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ESCI 241 – Meteorology
Lesson 20 – Optical Phenomena
Reading: Ahrens, Chapter 4
SNEL’S LAW
As the light passes from one medium into another, there is both reflection and
refraction.
The wave fronts bend as they cross from one medium into another. Therefore, a
ray will bend as it crosses from one medium into another.
The amount a ray is bent is given by Snel’s Law, which states that
sin θ 1 m2
=
sin θ 2 m1
where θ 1 is the angle of incidence (and reflection), and θ 2 is the angle of
refraction (see diagram below).
INDEX OF REFRACTION FOR AIR
The index of refraction for visible light in air depends on both the wavelength of
the light and on the temperature of the air.
The index of refraction for visible light in air decreases with increasing
temperature.
ο The diagram below shows m − 1 as a function of temperature for yellow light.
ο Though the change in index of refraction with temperature is slight, it is
enough to be important. Therefore
ο Light travels faster in warmer air, and slower in colder air.
ο Because the index of refraction is larger in colder air, from Snell’s law we
deduce that light bends toward colder temperatures.
The index of refraction for visible light in air decreases with increasing
wavelength.
ο The diagram below shows m − 1 as a function of wavelength.
ο The change of the index of refraction in air over the visible wavelengths is
very small, but shorter wavelengths do travel more slowly than do longer
wavelengths.
MIRAGES
Mirages are caused by the bending of light rays either upward or downward
near the ground.
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If the air near the ground is very hot, the light rays will bend upward
ο This causes objects far away to appear inverted, and underneath their actual
position
ο This results in an inferior image.
ο This is what causes hot asphalt to appear wet up ahead, because the sky
appears as an inferior image in the road.
If the air near the ground is very cold, the light rays bend downward
ο This causes far away objects to appear above their actual position
ο This results in a superior image
ο This can actually allow objects that are over the horizon to be seen.
PRIMARY RAINBOWS
Rainbows are caused by the refraction and reflection of light by rain and cloud
drops.
ο The index of refraction for water is much larger than that for air, and also
depends on wavelength.
ο The index of refraction is greater for shorter wavelengths, so therefore,
shorter wavelengths bend more.
The figures below shows the path of a light ray through a spherical water
droplet for a primary rainbow.
ο The angle θ i is the incidence angle.
ο The interior angle between the incoming and outgoing ray, θ ′, is called the
bending angle.
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ο The exterior angle, θ ′′ , is called the deviation angle.
Using Snel’s Law and geometry, the angle θ ′ for the primary rainbow can be
related to the incidence angle θ i, and this relationship is
1

sin θ i  − 2θ i .
m

θ ′ = 4 arcsin
Bending angle for primary rainbow
(1)
There are an infinite number of possible incidence angles, and therefore paths,
through the droplet. The plot below shows the deviation and bending angles as a
function of incidence angle.
ο Note that there is a certain incidence angle at which the deviation is a
minimum, and therefore, the bending angle is a maximum. At this incidence
angle, slight changes in the incidence angle θ i lead to no change in θ ′.
ο At this special incidence angle, the light beam will be very concentrated. At
other incidence angles, the light beam will not be as concentrated.
ο It is only at the this special incidence angle, where the deviation angle is a
minimum (and the bending angle is a maximum) that a rainbow will form.
To find this special incidence angle analytically, we need to differentiate
equation (1) with respect to θ i, set it equal to zero, and solve for θ i. Doing so
shows that for the primary rainbow
(
)
cos 2 θ i = m 2 − 1 3 .
Incidence angle for primary rainbow
Equations (1) and (2) are used together to determine the angle that a primary
rainbow subtends.
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(2)
The value of m depends on the wavelength of the light. The table below shows
these values for a few colors.
Color
violet
green
orange
red
λ(µm) 0.4047 0.5016 0.5893 0.7061
m
1.3427 1.3364
1.3330
1.3300
Since different colors have different indices of refraction, then each color will
have a slightly different angle after it leaves the droplet. Thus, the droplets act
as prisms to separate the colors.
For red light, the angle θ ′ is approximately 42°. This is the angle that a primary
rainbow subtends.
ο The shorter wavelengths have a smaller θ ′, because they have a larger index
of refraction. Therefore, red appears as the outer band of a primary
rainbow, with the other colors following in order of decreasing wavelength.
If the Sun angle is larger than 42°, a primary rainbow cannot be seen by a
ground-based observer.
SECONDARY RAINBOWS
It is also possible to have a path through the water droplet that has two internal
reflections, such as that shown below.
ο For this case, the equation for the bending angle is
1

