Why the rainbow is at 42 degrees
Tonight's idea
Light slows down as it passes from air into water, and that bends the ray. Measure angles from the normal, the line at right angles to the surface. Snell's law says
\(n_1 \sin\theta_1 = n_2 \sin\theta_2\), where \(n\) is the refractive index: 1 for air, about 1.33 for water.
So a ray that meets a raindrop at angle \(i\) to the normal travels inside at angle \(r\), where \(\sin i = 1.33 \sin r\).
Follow the ray. Entering, it turns through \(i - r\). At the back of the drop it reflects, turning through \(180° - 2r\). Leaving, it turns through \(i - r\) again. The total turn is the deviation
\[D = 180° + 2i - 4r.\]
Sunlight meets the drop at every \(i\) from 0° to 90°. The table below shows \(D\) falling, bottoming out near 138° when \(i\) is about 60°, and rising again. Near the bottom \(D\) hardly changes, so rays arriving at quite different angles leave in almost the same direction. That pile-up is the bow. You see it \(180° - 138° = 42°\) from the point opposite the sun.
Watch one: \(i = 60°\)
Snell's law gives \(\sin r = \dfrac{\sin 60°}{1.33} = 0.651\), so \(r = 40.6°\). Then \(D = 180° + 120° - 4 \times 40.6° = 137.5°\). A calculator does the same for every other angle:
| \(i\) | 0° | 10° | 20° | 30° | 40° | 50° | 60° | 70° | 80° | 90° |
|---|---|---|---|---|---|---|---|---|---|---|
| \(r\) | 0.0° | 7.5° | 14.9° | 22.1° | 28.9° | 35.2° | 40.6° | 45.0° | 47.8° | 48.8° |
| \(D\) | 180.0° | 170.0° | 160.4° | 151.7° | 144.4° | 139.3° | 137.5° | 140.2° | 148.9° | 165.0° |
The smallest entry is 137.5°, and no ray turns through less. No once-reflected ray reaches your eye from more than about 42° out, which is why the sky is darker just outside the bow.
Your turn
- A ray meets a flat pond at 30° to the normal. Find its angle to the normal under water.
- Find \(r\) and \(D\) for \(i = 45°\). Check that your \(D\) fits between its neighbours in the table.
- Find \(D\) for \(i = 55°\) and \(i = 65°\). Rays arriving anywhere between these two angles leave within how many degrees of each other? Compare with rays arriving between 20° and 30°.
- The refractive index depends slightly on colour: about 1.331 for red light and 1.343 for violet. Taking \(i = 59°\) for both, find \(D\) and the bow angle \(180° - D\) for each colour. Which colour is on the outside of the bow?
Stretch: the second bow
A ray that reflects twice inside the drop has \(D = 360° + 2i - 6r\). With \(n = 1.33\), find \(D\) for \(i = 60°, 65°, 70°, 72°, 75°, 80°\). What is the smallest \(D\), and how far is it from 180°? That is where the fainter second bow sits.
Answers
- \(\sin r = \sin 30° / 1.33 = 0.376\), so \(r \approx 22.1°\). The ray bends towards the normal.
- \(\sin r = \sin 45°/1.33 = 0.532\), so \(r \approx 32.1°\) and \(D = 180° + 90° - 128.5° = 141.5°\). That lies between 144.4° (at 40°) and 139.3° (at 50°).
- \(i = 55°\): \(r = 38.0°\), \(D = 137.9°\). \(i = 65°\): \(r = 43.0°\), \(D = 138.2°\). Between them \(D\) dips to 137.5°, so all these rays leave within about 0.7° of each other. From 20° to 30°, \(D\) runs from 160.4° to 151.7°, a spread of 8.7°. The same 10° range of arrivals is squeezed more than ten times tighter near the minimum.
- Red: \(r = 40.1°\), \(D = 137.6°\), bow angle 42.4°. Violet: \(r = 39.7°\), \(D = 139.4°\), bow angle 40.6°. Red has the larger angle, so red is on the outside of the bow and violet on the inside. The colours separate because each has its own minimum.
Stretch. \(D = 236.2°, 232.3°, 230.3°, 230.1°, 230.6°, 233.4°\). The smallest is about 230°, which is 50° past 180°. The second bow sits about 50° from the point opposite the sun, outside the first.