University of Ljubljana

Faculty for mathematics and physics

Rainbow

Author: Miha Mihovilovi£

Mentor: doc. dr. Igor Poberaj

June 7, 2007

Abstract

In this seminar I will talk about rainbow. In the rst part, I will discuss geometrical optical

theory of the rainbow and with it try to explain some basic properties of this meteorological

phenomena such as sequence of colours, the secondary bow, Alexander's black band... In the

second part of this seminar, I will take a look of the diraction theory of the rainbow and prove

the presence of supernumerary bows.

Contents

1 Introduction 1

2 Observations 2

3 Geometrical Optics 3

3.1 Descartes theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33.2 Intensity of emerged light . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

3.2.1 Reection and Refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2.2 Geometrical Factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.3 Monte Carlo Computer simulation of the rainbow. . . . . . . . . . . . . . . . . . . . 10

4 The diraction theory of the rainbow 12

4.1 Colours in rainbow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

5 Raindrop size distribution 17

6 Conclusion 17

7 Thanks 17

1 Introduction

The rainbow is probably the best-known and the most interesting phenomenon of atmospheric opticsand has excited attention from earliest times. The rainbow has an important position in physics.Some of the most powerful tools of mathematical physics were devised explicitly to deal with the

1

problem of the rainbow and with closely related problems. The rainbow has served as a touchstone fortesting theories of optics. It is often thought that the scientic description of the rainbow is a simpleproblem in geometrical physics, which was solved long ago. This is not so. A satisfactory quantitativetheory of the rainbow has been developed only in the last few years and involves much more thangeometrical optics[1]. The rst initial step in the theory of rainbow was taken by Descartes in 1637in an appendix to his Discours de la Methode[2] where he said:" The rainbow is such a remarkablemarvel of Nature... that I could hardly choose a more suitable example for the application of mymethod. "[1].

2 Observations

We see a rainbow in the sky usually only when the sun is behind us and we look at a cloud in frontof us from which rain is falling, and is illuminated by the sunlight in certain positions. Rainbows areusually formed by raindrops rather than cloud droplets, though this is not universally the case. Thebow itself appears as the arc of a circle, rarely complete, the centre of which is at the "anti-solarpoint", which represents the intersection of plane of the rainbow and the ray of light from the sunpassing through the eye of the observer (See gure 1).

Figure 1: First scheme shows the position of the sun and the observer so that he or she could see therainbow. In the gure is also denoted the anti-solar point and position of supernumerary bows(Figurefrom[3]). The second picture shows the cover of Descareses Discours de la Methode.(Figure from[4])

The rainbow contains all colours of the spectrum, more or less impure, with red lying on theoutside of the arc and violet on the inside. The outer edge is more sharply dened than is the violetinner one and within the arc, the sky is brighter than it is without.

The principal bow just described is known as the primary bow. Outside it is often formedanother bow, which is concentric to it. It is called the second bow and is usually much fainter thanthe primary. It consists of a similar range of colours as in the primary bow, but the order is reversed.Red appears on the inside and violet on the outside. The two rainbows are thus arranged red to redand between them, the sky is darker than it is either within the primary bow or outside the secondary.This dark band is often called the Alexander's dark band, from Alexander of Aphrodisias[5], whorst remarked upon it.

Within the primary bow, there are often seen pale bows, usually faint pinks and greens but theircolour is variable. These bands are called supernumerary bows(See gure 2.).

2

Figure 2: Full featured rainbow in Wrangell-St. Elias national Park, Alaska.(Figure from[4])

3 Geometrical Optics

3.1 Descartes theory

A basic analysis of the rainbow can be obtained by applying the laws of reection and refraction[6] tothe path of a ray through a droplet. Because the droplet is assumed to be spherical, all directions areequivalent and there is only one signicant variable: the impact angle θ (See gure 3) or alternativelythe impact parameter, which is dened as b = sin θ. At the surface of the droplet the incident rayis partially reected and this reected light we shell identify as the scattered rays of Class 1. Therest of light is transmitted with a change in direction caused by refraction. At the next surfacelight is partially transmitted (we get scattered rays of Class 2) and partially reected. At the nextboundary, the reected ray is again split into reected and transmitted components, and processcontinues indenitely (See gure 3). The Class 1 rays represent direct reection by the droplet andthose of Class 2 are directly transmitted through it. Rays of Class 3 are those that escape the dropletafter one internal reection, and they make up the primary rainbow. The class 4 rays, which maketwo internal reections, give rise to the secondary rainbow. Rainbows of higher order are formed byrays making more complicated passages, but they are not ordinarily visible.

