Wave in inhomogeneous medium (ray equation)

[deedsy]

Homework Statement

A wave travels in a stratified medium whose index of refraction is a function of the coordinate y. Show that the angle $\theta$ between a ray and the y-axis obeys the following law:
$\frac{d\theta}{ds} = \frac{-(dn/dy) sin(\theta)}{n}$ , where the distance s is measured along the ray.

Using this result, you can verify the ray equation $\frac{d}{ds} (nt) = \nabla n$ , where t is a unit vector tangent to the ray at a point where the index of refraction is n.

Hint: Select your y-axis along $\nabla n$ and your x-axis in the plane of incidence.

Homework Equations

Snells Law --> $n_1 sin(\theta_1) = n_2 sin(\theta_2)$

The Attempt at a Solution

I need an equation that relates $\theta, s, y,$and $n(y)$, which shouldn't be hard but I can't seem to figure it out.

I attached a simple figure of how I am setting this up.
From the figure, s and y are related by
$cos(\theta) = \frac{dy}{ds}$
$\theta = \cos^{-1}(\frac{dy}{ds})$

n and $\theta$ are related by Snell's Law, which for this problem, can be applied to a large number of thin y-axis layers with varying refractive indexes
$n_1 sin(\theta_1) = n_2 sin(\theta_2) = n_3 sin(\theta_3) = A$, where A is just a constant
$n = \frac{A}{\sin(\theta)}$
$\frac{dn}{d\theta} = \frac{-A cos(\theta)}{sin^2(\theta)}$

With these expressions for $dn, dy, ds, d\theta$, I've just been manipulating these expressions trying to match what the answer should be, but I'm not making any progress. Am I on the right track trying to solve this using only the geometry of the problem and Snell's law, or is there another way someone can reccomend? I've seen the ray equation derived by using phase differences and fermat's principle, but those didn't yield what the first part of this question is asking for.

 

[TSny]

$\frac{dn}{d\theta} = \frac{-A cos(\theta)}{sin^2(\theta)}$

Try invoking the chain rule to re-express $\frac{dn}{d\theta}$.

 

[deedsy]

thanks TSny, that tip led me to the right answer!

Now I'm trying to verify the ray equation $\frac{d}{ds} (nt) = \nabla n$ with that result; so far,
$\frac{d\theta}{ds} = \frac{-(dn/dy) sin(\theta)}{n}$
since n is only dependent on the y direction, $\nabla n = \frac{dn}{dy}$, so
$\frac{-d\theta}{ds} \frac{n}{sin(\theta)} = \nabla n$
I then tried taking $\frac{d}{ds}$ of both sides, but couldn't reduce it to the ray equation

so, somehow $\frac{-d\theta}{ds} \frac{n}{sin(\theta)}$ must equal $\frac{d}{ds} (nt)$ , where t is "a unit vector tangent to the ray at a point where the index of refraction is n".

 

[TSny]

$\frac{-d\theta}{ds} \frac{n}{sin(\theta)} = \nabla n$

This looks good assuming that here $\nabla n$ refers to the magnitude of the gradient: $|\vec{\nabla} n|$.

You can use this result to show that $\frac{d}{ds} \left ( n \hat{t} \right )$ reduces to $\vec{\nabla} n$, where here $\vec{\nabla} n$ is the gradient vector (not the magnitude).

I doubt if I have the most elegant way to do it, but I found it helpful to express the unit tangent vector $\hat{t}$ in terms of Cartesian unit vectors $\hat{i}$ and $\hat{j}$ and the angle $\theta$ between $\hat{t}$ and the y-axis.

[EDIT: Actually, it also works out nicely if you don't bother to express $\hat{t}$ in terms of $\hat{i}$ and $\hat{j}$. Either way, my hint would be to use the product rule on $\frac{d}{ds} \left ( n \hat{t} \right )$ and see if you can simplify it.]

 

[deedsy]

Okay, so I'm going to reduce $\frac{d}{ds} (n \hat{t})$ down to $\vec{\nabla}n$
$\frac{d}{ds} (n \hat{t})$
$\frac{dn}{ds} + \frac{d\hat{t}}{ds}$
Using a previous result...
$\frac{-A cos(\theta)}{sin^2(\theta)} \frac{d\theta}{ds} + \frac{d\hat{t}}{ds}$
Using another previous result...
$\frac{A cos(\theta)}{sin^2(\theta)} \frac{\nabla n sin(\theta)}{n} + \frac{d\hat{t}}{ds}$
$\frac{A cos(\theta) \nabla n}{n sin(\theta)} + \frac{d\hat{t}}{ds}$
$(\nabla n)(cos(\theta)) + \frac{d\hat{t}}{ds}$

Now, since $\hat{t}$ is always tangent to $ds$, I believe $\frac{d\hat{t}}{ds} = 0$

There is also the relation $cos(\theta) = \frac{dy}{\hat{t}}$

So,$(\nabla n)(cos(\theta)) + \frac{d\hat{t}}{ds} = \nabla n \frac{dy}{\hat{t}} = \vec{\nabla}n$?
I'm not sure about the last step there, though..