sin θ i  .
m

θ ′ = 180° + 2θ i − 6 arcsin 
Bending angle for secondary rainbow
As before, there are an infinite number of such paths. However, there is one
special path for which dθ ′ dθ i = 0 , which gives an incidence angle of
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(3)
(
)
cos 2 θ i = m 2 − 1 8 .
Incidence angle for secondary rainbow
(4)
For red light, the angle θ ′ is approximately 50°. This is the angle that a
secondary rainbow subtends.
The shorter wavelengths have a larger θ ′ in this case.
ο
Therefore, red appears as the inner band of a secondary rainbow, with the
other colors following in order of decreasing wavelength.
Secondary rainbows are much dimmer than primary rainbows, because every
time the light refracts and/or reflects, some of the radiance is lost out of the
beam.
If the Sun angle is larger than 50°, a secondary rainbow cannot be seen by a
ground-based observer.
PRISMS
A prism is a transparent wedge of material which allows light to pass through it,
and which separates the colors of the light.
There are many paths that the light can take through a prism, but only one path
will give a concentrated beam. This is the path that passes symmetrically
through the prism.
The deviation angle is related to the angle of the prism and the index of
refraction of the prism via
θ ′ = 2 arcsin[m sin ( A 2)] − A .
Minimum deviation angle through a prism
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REFRACTION OF LIGHT THROUGH ICE CRYSTALS
Ice crystals come in many different shapes, or habits.
ο The habit depends on the temperature at which the crystal formed.
However, all ice crystals are six-sided (hexagonal).
Optical phenomenon like halos and sun dogs are associated with columnar or
plate-like ice crystals, like that shown below.
There are two possible symmetric paths of the light through the column, and
these are shown below.
ο One path is for the light to pass laterally across the cyrstal, such as shown on
the left.
In this case, the crystal acts like a prism with an angle of 60°, and using
the index of refraction for ice (m = 1.31) equation (14) gives an angle of
bending of about 22°.
ο The other possible path is for the light to pass longitudinally along the
crystal, such as shown on the right.
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In this case, the crystal acts like a prism with an angle of 90°, and using
the index of refraction for ice (m = 1.31) equation (14) gives an angle of
bending of about 46°.
A third path (shown below) is impossible for ice crystals, because the index of
refraction for ice is too large, and therefore, equation (13) gives a nonsensical
result for sin θ i.
HALOS, SUN DOGS, AND SUN PILLARS
Halos, sun dogs (also called perihelia), and sun pillars are caused by the
reflection or refraction of sunlight from ice crystals.
Halos and sun dogs are displaced either 22° or 46° from the Sun, depending on
which path the light takes through the ice crystal (see below).
ο Halos are arcs around the Sun, and occur when the ice crystals are randomly
oriented.
ο Sun dogs only appear to either or both sides of the sun, and occur when the
ice crystals are primarily oriented vertically.
ο Sometimes, halos and sun dogs appear together.
ο Halos of other dimensions are sometimes observed, as well as tangent and
circumzenithal arcs, and are associated with oddly shaped ice crystals which
have a pyramidal cap, or to spinning ice crystal.
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Sun pillars are caused by reflection, rather than refraction, and appear as a
column of light above (and sometimes below) the sun.
ο For sun pillars, the ice crystals are more plate-like, rather than columnar,
and are oriented with the flat plate-like surfaces horizontally. They then act
together like a large mirror.
The diagram below shows some (though not all) of the various optical
phenomena associated with ice crystals.
ο a. are sun dogs
ο b. is the 22° halo
ο c. is the 46° halo
ο d. are sun pillars
ο e. is the upper tangent arc
ο f. is the circumzenithal arc
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EXERCISES
1. You are on a planet where the rain drops are made up of a liquid with an index
of refraction of 1.44. You notice a rainbow. Assuming it is a primary rainbow,
what is the bending angle?
2. If ice crystals were pentagonal (5-sided) columns, list the angles of all possible
haloes.
3. You are on a planet with strange clouds. You suspect the clouds are made up of
some sort of crystal, and other measurements lead you to believe that the index
of refraction of the crystal is 1.05. You have observed halos of 2.4°, 5.9°, and
16.9°. What would you guess the shape of the crystals are?
4. You are on a planet where the clouds are made up of droplets of an unknown
liquid. You notice a rainbow (assume it is a primary rainbow) with an angle of
15.0°. What is the index of refraction (to two decimal places) of the cloud
droplets?
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