-2 -1.5 -1 -0.5 0.5 1 1.5 2

-2

-1.5

-1

-0.5

0.5

1

1.5

2

Class 1

Class 2

Class 3

Class 4

Figure 3: Schemes show path of a monocromatic ray through a droplet.

3

Now we would like to calculate direction of class 3 rays in which they leave droplet. Thereforewe dene a quantity D, which tells us the deviation of outgoing ray (class 3) from direction of theincident ray. From gure 3 it is easy to see, that D equals:

D1 = θ − φ + π − 2φ + θ − φ = 2θ + π − 4φ(θ), (1)

where we must remember, that angle φ is determined with the angle θ and Snell law. The deviationD1 of scattered rays varies over wide range of values as a function of the impact parameter. Asthe impact parameter increases and incident rays are displaced from the center of the droplet, thescattering angle decreases. This trend, however, does not continue as the impact parameter isincreased to its maximum value. Instead, the deviation D passes through a minimum and thereafterstarts to increase (See gure 4).

The deviation of higher order scattered rays (Class 4...) can be determined in a similar way asD1 and can be written as:

Di = 2(θ − φ) + i(π − 2φ), i = 2, 3, 4, ... (2)

For rays of Class 4 the deviation D2 is zero when the impact parameter is zero. As the impactparameter increases so does D2 until it reaches maximum. After that, trend is reversed and scatteringangle starts to decraese with increasing impact parameter (See gure 4).

0.2 0.4 0.6 0.8 1b

0.5

1

1.5

2

2.5

3

D1,2Hb,nL

Alexander dark band

Primary Rainbow

Secondary Rainbow

D1

D2

Figure 4: Graph shows how deviations of class 3 (D1) and class 4 (D2) rays change with the impactparameter b.

In a sunlight the droplet is uniformly illuminated at all impact parameters at the same time.Therefore light is scattered virtually in all directions (See gure 7), but all scattered rays do notform a rainbow. If we want to see a rainbow, the intensity of the scattered light must be enhancedin one special direction (This special direction is called the rainbow angle and depends on the colorof light.). Therefore rays with dierent impact parameters must be scattered at almost same angles.This can happen were the scattering angle varies most slowly with the impact angle θ. That is ofcourse around the extreme deviation Di(b), where

dDi(b)dθ = 0. If we apply this condition to (1), we

get the condition for the primary rainbow:

Dextreme1 (n) = 2 arcsin

(√4− n2

3

)+ π − 4 arcsin

(√4− n2

3n2

), (3)

where is n(λ) the refractive index of water for dierent colours of light. A condition for the secondor higher rainbow can be derived in a similar manner from (2) and equals:

Dextremei (n) = 2π −

[iπ + 2 arcsin

(√(i + 1)2 − n2

(i + 1)2 − 1

)− 2(1 + i) arcsin

(1n

√(i + 1)2 − n2

(i + 1)2 − 1

)](4)

4

-2 -1 1 2

-1

-0.5

0.5

1

1.5

8Primary Rainbow<

-2 -1.5 -1 -0.5 0.5 1 1.5

-2

-1.5

-1

-0.5

0.5

1

8Secondary Rainbow<

Figure 5: Figures show scattered rays of class 3 and class 4, when droplet is uniformly illuminatedat all impact parameters at the same time.

With (3) and (4) we are now able to calculate scattering angles for dierent colour bands of theprimary and the secondary rainbow:

Colour λ nλ Dextreme1 (n) Dextreme

2 (n)Red 400nm 1.33141 139.57o 129.53o

Violet 700nm 1.24451 137.41o 126.14o

We can also calculate the width of both rainbows from dierence between these angles:

∆D1 = D1(nv)−D1(nr) = 1.88o (5)

∆D2 = D2(nv)−D2(nr) = 3.39o (6)

From results we can see, that the second rainbow is almost twice as broad as the primary rainbow.We must be aware, that these results would be true only if rays of incident sunlight were exactlyparallel. If we would want to be more precise, we would have to take in to account the apparentdiameter of the sun.

In gure 4 we can se that there are no rays of Class 3 and Class 4 scattered into the angularregion between 130 and 138 degrees. This region is called Alexander's dark band. If only rays ofClass 3 and Class 4 would exist this region would be quite black. However rays of Class 1 and ofhigher order are scattered into this region. Therefore Alexander's band is not totally black but stilldistinctly darker than the sky elsewhere (See gure 6).

3.2 Intensity of emerged light

Until now we have within geometrical optics shown how rays of light travel through a droplet andhow because of dierent paths of rays with dierent wavelengths the rainbow is formed. We shellnow take the theory of the rainbow a step further while still remaining within the same theory, andtry to tell something about the intensity of scattered light.

