# Homework Help: Optics: Find the index of refraction

1. Jul 1, 2010

### Niles

1. The problem statement, all variables and given/known data
Consider a cylindrical piece of material of a constant thickness d and radius r = a, but made out of a material with position-dependent index of refraction. How must the index of refraction vary with radial distance r such that the cylindrical piece act as a lens with a focal length of f. Assume d << a, i.e., a thin lens.

3. The attempt at a solution
Ok, first I remind myself that light follows the path, where the optical path length is stationary. So I write

$$\Gamma = \int {n\left( {r,\phi ,z} \right)} \,rdrd\phi dz.$$

I can integrate dz and dφ out, since we only want to look at the r-dependence. Now, I need to find dGamma/dr = 0, but that just gives me n(r)=0, which is clearly wrong. Can you give me a hint?

Best,
Niles.

Last edited: Jul 1, 2010
2. Jul 1, 2010

### zzzoak

May be this will be useful.

$$d(nr)/dr=rdn/dr+n=0$$

$$dn/n=-dr/r$$

$$lnn=-lnr+c$$

$$n=c/r$$

3. Jul 7, 2010

### PhillipKP

There is a relationship between the index, optical path length, and phase. The higher the index, the more the optical path length, and the more the phase delay a light will have as it traverses a piece of glass.

The point of a converging (aka positive aka convex) lens is to induce more optical path length (aka phase delay) in the middle and less optical path length on the edges.

Your first step is to find the optical path length induced by a lens of similar center thickness, d, as a function of radius r. From there you should be able to easily convert this optical path length into an index function of r for your thin cylindrical piece of glass.

The optical path length function of a lens of focal length f should be easily found in reference materials.

Another important concept: You can also derive the optical path length function of a lens (and thus your disk) by noting that the optical length for all rays whether traveling through the edge or through the center must be equal if it is imaging a point source to another point source.

By the way, this thing is called a GRIN lens and is very common in optical fibers.

Hope this helps.

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