Optics: Find the index of refraction

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SUMMARY

The discussion focuses on determining the position-dependent index of refraction required for a cylindrical lens to achieve a specified focal length, f. The key relationship derived is n(r) = c/r, indicating that the index of refraction decreases with increasing radial distance. This relationship is essential for creating a gradient-index (GRIN) lens, which is commonly used in optical fibers. The optical path length must be manipulated to ensure that light rays traveling through different parts of the lens converge correctly.

PREREQUISITES
  • Understanding of optical path length and its significance in lens design.
  • Familiarity with the concept of gradient-index (GRIN) lenses.
  • Knowledge of calculus, specifically integration and differentiation.
  • Basic principles of optics, including focal length and phase delay.
NEXT STEPS
  • Research the optical path length function for lenses with varying thicknesses.
  • Study the design and applications of gradient-index (GRIN) lenses in optical systems.
  • Learn about the mathematical derivation of lens equations and their implications in optics.
  • Explore the use of optical fibers and how GRIN lenses enhance their performance.
USEFUL FOR

Students and professionals in optics, optical engineers, and anyone involved in the design and application of lenses and optical systems.

Niles
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Homework Statement


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.

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

<br /> \Gamma = \int {n\left( {r,\phi ,z} \right)} \,rdrd\phi dz.<br />

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?Niles.
 
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May be this will be useful.

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

dn/n=-dr/r

lnn=-lnr+c

n=c/r
 
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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