The intensity of light which emerges from a raindrop after internal reections depends at least offour factors: The coecient of reection and refraction at a water surface corresponding to variousangles of incidence, the so called geometrical factor, the variation in the deviation in dependence ofincidence angle (dD

dθi) and diraction and interference. Of all this four factors we have in theory of

Descartes considered only the variation in the deviation. If we leave diraction and interference forlater discussion and consider only geometrical optics, we are left with rst two factors to discuss.

5

Figure 6: Figures show the primary and the secondary rainbow. It is also clearly seen, that the skybetween rainbows is much darker than sky within rst rainbow. This dark area is called Alexander'sdark band. (Figures from[3]).

3.2.1 Reection and Refraction

The coecient of reection of light at a water surface depends upon the polarisation of the light.It is therefore necessary to treat light which is polarized in the plane of incidence (TE polarization)and that which is polarized at right angles (TM polarization), separately.

Using Fresnel Law[6], the intensity of reected and transmitted light of TE polarization can bewritten as:

R⊥ =sin2(θr − θi)sin2(θr + θi)

=Ir

Ii(7)

T⊥ = 1− sin2(θr − θi)sin2(θr + θi)

=It

Ii(8)

From graphs of (7) and (8) shown in gure 7, we can see, that for water with n = 4/3 at zero impactangle (θi = 0o) 98% of the light is transmitted and only 2% reected. However for grazing incidence(θi = 90o) all light is reected.

0.25 0.5 0.75 1 1.25 1.5Θi

0.2

0.4

0.6

0.8

1

R,T 8TE Polarisation<

Fraction Reflected

Fraction Refracted

0.25 0.5 0.75 1 1.25 1.5Θi

0.2

0.4

0.6

0.8

1

R,T 8TM Polarisation<

Fraction Reflected

Fraction Refracted

BREWSTER ANGLE

Figure 7: Figures show reection (R) and transmission (T) coecients at dierent incident anglesθi for TE and TM polarisation of the sunlight.

When the incident light is polarized at right angles to the plane of incidence, Fresnel found that

6

the intensity of the reected and transmitted light was given by:

R‖ ∝ r2‖ =

tan2 (θi − θr)tan2 (θi + θr)

=Ir

Ii(9)

T‖ = 1− tan2 (θi − θr)tan2 (θi + θr)

=It

Ii(10)

Graphs of (9) and (10) are also shown in gure 7. At grazing incidence (i = 90o) again all the light isreected and none refracted. At normal incidence (i = 0o) again 2 per cent is reected, as with thelight polarized in the plane of incidence. In the case of light polarized at right angles to the plane ofincidence, however, a coecient of reection falls to zero when θi + θr = π

2 since tan(θi + θr) is theninnite. This phenomena is known as polarization by refraction[6] because, at that special Brewsterangle1 only light polarized in the plane of incidence is reected, that polarized at right angles isbeing totally refracted.

We may also notice, that the coecients of reection and refraction are unaltered if the anglesof incidence and refraction are interchanged.

With these ndings we are now able to calculate the intensity of light of the rainbow. Light whichfalls onto the primary rainbow is refracted on entry into the raindrop, is reected at the back surfaceand is refracted again on emerging (See gure 3). If the angle of incidence of the light on the frontsurface is θi and the angle of refraction is θr , then the angle of incidence under which the lightstrikes the back surface will be θr . Some light will be there refracted with the angle of refractionθi. The reected light will then fall again on spherical surface under the angle of incidence θr. Lightwhich will at that point come out of the droplet under refraction angle θi, will form the primaryrainbow. Intensities of emitted TE and TM polarized light are therefore given by:

I1⊥ =

(1− sin2(θr − θi)

sin2(θr + θi)

)2

︸ ︷︷ ︸Refraction on front and last surface

Reection on back surface︷ ︸︸ ︷sin2(θr − θi)sin2(θr + θi)

(11)

I1‖ =

(1− tan2(θr − θi)

tan2(θr + θi)

)2

︸ ︷︷ ︸Refraction on front and last surface

Reection on back surface︷ ︸︸ ︷tan2(θr − θi)tan2(θr + θi)

(12)

The total intensity of light which forms primary rainbow is sum of both partial intensities:

I1 = I1⊥ + I1

‖ (13)

The intensities of TE and TM polarized light for secondary bow can be calculated in the sameway. The only dierence is that there are two internal reections instead of one. This gives us:

I2⊥ =

(1− sin2(θr − θi)

sin2(θr + θi)

)2

︸ ︷︷ ︸Two refractions

(sin2(θr − θi)sin2(θr + θi)

)2

︸ ︷︷ ︸Two internal reections

(14)

I2‖ =

(1− tan2(θr − θi)

tan2(θr + θi)

)2

︸ ︷︷ ︸Two refractions

(tan2(θr − θi)tan2(θr + θi)

)2

︸ ︷︷ ︸Two internal reections

(15)

The total intensity is of course:

I2 = I2⊥ + I2

‖ (16)

1For water, with a refractive index of 4/3, the Brewster angle θB is of 53.13o.

