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## Homework Statement

We know, for constant

*n*,

[itex]\frac{\partial L}{\partial y} = \frac{d}{dx} (\frac{\partial L}{\partial y'}),[/itex]

where we define

*L*as being the optical light path length, and y' is the derivative of y with respect to x, and

[itex]L = L(y, y') = n \sqrt{1 + y' ^{2}}. [/itex]

However in a more complex medium, we can allow the index of refraction to vary as a function of position. This means that we can write n = n(y), and remember that y = y(x), so the index depends on position. With this idea, answer the following:

(a) Show that if

*n*is position dependent, it must obey it's own differential equation and will have the solution (where n

_{0}is a constant)

[itex]n = n_{0} e ^{\int \frac{y''}{1+y'^{2} } dy}.[/itex]

You will first need to derive the differential equation for

*n*, and then solve it to show that this is the solution. Notice that for a straight line (y = mx + b), we know that y'' = 0, so n must be constant.

(b) Light is seen to take an exponential path in a particular medium, so that [itex]y(x) = y_{0} e ^{-\alpha x},[/itex] where [itex]\alpha[/itex] and y

_{0}are constants. Find the position dependence of the index of refraction n(y) which causes light to take this path.

(c) Substitute your expression for n(y) back into the original equation [itex]\frac{\partial L}{\partial y} = \frac{d}{dx} (\frac{\partial L}{\partial y'})[/itex] and show that the function y(x) must satisfy the differential equation

[itex] y''(1 + \alpha^{2} y^{2}) - \alpha^{2}y(1 + y'^{2}) = 0[/itex]. The question then says "obviously this is NOT an easy equation to solve, but show that the exponential form of y(x) used in part (b) does indeed solve this equation."

## Homework Equations

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## The Attempt at a Solution

I have no idea how to start part (a). The question says "it's own differential equation"--what is that supposed to mean? The question wants me to "derive the differential equation for n", but where would I start?

In part (b), it seems as simple as taking the function [itex]y(x) = y_{0} e ^{-\alpha x},[/itex] and plugging it into the equation [itex]n = n_{0} e ^{\int \frac{y''}{1+y'^{2} } dy}.[/itex] Of course, the latter expression is integrated with respect to y, and I'm plugging in an expression that is a function of x, so is this not the way to approach this problem?

I won't even think about part (c) until I can at least figure out how to start parts (a) and (b). I just posted it so I wouldn't have to at a later point.

If someone could guide me in the right direction, I would really appreciate it. I am very lost and this is the first question on my assignment.