Refraction

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The apparent position of an object in the sky may be
changed by several different physical effects.

One of these is
refraction.

The speed of light changes as it passes through a
medium such as air.

We define the **refractive index** of any
transparent medium as 1/v,

where v is the speed of light in that
medium.

The speed of light in air depends on its temperature
and its pressure,

so the refractive index of the air varies in
different parts of the atmosphere.

Make a simple model of the atmosphere as *n*
layers of uniform air above a flat Earth,

with a different
velocity of light v*i *for
each layer (*i* from 1 to *n*).

Apply Snell's Law of
Refraction at each boundary.

At
the first boundary, sin(i_{1}) / sin(r_{1}) = v_{0
}/ v_{1 }.

At the next boundary, sin(i_{2})
/ sin(r_{2}) = v_{1 }/ v_{2 }, and so on.

But, by simple geometry, r_{1} = i_{2},
r_{2} = i_{3} and so on.

So we have

sin(i_{1}) = (v_{0 }/
v_{1}) sin(r_{1})

=
(v_{0 }/ v_{1}) sin(i_{2})

=
(v_{0 }/ v_{1}) (v_{1 }/ v_{2})
sin(r_{2})

=
(v_{0 }/ v_{2}) sin(r_{2})

=
..........

=
(v_{0 }/ v* _{n}*) sin(r

In other words, the refractive indices of the
intervening layers all cancel out.

The only thing that matters is
the ratio between v_{0}

(which
is c, the speed of light in vacuum)

and v* _{n}* (the
speed in the air at ground level).

Now r* _{n}* is the

and i

So sin(z) = (v

Refraction has no effect if a star is at the zenith
(z=0).

But at any other position, the star is apparently *raised*;
the effect is greatest at the horizon.

Define the **angle of refraction **R by **R = z -
z'**.

Rearrange this as z = R + z'.

Then sin(z) = sin(R)
cos(z') + cos(R) sin(z').

We assume R will be small, so, approximately,

sin(R)
= R (in radians), and cos(R) = 0.

Thus, approximately,

sin(z)
= sin(z') + R cos(z').

Divide throughout by sin(z') to get

sin(z)/sin(z') = 1 + R/tan(z')

which is to say,

v_{0}/v* _{n}*
= 1 + R/tan(z').

So we can write

R = (v

We write this as

where k = (v

Here v_{0} is c, the velocity of light in a
vacuum, which is constant.

But v* _{n}* depends on
the temperature and pressure of the air at ground level.

At "standard" temperature (0°C = 273K) and pressure (1000 millibars),

k = 59.6 arc-seconds.

The formula in the

k = 16.27" P/(273+T)

where P is in millibars, and T is in °C.

Exercise:

A star is at Right Ascension 5h
0m and declination +26°20'.

The latitude φ
is +56°20'.

Local Sidereal Time is 5h 0m.

Atmospheric
pressure is 1050 millibars,

and the temperature is +5°C.

What is the star’s true altitude?

How much will the star’s
image be shifted by atmospheric refraction,

and in which
direction?

What will be the star’s
Right Ascension and declination,

corrected for refraction?

Click here for the answer.

At large zenith
distances, this model is inadequate.

The amount of refraction
*near the horizon *is actually determined observationally.

At
standard temperature and pressure,

refraction at the horizon
(**horizontal refraction**) is found to be 34 arc-minutes.

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The Moon

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Sunrise,
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