Atmospheric refraction and slit transmission

S. Pedraz



Although the effects of atmospheric differential refraction on astronomical observations are well known, not always are enough considered when observing. This phenomenon is a source of errors in astrometric works, in measurements of relative line intensities, continuum profiles, etc ...To help observers to minimize the effect, or at least to have an idea of how much are affected the observations, we have developed this calculator to obtain the values of absolute or differential refraction and to find out which part of the light is transmitted through a slit, depending on the seeing value, wavelengths and on the width and orientation of the slit.

Atmospheric refraction could be determined theoretically by tracing the path of light through the Earth's atmosphere, wherein the refraction will be just the difference in the directions of the light before it enters the atmosphere, and as seen at the telescope. A good approximation is that the Earth's atmosphere is spherically symmetric, in which case, the atmospheric refraction R(λ) at wavelength λ is given exactly by the integral (Green 1985)

R(λ) = r0 n0 sin z0 ⌠ n0
⌡1		 dn
————————————
n(r2 n2 - r02 sin2 z0)1/2

where r0 and n0 are the geocentric distance and the index of refraction of air at the observing site. z0 is the apparent zenith distance of the star. At an arbitrary geocentric distance r along the incoming path of light, the refraction index of air is n. This index can be expressed in terms of the densities of air at that arbitrary point (ρ) and at the Earth's surface (ρ0)

n = 1 + (n0 - 1) ρ
——
ρ0

According to Stone(1996) this equation can be substituted in the previous one, and expanded into a power series

R(λ) = B1 tan z0 + B2 tan3 z0 + B3 tan5 z0 + ...

Retaining the first two terms, which is a very good approximation for z0 < 75 o

R(λ) = κ γ (1 - β) tan z0 - κ γ ( β - γ/2) tan3 z0

where

γ = n0 - 1 ,          β = H0
——
r0

H0 is the height of an equivalent homogenous atmosphere and is defined in terms of the air density at Earth's surface (ρ0) and the density (ρ) at a height (h) above it. Assuming to be exponential the radial density dependence, H0 will be the scale height h0 of the atmosphere

H0 = 1
——
ρ0 		⌠ ∞
⌡0		 ρ dh,         ρ = ρ0 e -h/h0

Calculating H0 by numerical approximations, β can be rewritten as

β = 0.001254 ( 273.15 + t
——————
273.15		 )

For a spherical Earth κ=1.0, but assuming the Earth's geoid or equipotential surface at sea level, κ is the ratio of the gravity g0 at the observing site (latitude φ) to the sea-level gravity g at the Earth's equator.

κ = g0/g = 1 + 0.005302 sin2 φ - 0.00000583 sin2 (2φ) - 0.000000315 h

Computing κ and β, we only need the refraction index of air at the observing site (n0) to calculate the atmospheric refraction at a wavelength λ and at a apparent zenith distance z0.
Using the empirical approximations given by Owens (1967), n0(λ) can be computed accurately from local atmospheric parameters as temperature (t), pressure (P) and relative humidity (RH).
First we need to differentiate between dry air pressure (Pa) and water vapour pressure (Pw). With temperature expressed in Kelvin (T = 273.15 + t), saturation pressure (Ps) can be calculated in millibars

Ps = − 10474 + 116.43 T − 0.43284 T2 + 0.00053840 T3 ; Pw = RH
———
100		 Ps ; Pa = P - Pw

Refraction index of air (n0), can be computed with Owen's formulae for a wavelength (λ) in microns (or wave number σ = 1 / λ) :

(n0(λ) − 1) × 108 = ( 2371.34 + 683939.7
——————
(130 - σ 2) 		+ 4547.3
——————
(38.9 - σ 2) 		) Da + (6487.31 + 58.058 σ2 − 0.71150 σ4 + 0.08851 σ6) Dw



Da = ( 1 + Pa ( 57.90 × 10-8 − 0.0009325
————————
T   		+ 0.25844
——————
T 2		 ) ) Pa
———
T  



Dw = ( 1 + Pw (1 + 3.7 × 10-4 Pw) ( − 2.37321 × 10-3 + 2.23366
————————
T   		− 710.792
————————
T 2 		+ 77514.1
————————
T 3 		) ) Pw
———
T  

These approximations are usable over the temperature range -23 < t < 47 oC , pressure 0 < P < 4 atm, relative humidity 0 < RH < 100% and wavelength 0,2302 < λ < 2,0586 µ.


