Refractive index question - light beam around the world

[jadejones]

1. The refractive index of the Earth’s atmosphere is
n = 1.01 + α(R − r),
where α is a constant, r is the radial distance from the Earth’s centre and
R = 6.4 × 10^6 m is the Earth’s radius. By considering a path comprising a series of
total internal reflections or otherwise, find a value of α for which a light ray emitted
horizontally close to the Earth’s surface would go around the Earth. (The effects of
absorption may be ignored and the Earth may be taken to be a perfect sphere of radius
R.)



2. Homework Equations
n1sin(x1)=n2sin(x2)

3. The Attempt at a Solution
I really have no idea what to do as I understand the refractive index is constantly changing?
Don't know where to start, any help would be much appreciated.

 

[rude man]

Think of a wavefront launched horizontally. Realize that the top of this wavefront will move faster than the bottom, since n is lower at higher altitudes ... yet the wavefront phasing across its front has to be the same for all heights - whar does that requirement impose on grad(n)?

 

[jadejones]

Think of a wavefront launched horizontally. Realize that the top of this wavefront will move faster than the bottom, since n is lower at higher altitudes ... yet the wavefront phasing across its front has to be the same for all heights - whar does that requirement impose on grad(n)?

Thank you very much for your reply, are you suggesting the light curves around the world? I considered this but thought that only really happened at black holes, or have I misunderstood your response?

 

[rude man]

Thank you very much for your reply, are you suggesting the light curves around the world? I considered this but thought that only really happened at black holes, or have I misunderstood your response?

I never heard of the herm 'light curves around the world' per se but that is what I had in mind. And I know next to zilch about black holes.

Can you go from there?

BTW the problem did not state that it was possible, it just asked what the n gradient had to be were it possible.

 

[paranormal]

As a scientist, it is important to approach this problem with a systematic and analytical mindset. First, let's break down the given information and identify the key components:

1. The refractive index of Earth's atmosphere is dependent on two variables: α and r.
2. The equation provided gives the relationship between the refractive index and the distance from the center of the Earth, r.
3. The Earth's radius, R, is given as 6.4 × 10^6 m.
4. The question asks for a value of α that would allow a light ray emitted horizontally close to the Earth's surface to go around the Earth.

Based on these components, we can start by setting up the problem using the Snell's Law equation, n1sin(x1) = n2sin(x2), where n1 and n2 are the refractive indices of the two mediums and x1 and x2 are the angles of incidence and refraction, respectively.

In this case, our two mediums are the Earth's atmosphere and the vacuum of space. Therefore, we can set n1 = 1.01 + α(R − r) and n2 = 1 (since the vacuum of space has a refractive index of 1). We can also assume that the light ray is emitted horizontally, so the angle of incidence, x1, would be 90 degrees.

Now, we need to consider the path of the light ray as it travels around the Earth. Since we are assuming a series of total internal reflections, we can imagine the light ray bouncing off the Earth's surface at each point of contact. This means that the angle of incidence at each point would be equal to the angle of reflection. We can use this information to set up a series of equations:

n1sin(x1) = n2sin(x2)
n2sin(x2) = n1sin(x3)
n1sin(x3) = n2sin(x4)
.
.
.
n1sin(xn) = n2sin(xn+1)

Note that xn+1 would be the angle of incidence at the next point of contact, and so on. We can also use the fact that the sum of all angles in a triangle is equal to 180 degrees, so we can write:

x1 + x2 + x3 + ... + xn + xn+1 = 180

Now, we can substitute the values from our previous equations into