How Does Refractive Index Variation Affect Mirage Visibility in a Desert?

[Quantum Sphinx]

Homework Statement

Question:[/B] Imagine you are in a massive desert. There is a tower of height H. A light source at the top of the tower can emit light in all directions. You are a person whose eyes are at a height h (h<H) above the ground. The refractive index of the air varies as μ=kd where k is a positive constant and d is the distance above the ground.

1. What is the minimum distance at which you will still be able to see the mirage.
2. What is the minimum possible distance between you and the mirage.
3. If the ray starts off making an angle α below the horizontal what is the condition that a mirage is created

(The 3 questions are independent)

2. Homework Equations :
snell's law

2. The attempt at a solution
My approach: I started off by assuming an element of height dx at an elevation x above the ground. Then I assumed that the ray of light is incident at an angle of θ and that the angle of refraction isθ+dθ. the refractive index changes from μ(x+dx) to μ(x). So,

μ(x+dx)⋅sin⁡(θ)=μ(x)⋅sin⁡(θ+dθ)

Solving this gave me weird results. I think it was because I was unable to incorporate Total Internal Reflection into it.

If somebody could either point me in the right direction for solving the question or point out an error (if there is one) in the question (because I made the question myself) then I would be grateful.

 

[haruspex]

Solving this gave me weird results.

Please post your working. It will be important to get the sign right in your first equation.

Edit: I get that the trajectory of the ray is an arc cosh function.

 

[Quantum Sphinx]

Please post your working. It will be important to get the sign right in your first equation.

Edit: I get that the trajectory of the ray is an arc cosh function.

Like I said:
I started off by assuming an element of height dx at an elevation x above the ground. Then I assumed that the ray of light is incident at an angle of θ and that the angle of refraction isθ+dθ. the refractive index changes from μ(x+dx) to μ(x). So,

μ(x+dx)⋅sin⁡(θ)=μ(x)⋅sin⁡(θ+dθ)

That is the first equation.

Then -
μ⋅x⋅sin(θ) + μ⋅sin(θ)⋅dx = μ(x)⋅sin(θ)⋅cos(dθ) + μ(x)⋅cos(θ)⋅sin(dθ)
( assumptions - cos(dθ) = 1 and sin(dθ) = dθ )

which gives -
sin(θ)⋅dx = x⋅cos(θ)⋅dθ

dx/x = cot(θ)⋅dθ

Integrating,
[ln x] = [ ln |sin(θ)| ]

The problem comes in putting the limits. What limit should I use over [ ln |sin(θ)| ]. For question 1, the limits over [ln x] should be from H to h(I think).

 

[haruspex]

The problem comes in putting the limits. What limit should I use over [ ln |sin(θ)| ]. For question 1, the limits over [ln x] should be from H to h(I think).

Very sorry - I was busy when I saw your reply and forgot to come back to it.

For a mirage to occur, what value of angle must be reached before what value of height?

Edit: just realized something about the problem statement. It implies the refractive index is zero at d=0. It cannot be less than 1, surely.
For the first two parts of the question, you need to look at how horizontal distance gets into the equation (x is your vertical distance).