# Index of refraction, position dependent

1. Feb 3, 2013

### stripes

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

We know, for constant n,

$\frac{\partial L}{\partial y} = \frac{d}{dx} (\frac{\partial L}{\partial y'}),$

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

$L = L(y, y') = n \sqrt{1 + y' ^{2}}.$

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 n0 is a constant)

$n = n_{0} e ^{\int \frac{y''}{1+y'^{2} } dy}.$

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 $y(x) = y_{0} e ^{-\alpha x},$ where $\alpha$ and y0 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 $\frac{\partial L}{\partial y} = \frac{d}{dx} (\frac{\partial L}{\partial y'})$ and show that the function y(x) must satisfy the differential equation

$y''(1 + \alpha^{2} y^{2}) - \alpha^{2}y(1 + y'^{2}) = 0$. 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."

2. Relevant equations

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3. 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 $y(x) = y_{0} e ^{-\alpha x},$ and plugging it into the equation $n = n_{0} e ^{\int \frac{y''}{1+y'^{2} } dy}.$ 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.

2. Feb 3, 2013

### TSny

Try substituting the given form of L into the Euler-Lagrange equation. What expression do you get for $\frac{\partial L}{\partial y}$?

What do you get for $\frac{\partial L}{\partial y'}$?

Finally, what do you get for $\frac{d}{dx} (\frac{\partial L}{\partial y'})$

3. Feb 3, 2013

### stripes

I have

$\frac{\partial L}{\partial y} = \frac{d }{dx} (\frac{dn}{dy} ) \sqrt{1 + \frac{dy}{dx} ^{2} }$

and

$\frac{\partial L}{\partial y'} = \frac{n(y) \frac{dy}{dx}}{ \sqrt{1 + (\frac{dy}{dx}) ^{2} } }$

finally,

$\frac{d}{dx} (\frac{\partial L}{\partial y'}) = \frac{4(\frac{dn(y)}{dx} \frac{dy}{dx} + n(y) \frac{d^{2} y }{dx^{2} }) \sqrt{1 + \frac{dy}{dx} ^{2} } - \frac{2 \frac{dy}{dx} \frac{d^{2} y }{dx^{2} } }{ \sqrt{1 + \frac{dy}{dx} ^{2} } }}{4(1 + \frac{dy}{dx} ^{2})}.$

So we have

$\frac{d }{dx} (\frac{dn}{dy} ) \sqrt{1 + \frac{dy}{dx} ^{2} } = \frac{4(\frac{dn(y)}{dx} \frac{dy}{dx} + n(y) \frac{d^{2} y }{dx^{2} }) \sqrt{1 + \frac{dy}{dx} ^{2} } - \frac{2 \frac{dy}{dx} \frac{d^{2} y }{dx^{2} } }{ \sqrt{1 + \frac{dy}{dx} ^{2} } }}{4(1 + \frac{dy}{dx} ^{2})}.$

I highly doubt I have done anything correctly. I just keep telling myself that I need to make sure I take implicit derivatives because n is a function of y is a function of x. In the unlikely event that I am correct, I just solve for n(y), but since this is obviously wrong, where have I made my mistake?

4. Feb 3, 2013

### haruspex

I don't see where you get that d/dx from.
One or two problems with that last step. The sign seems to be wrong; the last term in the numerator is missing a couple of factors (n, y").
It will be a lot easier to read and write if you use y' for dy/dx etc.

5. Feb 3, 2013

### stripes

I got that dy/dx from the Euler-Lagrange equation. I shouldn't have done that. It should be without the dy/dx.

So if I do the derivatives correctly, I should be able to find an explicit function n'(y, n)? Then I need to solve the differential equation and show the solution is the one I originally posted, and then parts (a) and (b) should follow. I will try finding the partial derivatives again in a while but thank you for your help.

6. Feb 4, 2013

### stripes

So I have been going at it for a while. I just keep getting complicated expressions with x's and y's everywhere. Should the differential equation cancel things nicely? Or does it look messy the whole way through?

I get

$\frac{d}{dx} (\frac{\partial L}{\partial y'}) = \frac{dn}{dx} (\frac{y'}{\sqrt{1 + y'^{2}}}) + n(\frac{y''\sqrt{1 + y'^{2} - \frac{y'^{2}y''}{\sqrt{1 + y'^{2}}}}}{1 + y'^{2}})$

and I have canceled a few things here and there but I don't get end up with anything that looks pleasant.

