Atmospheric Optics & Geometry
Rainbow Formation
Discover how the alignment of sun, rain, and observer forms spectacular primary and secondary rainbows. Toggle between raindrop cutaways, full arc landscape geometry, and garden hose backyard experiments.
Rainbow Formation Physics Lab
Analyze refractions, internal reflections, anti-solar axes, and primary/secondary angles.
Geometric Telemetry
θ = 4r - 2i ⇒ θ = 42° (Primary Bow)- Sun Altitude Angle
- 22.0°
- Rainbow State
- Visible Primary
- Primary Arc Radius
- 42.0° (Red Edge)
- Secondary Arc Radius
- 51.0° (Reversed)
- Anti-solar Point altitude
- -22.0° (Below Horizon)
The Physics of Rainbow Formation
A rainbow is a multi-colored circular arc formed in the sky when sunlight interacts with thousands of spherical water droplets suspended in the air. Far from being a physical object at a fixed location, a rainbow is an optical phenomenon that depends entirely on the relative geometry of the Sun, the raindrops, and the observer's eyes.
For a rainbow to form, two conditions must be met simultaneously: (1) The Sun must be shining behind the observer, and (2) rain or mist must be falling in front of the observer.
Light Path in a Raindrop
Each individual raindrop acts as a microscopic prism, refracting, dispersing, and reflecting the sunlight. The path follows three critical steps:
1. Refraction and Dispersion
As sunlight enters the curved face of the droplet, it bends (refracts) and splits into separate wavelengths because water has a slightly higher refractive index for violet than for red.
2. Total Internal Reflection
The rays travel through the drop and hit the back inner water-air boundary. Because the angle of incidence is larger than the critical angle of water (≈ 48.6°), the light is reflected internally.
3. Final Refraction
The reflected rays travel to the front-bottom surface and exit the droplet into the air, refracting once more and bending further apart.
The 42° Rainbow Angle
Using spherical ray-tracing geometry, Descartes derived that the emergence angle concentrates at a maximum deviation boundary:
Primary Bow Radius
θ = 4r - 2i ≈ 42° (Red)
Because the maximum concentration angle for red light is 42° and for violet is 40°, you see red on the outer edge and violet on the inner edge of the arc, centered on the anti-solar point.
Secondary Rainbows (Double Reflection)
Sometimes, a fainter, outer secondary rainbow is visible above the primary bow. It is created by a double internal reflection:
- Two Reflections: Light enters near the bottom of the raindrop, reflects TWICE off the back walls, and exits from the top.
- Reversed Colors: The second reflection crosses the rays again, reversing the spectrum. Violet is on the outside at 53°, and Red is on the inside at 50°.
- Fainter Arc: Since some light is lost at each internal reflection, the secondary bow is much fainter.
Step-by-Step Solved Problems
Practice math equations relating to anti-solar coordinate projections, primary rainbow angles, and secondary reflection paths.
Example 1 Problem Statement
A light ray enters a spherical raindrop at an incidence angle of 59.4°. The angle of refraction for red light in water is 40.2°. Find the angle of emergence relative to the incident direction for a primary rainbow.
View Step-by-Step Geometry Solution
- Identify the given values: Angle of incidence i = 59.4°, Angle of refraction r = 40.2°.
- Recall the primary rainbow angle relation relative to the anti-solar axis: θ = 4r - 2i.
- Substitute the values: θ = 4(40.2°) - 2(59.4°).
- Calculate each term: 4 × 40.2° = 160.8°, 2 × 59.4° = 118.8°.
- Find the final angle: θ = 160.8° - 118.8° = 42.0°.
Final Derived Answer: Emergence Angle (angular radius of the primary red arc) θ = 42.0°.
Example 2 Problem Statement
Determine the maximum angle of deviation for a secondary rainbow (two internal reflections, k=2) where the angle of incidence is 71.9° and the angle of refraction in water is 45.4°.
View Step-by-Step Geometry Solution
- Identify the parameters: Incidence angle i = 71.9°, Refraction angle r = 45.4°.
- Recall the secondary rainbow angle relation relative to the anti-solar point: θ = 6r - 2i - 180°.
- Substitute the values: θ = 6(45.4°) - 2(71.9°) - 180°.
- Evaluate each term: 6 × 45.4° = 272.4°, 2 × 71.9° = 143.8°.
- Compute the result: θ = 272.4° - 143.8° - 180° = 128.6° - 180° = -51.4°. By convention, the angular radius of the secondary bow is taken as positive: 51.4°.
Final Derived Answer: Secondary Rainbow Angular Radius θ = 51.4°.
Example 3 Problem Statement
An observer stands on flat ground looking at rain mist. The Sun is at an altitude of 35° above the horizon. Calculate the highest point of the primary red rainbow arc above the ground.
View Step-by-Step Geometry Solution
- Identify the parameters: Sun altitude angle H = 35°, Angular radius of the primary red rainbow R = 42°.
- Recall the geometry of the anti-solar point: The anti-solar point is located exactly H degrees below the horizon (i.e. at -35°).
- The rainbow forms a circle centered on the anti-solar point with radius R = 42°.
- The highest point of the arc above the horizon is given by: Height = R - H.
- Substitute values: Height = 42° - 35° = 7°.
Final Derived Answer: Highest point of the primary red arc is 7° above the horizon.
Self-Check Questions
Question 1
Why is a rainbow never visible at midday in mid-latitude regions when the Sun is high in the sky?
Show Answer & Explanation
A rainbow is only visible when the Sun's altitude is less than 42°. The center of the rainbow lies on the anti-solar point, which is as far below the horizon as the Sun is above it. If the Sun is higher than 42° (like at midday), the anti-solar point drops more than 42° below the horizon, pulling the entire 42° primary rainbow circle below the ground.
Question 2
Explain why the color order of a secondary rainbow is reversed compared to a primary rainbow.
Show Answer & Explanation
A primary rainbow light ray undergoes one internal reflection inside the raindrop, which deviates violet more than red but leaves red on the outside edge. A secondary rainbow undergoes two internal reflections. The second reflection crosses the paths of the exiting rays again, reversing the color order so that violet emerges at a larger angle (53°) than red (50°), putting violet on the outer rim.
Question 3
What is Alexander's dark band, and what causes it?
Show Answer & Explanation
Alexander's dark band is the dark region of the sky located between the primary and secondary rainbows. It is caused by the physics of minimum deviation: light rays refracting through raindrops cannot emerge at angles between 42° (primary limit) and 50° (secondary limit). Since no scattered light is directed into the observer's eye from these angles, the region appears dark.
Question 4
Why is it necessary to stand with your back to the Sun to see a rainbow?
Show Answer & Explanation
Sunlight must enter the raindrop, reflect off the back surface, and exit back out of the front surface towards the observer. Since the light is reflected back towards the side the Sun is on, the observer must have the Sun behind them and look towards the rain shower ahead to capture the reflected light.
Question 5
Can a rainbow be seen as a complete circle? If so, under what conditions?
Show Answer & Explanation
Yes, a rainbow is naturally a full circle. However, from the ground, the lower half is cut off by the horizon. An observer can see a complete 360-degree circle from high altitudes, such as from an airplane or a very tall mountain peak, where droplets exist below the observer's horizon.
Question 6
How does the size of the raindrops affect the appearance of a rainbow?
Show Answer & Explanation
Large raindrops (1-2 mm in diameter) produce bright, vivid rainbows with distinct, sharp colors. Small raindrops (less than 0.5 mm) cause diffraction to overlap the colors, producing a faded rainbow. Very tiny mist droplets (less than 0.05 mm) produce a white or nearly colorless arc called a fogbow.