7

0.25 0.5 0.75 1 1.25 1.5Θi

0.05

0.1

0.15

0.2

0.25

I 8Primary Rainbow<

IÈÈ

I¦

I

P

0.25 0.5 0.75 1 1.25 1.5Θi

0.02

0.04

0.06

0.08

0.1

I 8Secondary Rainbow<

IÈÈ

I¦

I

S

Figure 8: Graphs show how intensities of light of class 3 and class 4 rays depend on angle of incidence.With P and S are marked angles of incidence, which correspond to primary and secondary rainbow.

How intensities of light, which form the primary and the secondary rainbow depend of the angleof incidence θi is shown in gure 8. The angle of incidence corresponding to the primary rainbow is59.3o and is in gure marked by line labeled P. We can see, that the emergent light will consist of96.2 per cent of light polarized in the plane of incidence and only 3.8 per cent of light polarized atright angles. This means, that the light of primary rainbow is strongly polarized.

The angle of incidence for the secondary rainbow is 71.83o. From corresponding graph we cansee, that the light, which forms the secondary bow, consists of 90.3 per cent of TE polarized lightand 9.7 per cent of TM polarized light. That means, that the secondary bow is less strongly polarizedthan the primary bow, but still quite polarized.

Eventually we can also calculate the ratio between intensities of both rainbows:

I2(71.83o)I1(59.39o)

=0.03890.09123

= 0.4263 (17)

This result tells us, that the secondary rainbow is much fainter than the primary rainbow.

3.2.2 Geometrical Factor

We shell treat this factor seperately from reection and refraction. These two coecients will beadded in the end as factors to nal expression .

Let us consider a sunlight which falls upon the raindrop between angles of incidence θi and θi+δθi

(See gure 9). Between these angles of incidence the drop (with radius a) presents an area

δS = πa2 sin(2θi)δθi (18)

to the incident light. Suppose that on emergence these rays lie between the angles θE and θE + δθE ,measured from the direction from which the incident light comes (See gure 9). The solid angle intowhich the light is projected will be:

δΩ = 2π sin θEδθE (19)

If I0 is the intensity of light, which falls upon unit area at right angles to its direction of travel,then the intensity entering unit solid angle on emerging from the sphere after internal reection willbe:

I =I

δΩ= I0

a2 sin(2θi)sin θE︸ ︷︷ ︸

Geometric fac.

δθi

δθE︸︷︷︸Cartesian fac.

(20)

8

Figure 9: First gure shows the cross-section which is illuminated by rays, which fall upon the dropbetween angles θi and θi + δθi. Second gure shows the solid angle into which this light is projected.

This intensity will determine the brightness of the image formed in the eye. When δθEδθi

= 0, we havethe Cartesian condition for a rainbow.

Considering that angles of emergence of Class 3 and Class 4 rays are given by (See gure 10):

θClass 3 = 4θr − 2θi = θE1 (21)

θClass 4 = π + 2θi − 6θr = θE2, (22)

then intensity of light, which is emitted from droplet after one or two internal-reections, can bewritten as:

I1 = I0a2 sin(2θi)

sin θE1

14 cos θi√n2−sin2 θi

− 2(23)

I2 = I0a2 sin(2θi)

sin θE2

12− 6 cos θi√

n2−sin2 θi

. (24)

Figure 10: Figures show paths of Class 3, rays which have one internal reection, and of Class 4rays, which have two internal reections.

Now we have all the necessary parts (geometrical, reection and refraction factor) to write nalexpression for the intensity of light which is emitted after reections in the drop. If we multiply (23)by (13) we get the intensity of class 3 rays, which have one internal reection:

IClass3 = I0a2 sin(2θi)

sin θE1

14 cos θi√n2−sin2 θi

− 2

(I1‖ + I1

⊥

), (25)

9

where θE1 = 4 arcsin(

sin θin

)− 2θi. Similarly we can multiply (24) by (16) and get the intensity of

class 4 rays with two internal reections:

IClass4 = I0a2 sin(2θi)

sin θE2

12− 6 cos θi√

n2−sin2 θi

(I2‖ + I2

⊥

), (26)

where θE2 = π + 2θi − 6 arcsin(

sin θin

).

