Geometric optics and Fermat's principle

1. Oct 11, 2015

whatisreality

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
I've tried to work out the specific trajectory for the ray that grazes the x axis at xg. For x=xg, 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 xg = 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.

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2. Oct 11, 2015

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?

3. Oct 12, 2015

whatisreality

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 in to the d = expression, along with the given value of y!

4. Oct 12, 2015

andrewkirk

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 $y$?

5. Oct 13, 2015

whatisreality

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.

Last edited: Oct 13, 2015
6. Oct 13, 2015

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?

7. Oct 13, 2015

andrewkirk

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

8. Oct 14, 2015

whatisreality

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?