k-Distribution Lab

1. Using the Lorentz line shape, plot the absorption coefficient k as a function of wavenumber

in this interval. Subdivide the frequency range into a large number of equally spaced

subintervals such that < (the line half-width).

2. Divide the range in absorption coefficient into 50 equal intervals in log10

(k) and compute the

number of k in each intervals. Then compute the cumulative number n(0,k) in each interval

and plot n(0,k) as a function of log10

(k).

3. The cumulative probability function is defined by g(k) = n(0,k)/N, where N is the total number

such that g(0) = 0 and g(50logk) = 1. Plot k(g) in the g-domain (log10

(k) as a function of g).

4. Compute the spectral transmittance from the line-by-line approach employing an interval of

0.01 cm-1

and from the k-distribution method using quadrature points and weights from the

table below. Use a range in path length u from 10-5

to 10 g cm-2

. Compare the two results.

[Note that 1 g cm-2

is 2.24 104/M atm-cm where M is the molecular weight of the gas.]

Line-by-line

k-distribution

RT Results

The following slides are some of my radiative transfer

calculations and plots from Bohren and Clothiaux.

2-Stream Analytic = 1

1-D Monte Carlo = 1

3-D Monte Carlo, N = 106

= 1, o= 1

3-D Monte Carlo, N = 105

= 1, o= 1

F and F within cloudFo

Some Monte Carlo Results

Non-absorbing layer, transmittance

exponential

Two-stream

Monte Carlo, normal incidence

Monte Carlo, = 60°

Some Monte Carlo Results

Diffuse downward irradiance

Two-stream

Monte Carlo, normal incidence

Monte Carlo, = 60°

Some Monte Carlo Results

Upward and Downward Irradiance in cloud

Linear decrease

in F, F with

F > F0 !

But F,> F and F-

F is constant so

energy conserved

Recall 2-stream

solutions:

)1(*

*)1(

*)1(

g

CBF

CBF

Real (measured) irradiance in cloud

F- F nearly constant

with depth in cloud

FFLWC

Rainbows, Halos, and the Green Flash

• Read Ch. 8 in Bohren and Clothiaux

Halo

Halo

The Glory

Green Flash

The Primary Rainbow Angle

• The position of the primary rainbow is determined by the angle of minimum deviation of light scattered by a droplet after one internal reflection and refraction.

– Light rays scattered in the direction of the rainbow angle “pile up,” or focus.

• Measured from the anti-solar point (180° from the sun), it is about 42° for red light and about 40° for blue light.

Why does the rainbow form where it does?

Simple theory: rainbows and halos

• Geometric optics (x >>1): Angular distribution of scattered light can be derived from ray tracing

– Snel’s law of refraction

• Geometric optics predicts singularities, called caustics: points at which scattering for set of rays is infinite

• For both the rainbow and halo, caustics are associated with angles of minimum deviation of the scattering angle (with respect to the direction of incident beam):

• is scattering angle and i is angle of incidence. See figure on next slide.

.0

id

d

Drop Geometry for Rainbow

Snel’s law of refraction:

sin𝜗𝑖 = 𝑛sin𝜗𝑡

n is the refractive index, the ratio of the free-space speed of light, c, to the phase speed, 𝑣, of light in matter:

𝑛 =𝑐

𝑣

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Drop Geometry for Rainbow

𝜗is the scattering, or deviation angle. After one internal reflection and exit out of the drop, the ray has deviated by:

𝜗 = 𝜗𝑖 − 𝜗𝑡 + 𝜋 − 2𝜗𝑡 + (𝜗𝑖−𝜗𝑡)= 2𝜗𝑖 − 4𝜗𝑡 + 𝜋

Solve for extrema in 𝜗:𝑑𝜗

𝑑𝜗𝑖= 0;

cos𝜗𝑖 =𝑛2 − 1

3; 𝜗𝑖 ≈ 60for 𝑛 = 1.33

Primary rainbow angle: 𝜗 ≈ 13820

The Secondary Rainbow Angle

• The position of the secondary rainbow is determined by the angle of maximum deviation of light scattered by a droplet after two internal reflections and refraction.

• Measured from the anti-solar point (180° from the sun), it is about 52° for red light and about 54.5° for blue light.

• There is little or no light scattered by drops between the primary and secondary rainbow

– Alexander’s Dark band

Drop Geometry for Rainbow

For secondary rainbow, after two internal reflections,

𝜗 = 6𝜗𝑡 − 2𝜗𝑖Solve for extreme, now a maximum:

cos𝜗𝑖 =𝑛2 − 1

8; 𝜗𝑖 ≈ 72for 𝑛 = 1.33

Secondary rainbow angle: 𝜗 ≈ 129

22

A different set of drops is needed to produce the secondary rainbow.

The Rainbow

Rainbow angles

Rainbow Geometry

• Rainbows are seen when the Sun is behind the observer.