From graph 11 we can see, how the intensity of Class 3 rays changes with θE1(θi). As the an-gle of incidence of the light increases from zero, the angle of emergence θE1, at rst increases (Thatcan be seen from gure 4 if we consider that θE1 = π −D(θi)). At small θE1 the intensity of lightis almost constant. When θE1 comes close to 40o, the intensity of light starts rapidly to increase.In rst order theory it reaches a theoretically innite value at θE1 = 4202′ for light of refractiveindex n = 4/3. This is the angle at which we see primary rainbow. As the angle of incidence, θi,continuous to increase, both θE1 and the intensity of emerged light start to decrease again. However,by the time, the angle of incidence reaches the extreme value of 90o, θE1 has not again fall to zero.The smallest angle at which light emerges for large angles of incidence is θmin

E1 = 14o.Similar curves can be drawn for Class 4 rays, which form the secondary rainbow. But in this

case all the light is emerged at angles which are bigger than the angle at which we see the secondaryrainbow. In gure 11 we can also see, that there is no emerged light between angles θE1 = 42o2′ andθE2 = 50o58′. This area is, as we already mentioned, known as Alexander's dark band.

0.2 0.4 0.6 0.8 1 1.2 1.4 1.6ΘE

0.2

0.4

0.6

0.8

1

I

Primary Rainbow

Secondary Rainbow

Alexander’s d.b.

Figure 11: Graph shows how the intensity of emerged light of Class 3 (red line) and Class 4 (blueline) rays, which form the primary and the secondary rainbow, depend of angle of emergence θE .There is also clearly seen the Alexander's dark band, between angles θE = 42o2′ and θE = 50o58′.

With knowledge that we have gained about Cartesian theory of the rainbow, we can simulatelight scattering on the droplet and try to get a rainbow. Results of the simulation at dierent numberof incident rays at dierent impact parameters are shown in gure 12.

3.3 Monte Carlo Computer simulation of the rainbow.

In the long history of scientic attempts to explain the rainbow, computer simulation is one of themost powerful modern methods[7]. Therefore it would be interesting, to make such a simulation,before going to the diraction theory of the rainbow. This simple problem is perfect for illustratingthe power of Monte Carlo techniques. The term "Monte Carlo" is derived from the famous casinoin Monaco and is applied to any numerical method that makes frequent calls to a random-numbergenerator.

The simulated rainbow is generated by plotting many single points in accordance with the lawsof reection and refraction applied to a large number of parallel rays of light incident on a sphere ofwater.

10

Figure 12: Pictures present 2D simulation of light scattering on a round droplet of water. Single rayscattering at b = 0.75 is shown in the rst picture. Multiple ray scattering is shown in the secondone.

Figure 13: Graph shows Monte Carlo computer simulation of the rainbow, produced by N = 100000rays.

11

Results of the simulation are shown in gure 13. We can see, that computer from simple lawsof geometric optics, acting on thousands of incidence rays, produces all features of actual rainbow.There are primary and secondary bow at correct angles with colours in the correct sequences, thediuse white illumination just inside the primary, and the dark band between the primary and thesecondary bow.

Once we have simulated a rainbow, made of water droplets, we can try to explore the eects onthe dispersion of the colours, the sequence of colours, the angular sizes of the primary and secondaryarcs, and the width of the dark band separating the primary and secondary arcs if droplets arenot made of H2O but of some other substance. How the rainbow is changed if drops are made ofconcentrated (70%) H2SO4, or concentrated (70%) HNO3 is shown in gure 14.

Figure 14: Graphs show Monte Carlo computer simulation of the rainbow, produced by N = 100000rays, where droplets are made of dierent substances.

4 The diraction theory of the rainbow

No rainbow is exactly like any other. Sometimes red or blue may be absent and the width of othercolor bands also variable. Sometimes we can not observe orange. In some bows the yellow bandmay be less broad then green and violet, whereas in others the yellow is much wider than green andviolet. There is also a noticeable variation in intensities of dierent colours. Usually the violet sideis the brightest but the colours there are more admixed with white. The total width of the rainbowis also not a constant.

Supernumerary bows, which are commonly seen within the rainbow, also dier from case to case.Sometimes many can be seen. Up to six have been observed. In some cases they are contiguous withthe primary bow and in the other they are separated from it by a dark band. Their colours alsovary. Green and pink are most common, but yellow and violet may also be seen.

All of these variations between rainbows are result of diraction. The original diraction theoryof the rainbow was work of Sir George Airy in 1849 and Josef Pretner whose contribution to thesubject was the detailed application of Airy's calculation to the rainbow so as to disentangle thevarious colour formations predicted by the theory.

