00:01
For this problem on the topic of propagation of light, we are told that a secondary rainbow is formed when light incident on a spherical drop of water undergoes two internal reflections as shown in the diagram.
00:13
We want to find the angular deflection of the ray in terms of the incident angle theta a at a and the refractive index in of the spherical water drop.
00:24
We want to know the incident angle theta 2 for which the derivative of the angular deflection with respect to the incident angle is 0.
00:35
And then we want to find theta 2 and the angular deflection delta for violet and red light.
00:45
Now the total angular deflection of the ray is delta and we can see delta is equal to theta a at a minus deta b at a, plus pi minus two theta b at a, plus pi, minus two theta b at a, plus pi minus two theta b at a, plus tata a, plus theta a at a, minus theta b at a.
01:32
And this simplifies to two theta a at a, minus 6 theta b at a plus 2 pi, where we have used here the fact from the previous problem that all the internal angles are equal and the two external angles are also equal.
01:59
Now we can use snells law relationship, and we have theta b at a to be the arc sign of 1 over n times the sign of theta a at a.
02:25
And so this means that the angular deflection delta is equal to 2 theta a at a minus 6 times the arc sign of 1 over n sine of theta a at a plus 2 pi.
02:54
And so we have an expression for the angular deflection delta in terms of n and theta a at a.
03:10
Next, for part b, we want to find the incident angle theta 2, for which the derivative of delta with respect to the incident angle theta a, a, is zero.
03:26
So, the delta, d theta a at a, is equal to zero...