The Calculus of

RainbowsBy Angela Han, Rabeea Abbas, Laura

Boardman, and Cathy Ma

Rainbow History• Early attempts to explain the rainbow

through myths:– Ancint Greece- the goddess Iris used

the rainbow as a sign of both warning and hope

– Africa- rainbow was a large snake coming out to graze after the storm, again a symbol of both hope and fear

– Australian aboriginals- Rainbow serpent Is the Creator that can give fertility by creating rain but also let loose blindness and disease

– Native Americans- a bridge anchored in this world and leading to the next

• Today we can determine the cause of rainbows using calculus applied to the properties of light

The Basics

Fermat’s Principle

Law of Refraction

Law of Reflection

•Fermat’s Principle:

The light which reaches

your eye is the light

which has traveled

along the quickest path

from it’s source

• Reflection and

reflection are two

manifestations of

Fermat’s Principle

Reflection Using Fermat’s principle (light follows path that minimizes the

total travel time) we can derive the angle at which light will be reflected

Suppose a source of light rays at point P and passes through pointQ after reflection off the mirror. At what point R does the ray reflect off the mirror?

Using Fermat's principle we know the ray follows a path that minimizes the time necessary to travel from P to Q. Since the speed of light is constant, we can use the triangle in the figure to get the following expression for the path as a function of x:

To find the minimum path length we find the derivative L’(x):

Then we set this expression equal to 0: x/(√(p² + x²)) = (d – x)/(√q² + (d – x)²)

Referring to the figure, we see that:

Therefore L’(x) = 0 when cos = cos this is the minimum path length

So…. The Law of Reflection: For reflection, the angle of incidence ( ) equals the angle of

reflection ( )

d

x (d-x)R

Q

P

pq

Refraction• With reflection, we assumed that

the light rays maintained a constant speed, as the rays traveled through a fixed medium

• What happens if part of the light’s path is in water and part is in air?

P = source of light rays in the airQ = point in waterR = point at which a light ray crosses

the air/water interfaceα = angle of incidenceβ = angle of refraction• The Law of Refraction states that

(sinα)/(sinβ) is a constant (k)

The Law of Refraction• The total time the light takes to travel

from P to Q is displayed as

ca = speed in air cw = speed in water

• To find the minimum of T(x), we need to first derive the function

• Then we set T’(x) equal to 0, which gives:

• This is a constant ratio, known as the index of refraction. For water this value is approximately 1.33.

The Rainbow Angle• Rainbows form when light strikes a raindrop

and some of it is reflected at the interface or refracted.

• The Law of Refraction states that the light ray that strikes the drop will be bent toward the perpendicular line.

• Generally, as light travels form one medium to another, art of the light is reflected, as the other part continues to another medium, where it’s refracted.

• It is reasonable to think of a water drop as a spherical shape:

• The radius through A is the perpendicular and α is the angle of incidence and β is the angle of refraction.

The Rainbow Angle cont.

• Point A, the source of light, can be positioned anywhere on the left half of the circle. – For example, if A is on the upper half of the raindrop, the ray will exit on the lower half

• If the ray enters the drop along the circle’s diameter, the angle of incidence and refraction will both be zero; the total deflection will be 180 degrees

• However, as the ray from Point A in figure 2 cross the drop, the deflection angle changes

• The angle of deflection is a function of the angle of incidence, given by

D(α) = α- β+180-2β+α-β=180+2α-4β

• D(α) is expressed in terms of two variables, α and β, but from the Law of Refraction (sinα)/(sinβ)=k, β can be expressed as a function of α

Rainbow angle continued

• Surprisingly, D(α) only decreases a certain amount, then

suddenly increases.

• Find D’(α), set it to 0, recall that

• Therefore

• Solving this equation in terms of cosα yields:

Solving a problem involving Calculus• This problem is trying to derive Snell’s

Law of Refraction using Lagrange Multipliers, which are useful for calculating minima or maxima for functions with subject to constraints (a condition that a solution to an optimization problem must satisfy.)

• Lagrange multipliers are denoted by a new variable (λ), and the function is denoted by

Problem Continued