The diraction eects arising in connection with the rainbow are connected with the light whichemerges from a raindrop nearly in the direction of minimal deviation. As the point of entry, I, of thelight into the raindrop (see gure 10) moves round in a clockwise direction from the point S, so asto increase the angle of incidence, the point of emergence, E, rst moves in anti-clockwise direction,reaches an extreme position, and then moves back again. The point of interest to us in connectionwith the supernumerary bow is that the point at which E begins to retrace its steps lies beyond thepoint of minimum deviation[2]. For water of refractive index n = 4

3 , value of θi at which φ reachesits maximum value is θmax

i = 76.84o. For minimum deviation in the rainbow the angle of incidenceis, as we know, θD

i = 59.13o. That means, that point E passes the point of minimum deviation and

12

moves forward to the right in anti-clockwise direction until it reaches θmaxi .

-3 -2 -1 1 2 3

-1

1

2

Y X

aL

Figure 15: Figure a) shows how rays of light at dierent impact parameters are being scattered indroplet. Dashed lines represent virtual rays. We get them with backward extrapolation of emergingrays. With help of virtual rays we calculate the wave front of emerged light, as shown in gure b).

Let us now replace the wave front emerging from droplet with a virtual source (See gure 15a).The position of the virtual source is arbitrary and has been chosen in the position before virtualrays intersect. We can now try to determine the function, that will satisfactory approximate the realwave front.

In gure 15 b) DES represents the ray which emerges at minimum deviation. The rays both tothe right (θi < θD

i ) and to the left (iθDi < θi < θmax

i ) of it, will be deviated through larger angles, asrepresented by lines ending in arrows. The emerged wave front has to be perpendicular to the raysand is represented by the curve WEW'. The curvature is concave upwards on the left and concavedownwards on the right. Because the value of y changes sign with x, the expression for y will containodd powers of x only:

y(x) = fx + k′x3 + ... (27)

Since the curve WEW' is at right angles to the axis of y (consequently the slope of the curve WEW'there is zero) when x = 0, the coecient f of the linear term must be zero. To a rst order, therefore,the equation to the curve WEW' must be:

y(x) = k′x3. (28)

where k′ is constant. The linear dimensions of the curve will be proportional to the linear dimensionsof the raindrop from which the wave front emerges, i.e. to the radius a of the drop. The reciprocalvalue of the constant k′ must be of two dimensions in length, for dimensions of both sides of theequation to be the same. Therefore we can write:

y(x) =k

a2x3 (29)

where k is now dimensionless constant. Using (29), the complex amplitude of the wave front may bethan written as:

Φ(x) = e−i 2πλ

σ(x), (30)

where σ(x) is the path dierence between the disturbances from M, the coordinates of which are x,y,and from O in a direction making an angle θ with that of minimum deviation (see gure 16):

σ(x) = OS = OR−RS = OR−NT = x sin θ − y(x) cos θ = x sin θ − k

a2x3 cos θ (31)

13

Figure 16: Figure shows a path dierence between the disturbances from M and O on given wavefront.

With given wave front we are now able to calculate the disturbance formed at given ξ (Seegure 16). If the observation point is far away from the droplet, we may use Frauhoer diractionformula[6] for our calculation. The disturbance at given ξ can be than written as:

Φ(ξ) =ei 2π

λL

iλL

∫ ∞

−∞e−i 2π

λσ(x)e−i 2π

λξxL dx, (32)

If we put (31) into (32), we get:

Φ(ξ) =ei 2π

λL

iλL

(λa2

4k cos θ

)1/3 ∫ ∞

−∞cos(π

2(u3 − zu

))du︸ ︷︷ ︸

Rainbow Integral

, (33)

where variable z is dened as:

z3 =16a2

λ2k

(sin θ + tan θ)3

cos θ(34)

Derived integral is known as Rainbow integral. When we calculate it, we get:

Φ(ξ) =ei 2π

λL

iλL

(4π2

3λa2

k cos θ

)1/3

Ai

(−(

π2

12

)1/3

z

), (35)

where Ai(x) is famous Airy function[8]. A graph of the rainbow integral is drawn in gure 17. Itresembles the curve for the diraction pattern near the edge of the shadow of a straight edge in avery striking way. The illumination at the position of minimum deviation at x = 0, correspondingto Descartes ray of the simple rainbow theory is 0.44 of the rst maximum. From graph we can seethat, as with the shadow of the straight edge, there is some illumination within the area from whichlight should be cut o according to geometrical optics. Within this area illumination falls rapidlywhen we go further away from the geometrical edge of illumination. Again, as with the diractionround a straight edge, in the area which is illuminated according to geometrical optics, there occursa series of fringes. As observed in the sky they lie within the ark of the primary rainbow, and thebright fringes give rise to the supernumerary bows. The rst bright fringe is the most intense andis origin of the primary bow. This fringe is than followed by a long series of others of only slowly

