# Light beam path

1. Apr 20, 2015

### skrat

1. The problem statement, all variables and given/known data
In plane parallel plate the refractive index is a function of coordinate $z$, so that $n=n_0 -{n}'z^2$ for ${n}'>0$. The origin of the coordinate system is in the middle of the layer, and $z$ is parallel to the normal of the layer. In paraxial approximation calculate the path of a light beam.

2. Relevant equations
If we use parametrization $s$:

$\frac{d}{ds}(n(z)\frac{d\vec r}{ds})=\nabla n$

3. The attempt at a solution
For $z$ coordinate: $$\frac{d}{ds}(n(z)\frac{d z}{ds})=\frac{dn}{dz}=-2{n}'z$$ $$\frac{d}{ds}([n_0-{n}'z^2]\frac{d z}{ds})=-2{n}'z$$ Now I hope I can use approximation that $dz\sim dx$ this would bring me to $$[n_0-{n}'z^2]\frac{d^2z}{dx^2}=-2{n}'z$$ and finally since $z$ is very small, than I can forget about $z^2$ term. $${z}''+\frac{2{n}'}{n_0}z=0$$ which brings me to my final solution $$z(x)=Asin(\sqrt{\frac{2{n}'}{n_0}}x)+Bcos(\sqrt{\frac{2{n}'}{n_0}}x)$$
Or is this completely wrong?

2. Apr 20, 2015

### TSny

Reconsider this. Paraxial means that the ray always makes a small angle to the z-axis.

Note that this is a vector equation. So, you will need to think about its components.

3. Apr 20, 2015

### skrat

Hmmm, ok firstly, let me apologize for a mistake in my first post. There should be $ds\approx dx$ and not $dz\approx ds$.
But you got me a bit confused now. I think you are trying to say that I should use approximation where $ds\approx dz$ but... this is in contrary with the problem. The problem says that the ray is only travelling close to the centre of the layer (where $z$ is very small). Or...?

The other two components only give me $$n(z)\frac{dx}{ds}=n(z)\sin\vartheta _x=C_x$$ and $$n(z)=\sin\vartheta _y=C_y.$$ Do they have any physical interpretation?

4. Apr 20, 2015

### TSny

OK, I guess I misinterpreted the question. Since they only mention a z-axis, I assumed that paraxial meant that the ray is traveling approximately in the z-direction as it passes through the plate. So, yes, I was thinking ds ≅ dz.

[EDIT: I agree with your solution.]

Last edited: Apr 20, 2015