Optics/ total internal reflection (non-trivial at least for me)

In summary, the refractive index of the Earth's atmosphere is 1.01+(alpha)(R-r) where (alpha) is a constant, r is radial distance from Earth's centre and R is the Earth's radius.
  • #1
ed1234
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Homework Statement


The refractive index of the Earth's atmosphere is:
n=1.01+(alpha)(R-r)
where (alpha) is a constant, r is radial distance from Earth's centre and R is the Earth's radius. By considering a path comprising a series of total internal reflections or otherwise, find a value of alpha for which a light ray emitted horizontally close to the Earth's surface would go around the Earth.

The Earth may be taken to be a perfect sphere radius R and the effects of absorption ignored).

Homework Equations


n1sin(theta1)=n2sin(theta2)

The Attempt at a Solution


I'm really fairly stuck on this on. I think we're looking for a gradient of refractive index such that the curvature of the light ray is equal to the curvature of the Earth but I'm not sure how to go about doing this quantitativly. Any hints/help would be appreciated.
Thanks
 
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  • #2
A very simplistic approach, but at least it gives you something.

I guess you take n1 at r = R.
 

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  • #3
Thanks for your suggestion, but since there's no single boundary (n changes gradually) I'm not sure that your picture is quite what the question wants.

If anyones got any further help I'd appreciate it.
Thanks
 
  • #4
With a gradual change in refractive index I doubt that a state of total internal reflection would occur, but I might be proven wrong. My take on the situation is that one needs a drastic change in refractive index to achieve total internal reflection. Unless the problem uses the term but actually means that the beam just keeps on gradually curling around the earth, which I find hard to believe, since you generally find such behaviour only in the vicinity of black holes!

A more realistic approach for the refractive indices in my drawing might be to take n1 at r and n2 = 1.0 (vacuum).

I am amazed that the astonomy people have'nt jumped onto this thread by now.

It seems you are trying to solve a quite advance problem http://www.journals.uchicago.edu/AJ/journal/issues/v119n5/200020/200020.html"
 
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1. What is total internal reflection?

Total internal reflection is a phenomenon that occurs when a light ray traveling through a denser medium reaches the boundary of a less dense medium at an angle greater than the critical angle. This causes all of the light to be reflected back into the denser medium, instead of being refracted into the less dense medium.

2. How is total internal reflection used in everyday life?

Total internal reflection is used in a variety of everyday applications, such as in fiber optic cables for telecommunication, endoscopes for medical procedures, and binoculars and telescopes for improved imaging. It is also used in reflective road signs and mirrors.

3. What is the formula for calculating the critical angle?

The formula for calculating the critical angle is sin θc = n2/n1, where θc is the critical angle, n1 is the refractive index of the denser medium, and n2 is the refractive index of the less dense medium.

4. Can total internal reflection occur with other forms of energy besides light?

Yes, total internal reflection can occur with other forms of energy, such as sound waves and water waves. In these cases, it is referred to as total internal reflection of waves.

5. How does the angle of incidence affect total internal reflection?

The angle of incidence plays a crucial role in total internal reflection. If the angle of incidence is less than the critical angle, the light ray will be partially refracted into the less dense medium. However, if the angle of incidence is equal to or greater than the critical angle, total internal reflection will occur, with all of the light being reflected back into the denser medium.

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