Rainbow Lab

Teacher Notes

Objectives

• Students will understand geometric properties of light by analyzing the passage of light through a raindrop.
• Students will experimentally determine the angle at which rainbows appear in the sky.

Background

Prerequisite Skills

This lab requires students to be familiar with:

Law of Reflection

Snell's Law

Graphing and interpreting data

Right triangle trigonometry (including arcsine)

Necessary Resources

To do this lab, students will need:

Computers with Internet Web access

Lab printouts

Graph paper & calculator

Necessary Internet Web Site

Build a Rainbow - for collecting data on light in water droplets
http://www.geom.umn.edu/education/calc-init/rainbow/rainbow.cgi

Optional Internet Web Sites

Circles of Light: The Mathematics of Rainbows - the original calculus lab
http://www.geom.umn.edu/education/calc-init/rainbow/

About Rainbows - excellent additional information and resources
http://www.unidata.ucar.edu/staff/blynds/rnbw.html

Assessment

Solutions

Part 1 - Preparation

Question 1

1. The speed of light in a medium with a large index of refraction is much slower than c.
2. Light travelling into a higher index of refraction bends towards the normal.
3. Light travelling into a lower index of refraction speeds up and bends away from the normal.

Question 2

1. See diagram.
2. The angle of incidence is equal to the angle the light ray leaves the crystal.
3. Set up an experiement to measure the distance, R, that the light ray travels perpendicular to the crystal's thickness, T. (See diagram.) Then the angle of refraction, b, can be calculated by b = arctan(R/T).

Question 3

1. b(45° + 5°) 18.52°.
2. b(50°) = 18.45°. The estimate has an absolute error of 0.07° and a relative error of 0.38%.
3. b(45° + 20°) 23.10°. b(65°) = 21.99°. The estimate has an absolute error of 1.11° and a relative error of 5.05%.

Part 2 - Simulating a Rainbow

A sample printout of the rainbow simulation is included at the end of the Teacher Notes.

Experiment #1

With a wavelength of 700 nm the deflection angle decreases (in a nearly linear fashion) from 180° when w = 0 to a minimum deflection angle of 137.8° when w 0.86 and then sharply increases. The relationship between the angle of incidence and the impact parameter is given by w = sin(a). So the value of a that corresponds to the minimum deflection angle is a arcsin(0.86) = 59°.

Experiment #2

The outgoing rays would be most concentrated for the set w = 0.75, 0.80, 0.85 and the most diffuse for the set w = 0.05, 0.10, 0.15.

Impact Parameter w	Deflection at w D(w)	Deviation from w w	Deflection at w + w D(w + w)	Relative Change in Deflection
0.10	174.3°	-0.05	177.1°	56°
0.10	174.3°	+0.05	171.4°	58°
0.55	149.2°	-0.05	151.8°	52°
0.55	149.2°	+0.05	146.6°	52°
0.80	138.6°	-0.05	140.1°	30°
0.80	138.6°	+0.05	137.8°	16°

Experiment #3

1. With a wavelength of 400 nm the deflection angle has the same behavior as before, only the deflection angle is slightly larger. The minimum deflection angle of 139.6° occurs again at w 0.86 and a 59°.
2. As the wavelength of light decreases, the minimum deflection angle increases.
3. For w = 0.9:

   Wavelength	Deflection Angle
   400 nm	140.3°
   450 nm	139.7°
   525 nm	139.1°
   580 nm	138.7°
   625 nm	138.5°
   700 nm	138.3°

Part 3 - Analysis of the Experiment

Question 4

The longest set of impact parameters corresponds to the deflection angles in the range 137.5° - 140.0°.

Question 5

1. D(0) = 180°. Since the light ray enters the water droplet along the normal, it does not bend. Upon hitting the other side of the droplet, the reflected ray returns along the same path, only the opposite direction (180° deflection).
2. The sketch of D(a) for 0° < a < 90° will decrease to a minimum value and then increase.
3. The limit of D(a) as a approaches 90° depends on the wavelength. For a wavelength of 700 nm the limit is about 165.4°. For a wavelength of 400 nm the limit is about 167.9°.
4. The minimum of D(a) on the sketch occurs at the previously calculated a 59°.

Question 6

1. The tightest possible output beam will occur when the relative change in deflection is minimized.
2. The tightest possible output beam will occur when a 59°.

Question 7

1. The light ray is deflected the following amounts:

   Across Point A	a - b
   At Point B	180° - b
   Across Point C	a - b

2. D(a,b) = 180° + 2a - 4b.
3. b = arcsin[sin(a)/n].
4. D(a) = 180° + 2a - 4arcsin[sin(a)/n].
5. The rainbow angle for red light is approximately 138°.

Part 4 - Conclusions

Question 8

Since the rainbow angle for red light is approximately 138°, an observer will see a bright spot of red sunlight reflected and refracted off a water droplet when there is 180° - 138° = 42° between the direction of the incident sunlight and the observer's line of sight. Therefore, as long as a water droplet is viewed along a line of sight that makes this 42° angle, a bright spot of red will be seen. All lines of sight would form a cone, and thus a "rainbow" of red light appears as circle of angular radius 42°. (A person on the ground cannot see the full circle because the earth is in the way.) Since the incident rays must be reflected into the observer's eye, the sun must be behind a person in order to see a rainbow.

Question 9

The rainbow angle of red light is about 138°, but the rainbow angle of violet light is about 140°. So as red light is seen at a 42° angle, violet light is seen at a 40° angle, and other colors between these. Thus when a rainbow is seen, each color of light is seen by looking at different raindrops. The violet bow of light can be seen as circle of angular radius 40°, which would be inside or "below" the red bow light.

Another way of approaching this is to look at the indices of refraction as being inversely related to the speed of the frequencies in the refracting medium. In water, the speed of red light is 225 000 000 m/s and the speed of violet light is 223 000 000 m/s. Thus compared to violet light, red is slowed down less so it is less refracted from the path of the original ray. Therefore red is "higher" on the rainbow than violet.