# Refraction in a medium with a gradient of refractive index

1. Apr 6, 2012

### Gobil

Hi Folks,

After a general search online, I have not yet found a simple description of refraction in a medium with an inhomogeneous refractive index.

For example: if we have a block of glass with a beam of light shining through it, and the block has a gradient in the real part of the refractive index perpendicular to the direction of beam propagation, what will be the angle it exits the block with?

I assume the fromula will depend on the gradient and the thickness of the block, but I don't know how.

Also an intuitive picture would help me here, the light has different velocities, but favors the faster route right, so why then does it tend towards the values with higher n? Where it is slower! See wiki GRIN lenses.

Many thanks!

2. Apr 7, 2012

### Staff: Mentor

You can model this as a large number of small changes in the refraction index and let this number go to infinity. If you are just interested in the angle and the modification of refractive index is linear in one coordinate: Use that sin(theta)*n is constant, where theta is the angle relative to this coordinate.

3. Apr 7, 2012

### haruspex

The light takes the fastest route to a given endpoint. Look at where the light started and finished, and you'll see that in between it curved so as to go through less dense medium before having to head off to its destination.

4. Apr 7, 2012

### Staff: Mentor

In addition, it takes the fastest way only locally, not globally. Light will still go through some meters of glass, even if there would be a bended way around the glass (at least without diffraction).

5. Apr 8, 2012

### Gobil

thanks for the responses guys.

so, ignoring density for a while, and just thinking about the refractive index gradient, lets assume it´s a linear change in space perpendicular to the beam propagation direction. The light bends towards the slower part, i.e. it will tens to bend to the places with higher refractive index. Is this because the tranverse fields are slowed down there, kinda causing it to bend?

Also the equation sin(theta)*n= constant doesn´t really help much, how is the gradient and thickness of the glass taken into account?

thanks!

6. Apr 8, 2012

### Staff: Mentor

In that case, refraction does not change the angle at all.
Refraction occurs because the phase velocity is different for different media, right.

That is the nice property of this equation: The details don't matter. You have some refractive index n1 outside (usually ~1), an angle between the light and the change of refractive index theta1 and some value n2 at point x in the glass, and you can calculate the angle of the light theta2 there.

7. Apr 8, 2012

### Gobil

From what I've read this is not true: wiki GRIN lenses. They focus light by having a refractive index gradient inside the material. The actual optic is cOmpletely flat, and the light enters perpendicular to the gradient. I just can't find an equation to describe this. Or a picture of what's going on microscoPically.

8. Apr 8, 2012

### Staff: Mentor

The light which actually enters perpendicular to the gradient is not bent, just the light which has a small angle. See the wikipedia article, there is a good image of that.

9. Apr 8, 2012

### sophiecentaur

I remember doing some work on modelling the ionosphere in connection with MF (mainly) Radio Wave propagation. The refractive index, in that case, depended on the level of ionisation at different heights. The ray tracing method worked pretty well for finding the path taken through the 'graded index, step by step.. The problem was, however, when the ray was almost horizontal and the difference in index at the two levels approached zero. The ray 'never got down again'. I found a solution by recognising when the height steps got too small and then did a simple reflection calculation at that very shallow angle. Then you could go down again with no trouble. It worked pretty well as a method for predicting MF skywave interference. Of course, just like the weather, lower down, predicting the actual levels of ionisation at different heights was a bit hit and miss - but that was someone else's problem.

10. Apr 9, 2012

### Gobil

I think this is incorrect, and it´s just the way the graphic looks.

for example see the attached image, from: http://www.grintech.de/gradient-index-optics.html

it shows all the perpendicular rays being bent, which is the principal of these optics

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11. Apr 9, 2012

### sophiecentaur

Assuming Snell's law works. Normal incident rays leave normally .
If a graph or a simulation appears to give odd results, suspect it rather than the basic theory. Imo, simulations can lead you into trouble if you are not careful.

Last edited: Apr 9, 2012
12. Apr 9, 2012

### Staff: Mentor

Ok, this image is very convincing in that respect. You are right, assuming the creators of the images knew what they did (I think this is true).
It is not a contradiction to my formula - sin(theta) is 1 at the points where the light "turns", this allows to calculate the angle at higher refractive indices. For lower refractive indices, sin(theta) would have to be larger than 1, which is impossible. But it might be a bad approximation very close to these turnaround points.