• The angular positions of rainbows relative to the observers line of sight are: Primary: 180 - 138 = 42

Secondary: 180 - 129 = 51

• Color separation is caused by dispersion, the dependence of refractive index on wavelength. n = 1.343 for violet (400 nm) and 1.331 for red (650 nm)

Angular separation of primary rainbow is 1.7 and for secondary, 3.1.

Because n is larger for blue than red, blue is deviated more than red: 𝜗𝑏𝑙𝑢𝑒 > 𝜗𝑟𝑒𝑑. Relative to viewer’s line of site, subtract from 180 so blue on inside for primary.

Reversed for secondary

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The Rainbow

blue light deviated more than red

The primary rainbow is a result of one internal reflection in a water drop.

The Rainbow

The Rainbow•anti-solar point is 180 (opposite direction) from the sun

•If sun is on horizon, top of rainbow is 42above horizon.

•What is the maximum possible elevation angle (angle measured between sun and horizon) of the sun to see a rainbow? 42

The Rainbow

It takes a collection of many drops to produce the full bow and to produce different colors.

The Rainbow

Blue light is deviated more than red but after turning around inside drop it appears lower in sky.

The secondary rainbow is a result of two internal reflectionsin a water drop..

The Rainbow

The Rainbow

Blue light appears on top in the secondary rainbow, after turning around twice inside the drop.

The Rainbow

The secondary rainbow is fainter (less intense) than the primary.

Supernumerary Bows

• Supernumerary bows cannot be explained without invoking interference

• Except at the rainbow angle, a horizontal line intersects the curve of scattering angle versus angle of incidence in two points

• Two rays may exit the drop in the same direction but follow different trajectories in a drop.

• the two waves corresponding to the two rays in the same direction are different in phase and hence interfere.

• interference depends on drop size (unlike the positions of rainbow)

a

x

singularity (caustic) at Dmin

Descartes: the rainbow is a caustic of deflected rays

supernumerary rainbowsinterference fringes near an angular caustic

For deflection D, droplet radius a, refractive index n:

intensity D a

1/ 3

Ai2Dmin D

2a

3

2 / 3 n21

1/ 2

4 n2 1/ 6

Supernumerary Bows

• George Biddell Airy's paper "On the intensity of light in the neighborhood of a caustic" was published in 1838.

• Showed that the intensity of light in a rainbow could be modeled using a cubic wave-front.

Polarization

• Electric and magnetic fields are transverse waves: they oscillate perpendicular to the direct of propagation (direction of energy transport).

• If you could follow the tip of the electric vector as it traces out its trajectory in a plane perpendicular to propagation direction, it would trace out an ellipse.

• Unpolarized light can be broken up into two perpendicular components of linearly polarized light.

• If E is the time harmonic electric field vector, we can write:

𝐄 = 𝐄 + 𝐄where 𝐄and 𝐄are the orthogonal perpendicular and parallel components of the field, respectively.

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Polarization of the Primary Rainbow:Polarization upon reflection

100% polarization occurs at the Brewster angle:

From diagram, when 𝜗𝑟 = 𝜗𝑖 + 𝜗𝑡 =𝜋

2,

reflected light is 100% polarized.

Using this relationship and Snel’s law, sin𝜗𝑖 =𝑛sin𝜗𝑡, we can derive Brewster angle, 𝜗𝐵, the angle of incidence for 100% polarization upon reflectance:

tan𝜗𝑖 = 𝑛

For water in air, m = 1.33, 𝜗𝐵= 53

This is close to rainbow angle of incidence; therefore, the rainbow is partially polarized. 𝜗𝑖 = 60 for primary rainbow,

not far from 𝜗𝐵.

n

𝜗𝑖 = 𝜗𝑟

𝜗𝑖 𝜗𝑟

𝜗𝑡

(unpolarized)(polarized)

(partially polarized)

The Halo

• The position of the halo is determined by the angle of minimum deviation of light refracted by a hexagonal ice crystal.

• Refraction through side faces (a 60° prism) make the 22° halo.

• Refraction through perpendicular faces (a 90° prism) make the 45°halo.

Note that to view the halo you need to be facing the sun.

Halo angles

The Halo

22° halo 45° halo

60°

90°

Geometry of the Halo

46

𝜗𝑖

𝜗𝑡𝜗𝑖

′ 𝜗𝑡′

𝛼𝜗is the scattering, or deviation angle; 𝛼 is the prism angle . After exit out of the crystal, the ray has deviated by:

𝜗 = 𝜗𝑖 − 𝜗𝑡 + (𝜗𝑡′ − 𝜗𝑖

′)𝐬𝐢𝐧𝝑𝒊 = 𝒏𝐬𝐢𝐧𝝑𝒕, 𝐬𝐢𝐧𝝑𝒕′ = 𝒏𝐬𝐢𝐧𝝑𝒊′, and 𝝑𝒊

′ = 𝜶 − 𝝑𝒕

Solve for extrema in 𝜗:𝑑𝜗

𝑑𝜗𝑖= 1 −

cos𝜗𝑖′cos𝜗𝑖cos𝜗𝑡′cos𝜗𝑡

= 0;

⟹ 𝜗𝑖= 𝜗𝑡′ and 𝜗𝑡 = 𝜗𝑖′

Minimum deviation angle:

𝜗𝑚 = 2sin−1 𝑛 sin𝛼

2− 𝛼

𝜗𝑚 ≈ 22 for 60 prism and 46 for 90 prism

The Halo

• The position of the halo is determined by the angle of minimum deviation of light refracted by a hexagonal ice crystal.

