# Index of refraction, position dependent

• 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... advice.Yes, that's right. You should be able to find an explicit expression for n'(y,n) and then plug it into the Euler-Lagrange equation to get an explicit expression for n. Then you can
stripes

## Homework Statement

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."

--

## 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.

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

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

we can write n = n(y), and remember that y = y(x), so the index depends on position.
(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)

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

## 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?

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'})##

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?

stripes said:
$\frac{\partial L}{\partial y} = \frac{d }{dx} (\frac{dn}{dy} ) \sqrt{1 + \frac{dy}{dx} ^{2} }$
I don't see where you get that d/dx from.
$\frac{\partial L}{\partial y'} = \frac{n(y) \frac{dy}{dx}}{ \sqrt{1 + (\frac{dy}{dx}) ^{2} } }$
finally,
$\frac{d}{dx} (\frac{\partial L}{\partial y'}) = \frac{4(\frac{dn(y)}{dx} \frac{dy}{dx} + n(y) \frac{d^{2} y }{dx^{2} }) \sqrt{1 + \frac{dy}{dx} ^{2} } - \frac{2 \frac{dy}{dx} \frac{d^{2} y }{dx^{2} } }{ \sqrt{1 + \frac{dy}{dx} ^{2} } }}{4(1 + \frac{dy}{dx} ^{2})}.$
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.

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.

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??

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}##

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

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?

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.

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:
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.

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

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

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

## 1. What is the index of refraction?

The index of refraction is a measure of how much light bends when it passes through a material. It is represented by the symbol "n" and is calculated by dividing the speed of light in a vacuum by the speed of light in the material.

## 2. How does the index of refraction vary with position?

The index of refraction can vary with position in different materials, as the density and composition of the material may change. This can lead to changes in the speed of light and therefore the index of refraction.

## 3. How is the index of refraction measured?

The index of refraction can be measured using a variety of techniques, such as using a spectrometer or a refractometer. These instruments measure the angle of light as it passes through a material and can calculate the index of refraction based on this data.

## 4. What factors can affect the index of refraction?

The index of refraction can be affected by several factors, such as temperature, pressure, and the wavelength of light. Changes in these factors can lead to changes in the speed of light and therefore the index of refraction.

## 5. How does the index of refraction affect the behavior of light?

The index of refraction plays a crucial role in how light behaves in different materials. It determines the angle at which light bends when passing through a material and also affects the speed of light. This can impact the appearance of objects and how light is transmitted through different media.

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