Am I doing my derivatives correctly??

7. Feb 4, 2013

### TSny

Things will get messy for a while and then simplify quite a bit when you combine terms.

So, you have shown $\frac{\partial{L}}{\partial{y'}} = \frac{ny'}{\sqrt{1+y'^2}}= n\; y' \frac{1}{\sqrt{1+y'^2}}$

When you take the derivative of this with respect to $x$, you will use the product rule with three factors:

$\frac{dn}{dx}\; y' \frac{1}{\sqrt{1+y'^2}}+ n\; \frac{dy'}{dx} \frac{1}{\sqrt{1+y'^2}}+ n\; y'\; \frac{d}{dx}(1+y'^2)^{-1/2}$

Note that $n$ is a function of $y(x)$, so you will use the chain rule to evaluate $\frac{dn}{dx}$

8. Feb 4, 2013

### haruspex

I hope you mean $\frac{dn}{dx} (\frac{y'}{\sqrt{1 + y'^{2}}}) + n(\frac{y''\sqrt{1 + y'^{2}} - \frac{y'^{2}y''}{\sqrt{1 + y'^{2}}}}{1 + y'^{2}})$
After working out dn/dx, as TSny says, you'll see some cancellation.

9. Feb 5, 2013

### stripes

I have

$\frac{\partial L}{\partial y} = (\frac{dn}{dy} ) \sqrt{1 + y' ^{2} }$

and

$\frac{\partial{L}}{\partial{y'}} = \frac{ny'}{\sqrt{1+y'^2}}$

And finally,

$\frac{\partial L}{\partial y} = \frac{d}{dx} (\frac{\partial L}{\partial y'}).$

This means

$(\frac{dn}{dy} ) \sqrt{1 + y' ^{2} } = \frac{d}{dx} (\frac{ny'}{\sqrt{1+y'^2}}).$

So let's find $\frac{d}{dx} (\frac{ny'}{\sqrt{1+y'^2}})$.

Use the quotient and product rules:

$\frac{d}{dx} (\frac{ny'}{\sqrt{1+y'^2}}) = \frac{ \frac{d}{dx} (ny') ( \sqrt{1+y'^{2}} ) - \frac{d}{dx}(\sqrt{1+y'^2})(ny') } {1 + y'^{2} }$

$= \frac{ ( \frac{dn}{dx}y' + ny'')(\sqrt{1+y'^2}) - \frac{n y'^{2} y''}{\sqrt{1+y'^{2}}} }{1 + y'^{2}}$

$= \frac{ (\frac{dn}{dx} y' + ny'')(\sqrt{1+y'^2})}{1 + y'^{2}} - \frac{\frac{n y'^{2} y''}{\sqrt{1+y'^{2}}}}{1 + y'^{2}}$

$= \frac{ ( \frac{dn}{dx}y' + ny'')(\sqrt{1+y'^2})}{1 + y'^{2}} - \frac{n y'^{2} y''}{ \sqrt{1+y'^{2}} (1 + y'^{2}) }$

which means

$\frac{dn}{dy} = \frac{( \frac{dn}{dx}y' + ny'')}{1 + y'^{2}} - \frac{n y'^{2} y''}{ (1+y'^{2}) (1 + y'^{2}) }$

Now where do I go from here?

10. Feb 5, 2013

### TSny

Find least common denominator of right hand side and combine into one fraction.

You will also need to write out dn/dx using chain rule to express it in terms of dn/dy.

11. Feb 5, 2013

### stripes

So we have

$\frac{dn}{dy} = \frac{\frac{ny''}{(1+y'^{2})}}{1 - \frac{y'^{2}}{(1+y'^{2})^{2}} - \frac{y'^{4}}{(1+y'^{2})^{2}}}$

no way I was interested in doing the algebra, so I used Mathematica to simplify it; it would have taken me all night,

$\frac{dn}{dy} = \frac{n y''}{1 + y'^{2}}$

and I'm not sure how to solve that...

Last edited: Feb 5, 2013
12. Feb 5, 2013

### stripes

Nope I think I did something wrong. This is absolutely ridiculous, if I have to keep track of all this stuff then I might as well not do the question because I have more assignments due tomorrow and this is only one question.

13. Feb 5, 2013

### TSny

Rearrange as

$\frac{dn}{n} = \frac{y''dy}{1 + y'^{2}}$ and integrate both sides.