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

Click For Summary

Homework Help Overview

The problem involves the refractive index of the Earth's atmosphere and its effect on light rays emitted horizontally near the Earth's surface. The original poster seeks to determine a value of a constant (alpha) that allows a light ray to travel around the Earth through a series of total internal reflections, considering the Earth's curvature and the gradual change in refractive index.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to relate the curvature of the light ray to the curvature of the Earth but expresses uncertainty about how to quantitatively approach the problem. Some participants suggest considering the refractive index at the Earth's radius, while others question the feasibility of total internal reflection given the gradual change in refractive index.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the refractive indices, but there is no clear consensus on the approach to take, particularly concerning the conditions for total internal reflection.

Contextual Notes

Participants note the complexity of the problem and the lack of a single boundary for the refractive index, which may affect the conditions for total internal reflection. There is also mention of the problem's advanced nature, suggesting it may require deeper insights into optics.

ed1234
Messages
2
Reaction score
0

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
 
Physics news on Phys.org
A very simplistic approach, but at least it gives you something.

I guess you take n1 at r = R.
 

Attachments

  • around earth.gif
    around earth.gif
    12.5 KB · Views: 586
Last edited:
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
 
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"
 
Last edited by a moderator:

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
11K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
4
Views
2K
Replies
9
Views
5K