

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Absorption

How does absorption in the atmosphere affect the apparent brightness of objects?

We model the atmosphere as a series of plane-parallel slabs.

Consider a thin slab of thickness and radiation of intensity incident

perpendicular to the slab

We model the

absorption inthe slab as:

ld

dII+

ld

ldIdI =

I

(5.1)

Absorption coefficient, which is not in

general constant, but depends on depthin the atmospherel

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Absorption

Re-arranging, and integrating from the source (at distance , emitting radiation ofintensity ) to the Earths surface

We define the right hand integral as the optical depth, denoted by .

Thus or

=DI

I

dI

dI

0

obs

0

l (5.2)

0I

=

0

obslnI

I

(5.3)

= eII 0obs

(5.4)

Although we are thinking mainlyabout atmospheric absorption inthis section, the same formula candescribe e.g. interstellarabsorption along line of sight

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Absorption

If we describe the atmosphere as transparent and

If we describe the atmosphere as optically thin and

If we describe the atmosphere as optically thick and

Expressing in terms of apparent magnitude

We can write

So

or

0= 0obs II =

0obs II 1

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Absorption

Suppose we observe a star at zenith angle

Path length through slab of thickness

is

So this introduces an extra factor

of in eqs. 5.1 5.3.

Hence

ld

ds

ld

seccos

ll

dd

ds == (5.8)

sec

secsec09.1 mm = (5.9)

Zenith extinction

This treatment is only an approximation, as the lightray is also refracted by the atmosphere, thus changing along its path.

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Suppose we break the atmosphere into fourparallel slabs, each with uniform refractiveindex: 30 L

GROUND

Top of atmosphere

0

1

2

3

0

1

2

3

A

B

C

D

Refraction

We model the atmosphere as plane-parallel and consider a light ray incident atzenith angle on the top of the atmosphere.

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Suppose we break the atmosphere into fourparallel slabs, each with uniform refractiveindex:

Applying Snells Law,

at level A:

at level B:

at level C:

Finally, at level D:

Putting these together:

30 L

33 sinsin =

00 sinsin =

2233 sinsin =

1122 sinsin =

0011 sinsin =

GROUND

Top of atmosphere

0

1

2

3

0

1

2

3

A

B

C

D

Refraction

We model the atmosphere as plane-parallel and consider a light ray incident atzenith angle on the top of the atmosphere.

Extends to an arbitrary number of slabs: to correct the observed zenith anglefor refraction, we need only to know at ground level

0

(5.10)

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Suppose we break the atmosphere into four

parallel slabs, each with uniform refractiveindex:

Applying Snells Law,

at level A:

at level B:

at level C:

Finally, at level D:

Putting these together:

30 L

33 sinsin =

00 sinsin =

2233 sinsin =

1122 sinsin =

0011 sinsin =

GROUND

Top of atmosphere

0

1

2

3

0

1

2

3

A

B

C

D

Refraction

We model the atmosphere as plane-parallel and consider a light ray incident atzenith angle on the top of the atmosphere.

Plane-parallel treatment valid for . At larger zenith angles a more exact

treatment that includes the curvature of the Earth is needed.

o600 (5.10)

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Scattering

Air molecules, dust and water vapour all scatter light. However, their

different sizes cause different effects on light.

3 regimes:-

1. Particle size, particles scatter all wavelengths equally

Examples: water droplets

This is why clouds and mist

appearwhite

opt>>a

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Scattering

Air molecules, dust and water vapour all scatter light. However, their

different sizes cause different effects on light.

3 regimes:-

2. Particle size, scattering power

Examples: fine dust,

cigarette smoke

This is why e.g. smoke ringshave a bluish tinge: blue light

is scattered by the smoke

particles more than red light.

opt~ a 1

American magician Harry Garrison

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Scattering

Air molecules, dust and water vapour all scatter light. However, their

different sizes cause different effects on optical light.

3 regimes:-

3. Particle size, scattering power

Examples: air molecules

Explains why the daytime sky is

blue, and why the sun appearsred at sunset (blue light

scattered out of line of sight)

opt

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Scattering

Air molecules, dust and water vapour all scatter light. However, their

different sizes cause different effects on optical light.

3 regimes:-

3. Particle size, scattering power

Examples: air molecules

Explains why the daytime sky is

blue, and why the sun appearsred at sunset (blue light

scattered out of line of sight)

Rayleigh scattering is anisotropic sky light is polarised

opt

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Scattering

We can analyse the loss of intensity due to scattering in the same way as for

absorption.

ldIdI =Scattering coefficient,

for Rayleigh

scattering by air molecules

4

Stars appear reddened

=D

d0

)( ll is larger forblue light

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Scattering

At visible wavelengths, the scattering of sunlight makes the sky so bright

that we must observe at night.

In the Far Infra-Red (and beyond) on the other hand, scattering makesa small enough contribution that we can observe during the day too.

(As we saw in previous sections, however, the thermal emission from the

daytime sky may be a problem at FIR wavelengths).

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Scintillation

Scintillation, or twinkling of

starlight is caused by turbulence

in the atmosphere. Air cells of

varying density and hencerefractive index are continually

passing across the line of sight to

a star, and changing the pattern of

illumination from the star whichreaches ground level. GROUND

r0 ~ 10cm

Turbulent cells

h ~ 7km

Incoming parallel rays

from point source

Overdense cells refract parallelrays; smears out light from

point source over seeing disk

Brighter patch

Darker patch

Typical scale length for the cells is r0 ~ 10cm, at a height of ~ 7km.

Hence, illumination from the star at any instant will not be uniform, but willconsist of brighter and darker patches, typically also ~ 10cm across. Cells

smear out the light from a point source over seeing disk:

Angular radius of seeing disk ~ = 3 arcsec.10 cm7 km

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Scintillation

The cells are continually moving across the line of sight, with a transverse

speed of about 10m/s.

If the telescope aperture, D ~ r0

We see rapid variations in position and brightness of the image as individualcells cross the line of sight.

If the telescope aperture, D >> r0

We see an image formed from many cells added together

Average brightness of the image is ~ constant, but there are rapid variations in

the position, size and shape of the seeing disk

Can be corrected using Adaptive Optics (see Honours Astronomy)

Scintillation timescale of variations ~ = 0.01 sec.10 cm

10 m/s

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ESO 3.6m telescope at La Silla, Chile

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Scintillation

Radio observations are also affected by scintillation this time not fromthe Earths atmosphere, but from turbulence in the interstellar and

interplanetary medium.

The typical size of turbulent cells is much larger (as is the wavelengthof the radiation).

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Example

At 700nm the zenith extinction (m700) is 0.08 magnitudes. Estimate theextinction (m400) at 400nm.

After correction for the atmosphere, a star is found to have a truecolour index (B-V)0 = -0.13. At a particular observatory, the zenithextinction (m) for the B band is 0.29 and for the V band is 0.17.

At what zenith distance would the star have the same apparentmagnitude in the two bands?