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−10 −8 −6 −4 −2 0 2 4 6 8 100

0.2

0.4

0.6

0.8

1

1.2

1.4

X: 0Y: 0.4432

x

I(x)

Diffraction of raindrop

−10 0 100

0.5

1

1.5

z

I(z)

Diffraction of straight edge

0 2 4 6 8x

Intensity

0 2 4 6 8x

Figure 17: Graphs show the square of the rainbow integral. It resembles the curve for the diractionpattern near the edge of the shadow of the straight edge.

diminishing intensity. As one proceeds into the illuminated area the minima become closer togetherand the resulting coloured arcs will become narrower. In contrast to the pattern produced by astraight edge, there is no illuminationin the dark fringes. This corresponds to the fact that at largedistances from the rainbow the intensity is low.

As mentioned, the derived equation for the emergent wave front is only a rst approximation andwill be valid only for points near the ray with minimal deviation. While applying our theory, we havemade an assumption that diraction arises from points on the wave front near to the Descartes ray.Therefore Airy's theory holds only if the drops are large compared to the wavelength of light. Raindrops which form rainbow normally satisfy this condition2. However, even in the case of smallestcloud droplets their diameter is of order of ten wavelengths so that even here a useful quantitativepicture should still be obtainable.

The rainbow integral, which we have been discussing so far, gives the intensities to be expectedin various directions, as long as the drop size remains the same and the wavelength of the light isalways the same. The constant factor:(

4π2

3a2

kλ2 cos θ

)1/3

, (36)

by which the Airy function in (35) is multiplied, suggests that as the wavelength and the radius ofthe drop are altered, the amplitude will vary as ( a

λ)2/3, considering that θ is always small and itscosine approximately equal to unity.

The described calculation of Airy's theory has been so far made in two dimensions. However, thereal problem is, of course, one in three dimensions. If the radius of the drop from which the lightis emerged is large compared with the distance along the wave front over which the integration hasto be preformed in order to calculate the intensity of the diracted light, the integral itself will notbe aected. But it is necessary to multiply the constant factor (36) by a further

√aλ (See[2]). The

intensity, which varies as Φ2 will, therefore, be proportional to:

I ∝(a

λ

)7/3(37)

2Raindrops have generaly diameter greater than 0.5mm and range in size up to few millimeters.

15

This result will be required, when we will compare intensities of various colours ( i. e. with dierentλ ) and when we will take a look of the eect of variations in the sizes of the drops.

4.1 Colours in rainbow

In our diraction theory of rainbow we have been dealing so far only with a monochromatic light.We know that in the rainbow all spectral colours are present and superimposed upon each other.Each colour will possess its own characteristic angle of minimum of deviation, its own characteristicintensity in the spectrum and consequently its own diraction curve, in its own position, and withits own characteristic amplitude. The problem in providing a theoretical analysis for the rainbow, isthus to work out the eect of such superimposition.

In my own analysis I have followed the investigation of J.M. Pernter, who rst solved this problemand made an important contribution to the optics of the rainbow. I have started my analysis withcalculation of the angle of minimum deviation for each colour (wavelength). With that informationI was than able to determine the positions in which the various Airy's curves have to be arranged.

500 1000 1500 2000Λ

0.0002

0.0004

0.0006

0.0008

0.001

0.0012

djdΛ

-0.02 -0.01 0.01 0.02 0.03 0.04 0.05z

0.5

1

1.5

2

I

Figure 18: First gure shows Planck's function for a blackbody radiation. Second one shows variousAiry curves for dierent wavelengths, when the sun radiates as a blackbody.

I have also taken into account that the Amplitude of each Airy's curve must be proportional to thesquare root of the intensity of the light of the corresponding colour in the sunlight. I have presumedthat the sun radiates as a blackbody with T = 5780K and, therefore, obeys Planck radiation[9]formula:

dI

dλ=

2πhc2

λ5

1

ehc

λkBT − 1(38)

Results I have got are shown in gure 18. We can see, that in case of the sun beeing a blackbodythe violet light has the greatest intensity in the rainbow. From observations we know, that this isnot true. To get more accurate results, we have to consider the scattering of sunlight o particles ofthe atmosphere. The intensity of the transmitted light is given by:

I(λ) = I0(λ)e−Aλ4 , (39)

where I0(λ) is the intensity of the sunlight before scattering in the atmosphere. The parameter Adepends on the mean thickness of the atmosphere, the speed of light and of the total cross-sectionof the scattering of light o the particles in the air. From (39) we can see, that strong wavelengthdependence of scattering enhances the long wavelengths and reduces the short wavelengths. Whenwe multiply the amplitude of each Airy's curve with this additional factor, we get results shown ingure 19.

