### Exercise 1: Obtain and use information from peer-reviewed literature

The plot for the refractive index of water should look like:

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The plot for the deviation of the refractive index of air from unity should look like:

### Exercise 2: Deflection angle for a light ray entering a spherical raindrop

To calculate the deflection angle, one needs to know the refractive index of the air and the water at the particular wavelength and also the angle of incidence of the incoming ray. The laws of refraction and reflection are then applied to obtain the angles of refraction and the angle of reflection for the internal reflection. Students should be able to show that the equalities shown in the figure illustrating the ray/raindrop geometry.

### Exercise 3: Compute the deflection angle versus incident angle

The solution for Exercise 3 is:

The minimum of the deflection function for λ=400 λ=400 nm (λ=650 λ=650 nm) is δ=139.3∘δ=139.3∘ (δ=137.6∘)(δ=137.6∘). This means that you must look (with the sun at your back and the distant raindrops in front of you) at an angle of 180∘−139.3∘=40.7∘180∘−139.3∘=40.7∘ (180∘−137.6∘=42.4∘)(180∘−137.6∘=42.4∘) relative to the horizontal to see the bright violet (red) band of the rainbow. So the red band appears above the violet band. (It really does!) It is worth noting that the rays producing the primary rainbow enter the top half of the raindrops.

### Exercise 4: Where is the double (secondary) rainbow?

The solution for Exercise 4 is:

The minimum of the deflection function for λ=400 λ=400 nm (λ=650 λ=650 nm) is δ=233.3∘δ=233.3∘ (δ=230.2∘δ=230.2∘). These rays are directed back up into the sky and the ground-based observer wonâ€™t see them. However, if you realize that you have just computed the deflection angles for rays entering the bottom half of the raindrop, you realize that you must look (with the sun at your back and the distant raindrops in front of you) at an angle of 233.3∘−180∘=53.3∘,,(230.2∘−180∘=50.2∘)233.3∘−180∘=53.3∘,,(230.2∘−180∘=50.2∘) relative to the horizontal to see the bright violet (red) band of the rainbow. So the red band of the secondary rainbow appears below the violet band. (It really does!). Thus the rays producing the secondary rainbow enter the bottom half of the raindrops and the order of the color bands is reversed relative to the primary rainbow. Comparing these angles to those from Exercise 3, we see that the secondary rainbow appears above the primary rainbow. The reduced brightness of the secondary rainbow is due to the additional internal reflection and the corresponding loss of irradiance during that reflection. It is posible to extend Exercises 3 and 4 to account for the change in intensity occurring for every refraction or internal reflection. For fun, it must be noted that fascination with the double rainbow became an Internet meme in 2010 when a person living on the border of Yosemite National Park recorded his reaction to a spectacular double rainbow. At the time of writing, the video can be seen at: https://www.youtube.com/watch?v=OQSNhk5ICTI).

### Exercise 5: Crude estimate of the irradiance versus deflection angle for a single wavelength: Primary Rainbow

The solution for Exercise 5 is:

Note that irradiance is distributed over a large range of detection angles, and that the irradiance has a peak near (but not at) the minimum of the deflection function for these chosen parameters. If we use an angular width smaller than 1∘1∘ for the assumed irradiance distribution of a ray, the irradiance peak approaches the minimum in the deflection angle and the “background irradiance” decreases. The main result is that irradiance accumulates in the direction specified by the minimum in the deflection function because several rays are deflected into essentially the same direction. This is known as rainbow scattering, and it is also observed in scattering events such as atomic/molecular collisions or ion collisions with ordered surfaces.