 

[TSny]

Okay, so I'm going to reduce $\frac{d}{ds} (n \hat{t})$ down to $\vec{\nabla}n$
$\frac{d}{ds} (n \hat{t})$
$\frac{dn}{ds} + \frac{d\hat{t}}{ds}$

This isn't correct. The product rule for derivatives gives $\frac{d}{ds} (n \hat{t}) = \frac{dn}{ds}\hat{t} + n \frac{d\hat{t}}{ds}$

But you are right that $\frac{dn}{ds} = |\vec{\nabla} n| \cos \theta$. Another way to get this is to write $\frac{dn}{ds} = \frac{dn}{dy} \frac{dy}{ds}$ and note that $\frac{dn}{dy} = |\vec{\nabla} n|$ and $\frac{dy}{ds} = \cos \theta$

Now, since $\hat{t}$ is always tangent to $ds$, I believe $\frac{d\hat{t}}{ds} = 0$

This is not zero. As you move along the ray, the unit tangent vector will change direction.

There is also the relation $cos(\theta) = \frac{dy}{\hat{t}}$

You can't divide by a vector.

 

[deedsy]

wow, yeah, so after the real product rule, I have
$\frac{d}{ds} (n\hat{t}) = \nabla n cos(\theta) \hat{t} + n \frac{d\hat{t}}{ds}$

so I need an expression that can show how $\hat{t}$ changes with $s$, but I'm having trouble with that.
You said before you found an expression between $\theta$, $\hat{t}$ and the y-axis; is this the expression that can be useful for this part? I don't see what else this could be besides $cos(\theta) \hat{t} = dy$, but that doesn't work because you can't divide by a unit vector.

 

[TSny]

wow, yeah, so after the real product rule, I have
$\frac{d}{ds} (n\hat{t}) = \nabla n cos(\theta) \hat{t} + n \frac{d\hat{t}}{ds}$

OK

so I need an expression that can show how $\hat{t}$ changes with $s$, but I'm having trouble with that.

Try the chain rule again. $\frac{d\hat{t}}{ds} = \frac{d\hat{t}}{d \theta} \frac{d\theta}{ds}$

You said before you found an expression between $\theta$, $\hat{t}$ and the y-axis; is this the expression that can be used for this part?

Yes. How would you express the unit tangent vector $\hat{t}$ in terms of $\hat{i}$, $\hat{j}$, and $\theta$?

 

[deedsy]

$\frac{d\hat{t}}{ds} = \frac{d\hat{t}}{d\theta} \frac{d\theta}{ds}$
$\frac{d\hat{t}}{d\theta} \frac{d\theta}{ds} = \frac{-d\hat{t}}{d\theta} \frac{\nabla n sin(\theta)}{n}$
$\frac{d\hat{t}}{ds} = -\frac{\nabla n sin(\theta)}{n} \frac{d\hat{t}}{d\theta}$

so now I need my expression relating $\theta$ and $\hat{t}$

$\hat{t} = (ds) sin(\theta) \hat{i} + (ds) cos(\theta) \hat{j}$
$\frac{d\hat{t}}{d\theta} = ds cos(\theta) \hat{i} - ds sin(\theta) \hat{j}$

$\frac{d\hat{t}}{ds} = -\frac{\nabla n sin(\theta)}{n} [ds cos(\theta) \hat{i} - ds sin(\theta) \hat{j}]$

Does this look right so far? If so, next I'll plug it into
$\frac{d}{ds} (n\hat{t}) = \nabla n cos(\theta) \hat{t} + n \frac{d\hat{t}}{ds}$

 

[deedsy]

actually, I'm not sure i need the (ds) in the unit vector equation

 

[deedsy]

ok, without including the (ds) I was able to get the answer!
$\frac{d}{ds} (n\hat{t}) = \nabla n\hat{j}$

Thanks so much for your help TSny

 

[TSny]

ok, without including the (ds) I was able to get the answer!
$\frac{d}{ds} (n\hat{t}) = \nabla n\hat{j}$

That looks right. Good work! (Yes, the ds should not be in the expression for the unit tangent vector.)