Once I had all individual Airy curves for every color, I superimposed them upon each other andgot the nal theoretical picture of the rainbow. Results of my analysis for dierent sizes of droplets

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-0.02 -0.01 0.01 0.02 0.03 0.04 0.05z

0.1

0.2

0.3

0.4

0.5

I

Figure 19: Figure shows various Airy curves for dierent wavelengths, when scattering of light othe particles in the air is considered. I have set parameter A to A = 4× 1010nm4

are shown in gure 20. From pictures we can see how properties of the rainbow (width of thebow, colours, number of visible supernumerary bows...) depend of the radius of droplet. In simpleDescartes theory the rainbow was independent of the radius of raindrops, but now we can see (fromAiry's theory) that the size of the drops matters and is an important parameter of the rainbow.

5 Raindrop size distribution

In the end it would be right to say a few words about the distribution of raindrop size. It is describedby empirical Marshall-Palmer distribution, which is a negative exponential function:

N(D) = N0e−Λ(R)D. (40)

N(D) tells us the number of drops with diameter D and N0 is the number of intercepted raindrops.The empirical parameter λ(R) determines the slope of exponential curve and is given by:

Λ(R) = 41R−0.21, (41)

where R is rainfall rate in millimeters per hour. Graphs of Marshall-Palmer distribution for dierentrainfall rates are shown in gure 21.

6 Conclusion

The theory of the rainbow described in this seminar explains formation of the rainbow in terms ofclassical rays of light and Airy theory which is based on the interference between two rays of lighttraveling in the same direction. In terms of these theories we could also explain similar phenomenonlike white bow, rainbow by reected light, moonbow and haló[2]. However, these theories are notthe whole story of light scattering by water droplets, and they cannot begin to explain anotherphenomenon involving light scattering from mist: the glory. To understand glory, the full scatteringtheory of light by a dielectric sphere, Mie theory, is needed. In spite of that, the beauty of thephysics of the rainbow is that so much can be understood in terms of rays and simple wave theory.Rainbows open our eyes to some of the fundamental properties of light.

7 Thanks

I would like to thank to my mentor doc. dr. Igor Poberaj for all support and help he has given me,while I have been preparing this seminar.

17

0 2 4 6x

Intensity&Color

0 2 4 6

9 a!!!!k

= 0.8mm=

0 2 4 6x

Intensity&Color

0 2 4 6

9 a!!!!k

= 0.5mm=

0 2 4 6x

Intensity&Color

0 2 4 6

9 a!!!!k

= 0.2mm=

0 2 4 6x

Intensity&Color

0 2 4 6

9 a!!!!k

= 0.1mm=

0 2 4 6x

Intensity&Color

0 2 4 6

9 a!!!!k

= 0.08mm=

0 2 4 6x

Intensity&Color

0 2 4 6

9 a!!!!k

= 0.05mm=

Figure 20: Figures show theoretical rainbows calculated from Airy's rainbow theory for dierentraindrop sizes.

18

0.2 0.4 0.6 0.8 1D

0.2

0.4

0.6

0.8

1

N HDL

N0 8Marshall-Palmer dist.<

Increasing R

Figure 21: Figure shows Marshall-Palmer distribution of raindrop size for dierent values of theparameter R (rainfall rate).

References

[1] H. Moysés Nussenzveig: The theory of the Rainbow, Scientic American, 236 (1977) 116

[2] R.A.R. Tricker : Introduction to Meteorological Optics, Mills and Boon , London (1970) .

[3] HyperPhysics: Rainbows, http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html,7.6.2007

[4] Wikipedia: Rainbow, http://en.wikipedia.org/wiki/Rainbow, 7.6.2007

[5] Alexander III 'The Great', http://www.seleukids.org/Alexander.htm, 21.12.2006

[6] Eugene Hecht : Optics, Second Edition, Addison Wesley, New York (1987).

[7] Donald Olson: Monte Carlo Computer Simulation of a Rainbow, The Physics Teacher, April(1990) 226

[8] M. Abramowitz and I.A. Stegun : Handbook of Mathematical Functions,with Formulas, Graphsand mathematical Tables, National Bureau of Standards, Washington (1964).

[9] Janez Strnad : Fizika, etrti del, 2.Spremenjena izdaja, DMFA, Ljubljana (2000).

[10] J. Rakovec in T. Vrhovec : Osnove Meteorologije za naravoslovce in tehnike, 2. popravljenaizdaja, DMFA, Ljubljana (2000).

[11] O. Davies et al.: The rainbow, Europhysics News, Number 1, Volume 37 (2006), 21

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