Optics: Find the index of refraction

In summary, the problem involves a cylindrical piece of material with a constant thickness and radius, but with a position-dependent index of refraction. The goal is to determine how the index of refraction must vary with radial distance in order for the cylindrical piece to act as a lens with a focal length of f. By considering the optical path length and phase delay, it can be determined that the index function of r for the thin cylindrical piece of glass should be n=c/r, where c is a constant. This type of lens is commonly known as a GRIN lens and is commonly used in optical fibers.
  • #1
Niles
1,866
0

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

[tex]
\Gamma = \int {n\left( {r,\phi ,z} \right)} \,rdrd\phi dz.
[/tex]

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.
 
Last edited:
Physics news on Phys.org
  • #2
May be this will be useful.

[tex]d(nr)/dr=rdn/dr+n=0[/tex]

[tex]dn/n=-dr/r[/tex]

[tex]lnn=-lnr+c[/tex]

[tex]n=c/r[/tex]
 
  • #3
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.
 

What is the index of refraction?

The index of refraction is a measure of how much a material can slow down the speed of light passing through it. It is a dimensionless quantity and is denoted by the symbol "n".

How is the index of refraction measured?

The index of refraction is typically measured using a device called a refractometer. This device measures the angle at which light bends as it passes from one medium to another, and this can be used to calculate the index of refraction.

What factors affect the index of refraction?

The index of refraction of a material is affected by its composition, density, and temperature. Generally, materials with a higher density and higher refractive index will have a higher index of refraction.

What is Snell's Law and how is it related to the index of refraction?

Snell's Law is a mathematical equation that describes the relationship between the angle of incidence and the angle of refraction for light passing through different mediums. The index of refraction is used in this equation to calculate the angle of refraction.

How is the index of refraction used in practical applications?

The index of refraction is used in many practical applications, including the design of lenses for glasses and cameras, the creation of optical fibers for telecommunication, and in the study of light and its behavior in different materials.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Special and General Relativity
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
810
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Advanced Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Back
Top