Geometric optics and Fermat's principle

[whatisreality]

Homework Statement

A ray travels as shown in the image attached below. In this case, Fermat's principle may be written as

$A =\frac{n(1+ay)}{\sqrt{1+(y')^2}}$

Where y' is dy/dx, n is the index of refraction and A is a real constant.

The trajectory of a ray of light is given by

$y = -\frac{1}{a}$ $+ \frac{A}{na}$ $\cosh{\frac{na}{A}(x-x0)}$

When the ray is observed from a height y above the x axis, it seems to come from the ground at horizontal distance d from the observer (see figure). Determine d if a(x−x0) is small. Write your answer in terms of y and a.

Homework Equations

The Attempt at a Solution

I've tried to work out the specific trajectory for the ray that grazes the x-axis at x_g. For x=x_g, dy/dx is 0 and y=0, so at that point A = n. Although there is a sqrt involved, take positive n because n=c/v is always positive as far as I know.

Sub that into the second equation for y, taking y=0 again, then I got x_g = x₀.

So the specific trajectory of this ray would then be

$y = -\frac{1}{a}$ $+ \frac{1}{a}$ $\cosh{a(x-xg)}$

Really have no clue what to do from there! I did try finding the tangent to the curve to maybe find an expression for the length of the hypotenuse, but that didn't work. So I don't know.

 

[andrewkirk]

If the eye is at coordinates (x,y) and y' represents the gradient at that point then what expression can you write for d in terms of y and y'?

If you can do that, you just need to get rid of y'. Could the Fermat formula be of any help?

 

[whatisreality]

If the eye is at coordinates (x,y) and y' represents the gradient at that point then what expression can you write for d in terms of y and y'?

If you can do that, you just need to get rid of y'. Could the Fermat formula be of any help?

Oh yeah! An expression such as y'²= d²+y²? Rearranged to make d the subject, then rearrange the fermat formula to make y' the subject and sub into the d = expression, along with the given value of y!

 

[andrewkirk]

An expression such as $y'^2= d^2+y^2$?

Yes, but not that particular expression. $y'$ is the gradient, which is the tangent of the angle of the hypotenuse to the horizontal. What is the relation of that to $d$ and $y$?

 

[whatisreality]

Yes, but not that particular expression. $y'$ is the gradient, which is the tangent of the angle of the hypotenuse to the horizontal. What is the relation of that to $d$ and $y$?

Oh, ok... so tan(θ) = $\frac{y}{d}$=$y'$? So in fact, d = y/y' then?

I don't understand why not the first expression isn't right though. Because the gradient is dy/dx, and dx at the point we're looking at is d, isn't it? I thought the hypotenuse would be represented, when squared, by $d^2 + y^2$.
Although I do know why dy/dx is the tan of the angle of the hypotenuse to the horizontal.

 

[whatisreality]

But then, if I use

$y = -\frac{1}{a}$ $+ \frac{1}{a}$ $\cosh{a(x-xg)}$

Then since I've been given that a(x-xg) is small, cosh(a(x-xg)) = 1. So I'd get that y=0 and therefore d = 0, wouldn't I?

 

[andrewkirk]

But then, if I use

$y = -\frac{1}{a}$ $+ \frac{1}{a}$ $\cosh{a(x-xg)}$

I wouldn't use that, because it has x in it and you are looking for a formula for d in terms of only y and a. Try instead using the first formula in the OP:
$$
A =\frac{n(1+ay)}{\sqrt{1+(y')^2}}
$$

 

[whatisreality]

I wouldn't use that, because it has x in it and you are looking for a formula for d in terms of only y and a. Try instead using the first formula in the OP:
$$
A =\frac{n(1+ay)}{\sqrt{1+(y')^2}}
$$

It does have x in it, but x disappears because I've been given that a(x-xg) is small. I know that y isn't zero! So why would I get that result?