13. Apr 9, 2012

### sophiecentaur

The actual tipping point for the ray - as it changes lateral direction - must be treated differently from all other parts of the ray. If you just apply Snells Law and a simple 'layered' model, the logic is, I think, that the ray will asymptote to stay parallel to the axis with nothing to turn it inwards again.
μ12 gets closer to 1 as the layers get thinner and angle C becomes closer and closer to 90 and can't ever start to get less?? So it is necessary to put a lower limit on the thickness of elemental slice and 'assume' straightforward reflection in the final slice (by making the assumption that C has been reached), I think.

14. Apr 9, 2012

### Staff: Mentor

Something along the line of this, yes.
Here is an approximation for light perpendicular to the diffraction index gradient:

Let's assume a diffraction index of n at position x=0 and a gradient of g along the x-axis. A light ray at x-position 0 therefore has the velocity c/(n) and a light ray at +dx has velocity c/(n+g*dx). Using the approximation g*dx << n, this velocity can be approximated as c/n (1-g*dx/n). This is equivalent to a circle with radius g/n.

It becomes a bit more tricky for light in arbitrary directions, but it should be possible to derive the path of the light there as well, using the same idea.

15. Apr 9, 2012

### sophiecentaur

That's smart. That method gives the angle of the wave front and, hence, the direction of the ray.

16. Apr 9, 2012

### haruspex

Consider a medium occupying the x>0, y>0 quarter plane. R.I at x,y is r = 1+k.x.
Ray enters in the y direction at x =a.
Dy/dx = tan θ.
Sin (θ+δθ)/sin (θ) = (1+k.x)/(1+k.x+k.δx)
From these I get
1+k.x = c.cosec(θ), some constant c.
y= (c/2k) ln ((1+cos θ) / (1 - cos θ))
Or something like that.

Last edited: Apr 9, 2012
17. Apr 10, 2012

### mikeph

Am I missing something here? If the gradient is perpendicular to the light then each beam coincides with a contour of n and travels in a straight line across the medium.

18. Apr 10, 2012

### Born2bwire

This is my take on the problem:

The confusion probably lies in the fact that most of us have only dealt with the case of homogeneous slabs. The best way to think of the problem is in terms of Fermat's Principle in which the path length of the light between any two points represents an extremum. So if the light strikes normally to the homogeneous slab, the extremum path for it to follow is the path of least time, going straight through without refracting. The other extremum path (path of longest time) would be to follow the surface of the slab as a surface wave which we reject for physical reasons.

But now we have light striking an inhomogeneous slab. Now the conditions for the extremum path differ. Let us take the parabolic profile that changes along the length of the slab where the maximum lies at the center of the slab and the minimum at the slabs edges. Now the path going straight through, given that the incident ray is normal to the slab, is no longer an extremum path. For example, if the light bends away from the center of the slab it will enter parts of the slab with decreasing n and thus increasing phase velocity. Bending away would represent a minimizing path. Bending towards the center would represent a maximizing path as now the index of refraction increases. So regardless of whether or not the physical path is the maximum or minimum path, it cannot go straight through the slab since we can quickly see that the unrefracted path cannot be a maximum.

So now, how do we choose which path represents the physical extremum path, the minimum or maximum? This depends upon the relative change in the index of refraction. Let us make the converging lens which has a parabolic profile that has a maximum index of refraction at the center, $n(x) = -x^2+n_0$. Now, taking the first order derivative, we see that the index of refraction changes as $n'(x) = -2x$. So the ray is going to bend towards the center of the lens. This we can see by the resulting sinusoidal path that is shown the previously posted graphic. Note that once the ray has crossed the center axis, it reverses the direction of refraction to go back towards it.

Likewise, if we choose a parabolic profile such that the center is the minimum index of refraction, then we would have a diverging lens because now the first order derivative of the profile is $n'(x) = 2x$.

19. Apr 16, 2012

### Gobil

Good post thanks!

something is still missing for me though, why, microscopically, does it go towards the slower regions? does the phase velocity suffer a drag there and hence tends towards that region?

20. Apr 16, 2012

### sophiecentaur

Fermat's Principle is what it's about.