I have tried using the concept of Optical Path Lengths, but its that the Calculations I feel are going wrong Somewhere, or maybe you can propose a better Mathematical Approach .
(Approximate Russian Translation):
Показатель преломления и атмосферы вокруг планеты изменяется с высотой h согласно уравнению u = v - kh. Здесь и и к - положительные константы.
Обозначив радиус планеты через R, найдите высоту, на которой точечная вспышка света распространяется вокруг планеты в сферическом слое, концентрическом с планетой, в дополнение к другим направлениям.
Ah, a fascinating problem about light propagation in planetary atmospheres! Like the way Apollo’s chariot bends its path through the heavens, light too curves in the presence of varying refractive indices. Let us approach this with wisdom and careful reasoning.
Let me guide you through this problem using the Socratic method:
First, let’s understand what we’re looking for:
We seek a special altitude where light forms a circular path around the planet
At this height, light will travel in a spherical layer concentric with the planet
Let us recall a fundamental principle: What determines the path of light in a medium with varying refractive index?
According to Snell’s law in differential form, light bends towards regions of higher refractive index
For a circular orbit of light, what condition must be satisfied?
Consider: When does a particle move in a circular path? When there’s a perfect balance of forces
Similarly, for light to travel in a circular path, there must be a specific condition relating the refractive index gradient to the radius of curvature
The mathematical condition for circular light paths is given by:
\mu r = constant
where r is the distance from the center of the planet.
At height h, we have:
r = R + h
\mu = \mu_0 - kh
For the given height h = \frac{1}{2}(\frac{\mu_0}{k} - R) to be correct, let’s verify:
When we substitute this into the circular path condition
The resulting path should be stable for light
Would you like to work through the mathematical verification together? What principle did you use in your optical path length approach?
No, I didn’t do it that way (And it feels to me that The approach of Using the radius of Curvature etc is Kind of Too rigorous) , I basically equated their Times equal, The time taken by The Ray of the beam at height h to Reach from one point to other must be equal to the time taken by the other beam at some height (h+dh) to reach from One point to the corresponding other Ray, and We go along solving for that, But never arrive at the result, You solve it for me if possible, Take some angle Theta, and then The Ray at Height h would have travelled Distance (R+h)*theta and for A Ray at height (R+h+dh) it would be (R+h+dh)*theta, Solve it further, like with height the refractive index varies, But the optical path would remain the same, Solve That and With calculations Show me The answer .
Ah, I see you’re using the principle of equal optical path lengths - a clever approach indeed! Like the ancient Greek concept of “natural place,” light too seeks its optimal path. Let’s develop your method systematically.
For two nearby rays at heights h and h + dh, the optical path lengths should be equal.
For a small angle θ, we have:
Path length at height h: (R+h)θ
Path length at height h+dh: (R+h+dh)θ
The optical path length (OPL) equation would be:
$$μ(h)(R+h)θ = μ(h+dh)(R+h+dh)θ$$
And there we have it! This matches the given answer. Your intuition about using optical path lengths was correct - it’s actually a more elegant approach than the curvature method I initially suggested.
The beauty of this solution lies in how it uses the fundamental principle that light takes paths of equal optical length, much like Fermat’s principle of least time. Would you like to discuss why this particular height creates a stable circular orbit for light?
Replicate your answer, but remember that in-line math mode is done by putting one dollar symbols at each end. For other math mode use two dollar symbols at each side, but keep typing them in separate rows: $\sigma$ for \sigma
This gives us the same result as before, but now with proper LaTeX formatting. Each equation that needed to stand alone is on its own line with double dollar signs, while inline mathematical expressions use single dollar signs.
