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Question: Rainbows from square drops. Suppose that, on some surreal world, raindrops had a square cross section and always fell with one face horizontal. Figure shows such a falling drop, with a white beam of sunlight incident at θ=70.0°at point P. The part of the light that enters the drop then travels to point, where some of it refracts out into the air and the rest reflects. That reflected light then travels to point, where again some of the light refracts out into the air and the rest reflects.(a)What is the difference in the angles of the red light ( n = 1.331) and the blue light(n = 1.343) that emerge atpoint A? and(b) What is the difference in the angles of the red lightn = 1.331 )and the blue light (n = 1.343)that emerge atpoint B? (This angular difference in the light emerging at, say, pointwould be the rainbow’s angular width.)

Figure:

Short Answer

Expert verified

Answer

  1. The difference in angles of the red light and blue light that emerge at point A is 3.10
  2. The difference in angles of the red light and blue light that emerge at point B is 00

Step by step solution

01

Given

θ1=700

  1. Angle of incidence
  2. Index of refraction for bluen2b=1.343
  3. Index of refraction for radn2r=1.331
02

Understanding the concept

We can use Snell’s law for each color to find the angle of refraction for red and blue colors at point A and at B. From this, the difference in angles of the red light and blue light that emerge at pointand B can be calculated.

Formula:

n1sinθ1=n2sinθ2

03

(a) Calculate the difference in angles of the red light and blue light that emerge at point  

According to Snell’s law, for blue color,

nairsinθ1=n2bsinθ2bθ2b=sin-1nairsinθ1n2bθ2b=sin-11.0sin7001.343θ2b=44.4030

Now, according to Snell’s law, for red color,

nairsinθ1=n2rsinθ2rθ2r=sin-1nairsinθ1n2rθ2r=sin-11.0sin7001.331θ2r=44.9110

These are the refraction angles at the first surface.

Now, for the second refraction,

According to Snell’s law, for blue color,

n2bsin(900-θ2b)=n3bsinθ3bθ3b=sin-11.343sin(900-44.4030)1.0θ3b=73.6360

Now, according to Snell’s law, for red color,

θ3r=sin-11.331sin(900-44.9110)1.0θ3r=70.4970

Therefore,

∆θ=73.6360-70.4970∆θ=3.10

Therefore, the difference in angles of the red light and blue light that emerge at point A is 3.10.

04

(b) The difference in angles of the red light and blue light that emerge at point  

The angle of incidence and angle of emergence is the same. Therefore, the difference in angles of the red light and blue light that emerge at point B is 00.

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