Since we have the formulae to calculate the atmospheric refraction (absolute) that affects the position of objects in the sky for a given wavelength, we can calculate the differential refraction, i.e. the difference between the values of refraction for two different wavelengths at the same zenith distance (use the calculator ).
Taking into account these different apparent positions, we can calculate also how much it affect to spectroscopic observations. We can find out what is the fraction of the light transmitted by a slit according to the wavelength and slit size and orientation.
Frequently the object is centered in the spectrograph slit with the aid of a guiding camera or any acquisition system that use to be optimize in the optical range (i.e. ~5000 angs) and the effect is that a higher fraction of light is lost for shorter or longer wavelengths.
Because atmospheric differential refraction is perpendicular to the horizon, if the slit or rectangular aperture is oriented along this direction, the proportion of light transmitted through the slit is the same for any wavelength. But, that is not the case when rotating the slit.
When the star is on the meridian, to align the slit with the vertical, the position angle must be 0o (measured from north through the east). But for any other position of the star, the required angle of the slit is equal to the parallactic angle

To calculate the slit transmission we follow Filippenko (1982) reasoning. Considering that the convolution of guiding errors and seeing produces a point spread function whose surface brightness µ(r) is a Gaussian with dispersion σ

µ(r) =
1
————
2 π σ 2		    e -r2/2 σ 2

the fraction of the transmitted light, through a slit of width 2a and length 2b for the wavelength used to center de star is:
I(λ) = ⌠ b
⌡-b		 ⌠ a
⌡-a		 µ(r) dx dy =
1
————
2 π σ 2		 ⌠ b
⌡-b		 ⌠ a
⌡-a		 e -(x2+y2)/2 σ2 dx dy


I(λ) = (
1
————
(2 π)1/2 σ 		⌠ a
⌡-a		 e -x2/2 σ2 dx ) × (
1
————
(2 π)1/2 σ 		⌠ b
⌡-b		 e -y2/2 σ2 dy )


Due to differential refraction, the image of the star for a different wavelength will be displaced by xo arc seconds along the vertical. Lower or higher elevations for shorter or longer wavelengths. So, if the star is displaced from the slit center along the x direction and assuming that σ is independent of wavelength (which is not exactly correct), the transmitted light will be:

I(λ) = 1
——
2 		(
1
————
(2 π)1/2 σ 		⌠ (a-xo)
⌡-(a-xo)		 e -x2/2 σ2 dx   + 
1
————
(2 π)1/2 σ 		⌠ (a+xo)
⌡-(a+xo)		 e -x2/2 σ2 dx )   ×  (
1
————
(2 π)1/2 σ 		⌠ b
⌡-b		 e -y2/2 σ2 dy )

hence, for an arbitrary slit position angle, the difference between this angle and the parallactic angle will give us the value of the angle of the slit position relative to the vertical (ω). So, for a differential refraction of X arc seconds, the displacement of the image, from the center, along the slit is xo = X cos(ω), and perpendicular to it yo = X sin(ω). And the final integral we have to calculate, to obtain the transmission through the slit for a wavelength that is not the one used to center the star, will be:

I(λ) = 1
——
2 		(
1
————
(2 π)1/2 σ 		⌠ (a-xo)
⌡-(a-xo)		 e -x2/2 σ2 dx   + 
1
————
(2 π)1/2 σ 		⌠ (a+xo)
⌡-(a+xo)		 e -x2/2 σ2 dx )   ×  
1
——
2 		(
1
————
(2 π)1/2 σ 		⌠ (b-yo)
⌡-(b-yo)		 e -x2/2 σ2 dx   + 
1
————
(2 π)1/2 σ 		⌠ (b+yo)
⌡-(b+yo)		 e -x2/2 σ2 dx )


The atmospheric refraction and slit transmission calculator use these formulae and take the temperature, pressure and relative humidity from our local weather station. The user has to input the zenith distance (or airmass), set the slit position and size, and select two wavelengths.

Filippenko A.V., 1982 PASP 715_721
Green R.M., 1985, Spherical Astronomy (Cambridge University), p.87
Owens J.C., 1967, Appl.Opt. 6, 51
Stone R.C., 1996, PASP 108,1051