• Refraction through adjacent side faces (a 60° prism) make the 22° halo.

• Refraction through perpendicular faces (a 90° prism) make the 45° halo.

The Halo

• The primary difference between the rainbow and the halois the shape of the scatterer: spheres make the rainbow and hexagonal crystals make the halo. Composition (liquid water droplets versus ice crystals) also determines the rainbow and halo angles.

More information here.

The Green Flash

The Green FlashRarely observed but not rarely observable.

• Jules Verne’s novel Le Rayon-Vert (1982):

– One who has seen the Green Ray is incapable of being “deceived in matters of sentiment,” so that “he who has been fortunate enough once to behold it is enabled to see closely into his own heart and to read the thoughts of others.”

• Many more pictures here …

The Green Flash

• Rarely observed phenomenon but not rarely observable.

• consider the sun to be an infinite set of overlapping discs, one for each visible wavelength.

• When the sun is overhead discs coincide: sun is white

• as sun descends atmospheric refraction displaces the discs in angle by slightly different amounts, the red less than the violet

• Most of each disc overlap except for the discs at the extremes of the visible spectrum.

– the upper rim of the low sun is violet or blue, its lower

– Lower rim is rim red

– interior, the region in which all discs overlap, is still white.

The Green Flash

From Clouds in a Glass of Beer, C.F. Bohren, 1987.

The Green Flash

The Green Rim

• So why not the Blue Rim and Blue Flash?

• Selective scattering by atmospheric molecules and particles causes the spectrum of transmitted sunlight to shift toward longer wavelengths

– The perceived color of the Sun changes.

– Violet-bluish upper rim transformed to green

The Green Flash

• The separation of colors alone does not explain the flash or why it is observable

• The angular separation between the violet and red solar discs when the sun is on the horizon is about 0.01, too small to be resolved by the human eye. (~1 minute of arc resolution)

• In order to see the upper green rim of the sun requires binoculars or a telescope.

• depending on the temperature profile, the atmosphere can magnify the rim and yield a second image of it, a mirage, enabling it to be seen by the naked eye.– Green Flash simulation

• As the final sliver of the mirage-magnified upper green arc of the Sun crosses the horizon a flash of green light in the sky may be seen.

More Scattering by Water Drops

• All the colored atmospheric displays that result when water droplets or ice crystals are illuminated by sunlight have the same underlying cause:

– light is scattered in different amounts in different directions by particles larger than the wavelength of illumination.

– directions in which scattering is greatest depend on wavelength.

– When such particles are illuminated by sunlight, the result can be angular separation of colors even if scattering integrated over all directions is independent of wavelength.

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The Corona

• A series of concentric colored rings around the Sun or Moon

– Caused by thin cloud of droplets with narrow size distribution

– This corona is describe by the Fraunhofer diffraction theory, a simple approximation valid for particles large compared with the wavelength for scattering near the forward direction

– The resultant scattering has maxima (constructive interference) for 𝑥sinϑ= 5.137, 8.417, 11.62 where size parameter 𝑥 = 2𝜋𝑟 , and ϑ is the scattering angle.

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Edge of inner-most ring at

about 𝜗 ≈ 3 𝑥

Scattering Diagram for Diffraction by a Circular Disk

Scattering peaks locations:𝑥sinϑ= 5.137, 8.417, 11.62, ...

Coronal Color Dispersion

B&C Figure 8.22: Differential scattering cross section calculated by the Fraunhofer approximation for a sphere of diameter 10 μm. Green corresponds to 510 nm, red to 660 nm. Both cross sections are normalized to the value for green at 0.

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Problem

A corona is a colored ring of light around the moon or Sun when viewed through a thin cloud. The outer part of the disk often has a brownish-red tinge. If the diameter of the disk is 10 times that of the moon, what is the approximate radius of particles in the cloud?

http://www.atoptics.co.uk/droplets/corim1.htm

Corona Around Moon, March 2014

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Iridescent Clouds

• Iridescent clouds do not require a uniform, thin layer of cloud in front of Sun or Moon.

– Not necessarily arranged in any obvious geometrical pattern.

– Only organized into coronal rings when the droplet size is uniform right across the cloud.

• The thin edges of even thick clouds may be tinged with red and green, if they are in the direction toward the Sun.

• When you know where and how to look (usually toward the Sun), you can see it frequently.

• More images here.

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