What is the mathematical expression for the index of refraction in a GRIN lens?

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Homework Help Overview

The discussion revolves around calculating the focal length of a gradient-index (GRIN) lens, particularly focusing on the mathematical expression for the varying refractive index as a function of height in the x-y plane. The original poster is uncertain about how to express the refractive index mathematically for different heights, given specific values at certain points.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of infinite radii of curvature on the focal length and question the behavior of light rays in relation to the refractive index. There are discussions about the relationship between the refractive index and the optical path, as well as comparisons to phenomena like mirages.

Discussion Status

The discussion is ongoing, with participants providing insights and raising questions about the nature of the lens and the refractive index. Some participants suggest that traditional optics may not fully explain the behavior observed in GRIN lenses, while others offer mathematical expressions for the refractive index.

Contextual Notes

There is a mention of the original poster's need for a mathematical expression for the refractive index, as well as the constraints of homework rules that may limit the type of assistance provided. The discussion also touches on the challenges of understanding lens behavior in the context of geometrical optics.

fluidistic
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Homework Statement


I must calculate the focal length of a GRIN lens. I.e. a lens which radii of curvature are both infinite. I will consider the refractive index to be varying in function of y (the height in the x-y plane). Say I'm given that at y=0, n(0)=n_0. I'm not sure how to write n(y) mathematically for all y. Say at y=1, n(1)=\frac{n_0}{2} and n(2)=\frac{n_0}{4}.

Homework Equations



n(\vec r )=\int _{\vec r_1}^{\vec r_2 } n(\vec r ) d \vec r.

The Attempt at a Solution



Only thoughts. I must absolutely get an expression for n(y) to start with.
What I know is that all the rays of lights that goes perpendicular to the lens' surface must reach the focal point. I also know that all these rays must have been through the same optical path.
But I'm not sure how to get n(y) nor how to further proceed.
Any help is greatly appreciated.
 
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if the radii of curvature is infinite it has no focal length. it would be a plane parallel piece of glass. If you remember your old optics equations (myself included), the thin lens equation and or thick lens is

(1/f)=(n_2 - n_1)*((1/r_2)-(1/r_1))

if both radii are infinite and 1/infinity = 0.

Never been much for optics, so i hope this helps
 
Liquidxlax said:
if the radii of curvature is infinite it has no focal length. it would be a plane parallel piece of glass. If you remember your old optics equations (myself included), the thin lens equation and or thick lens is

(1/f)=(n_2 - n_1)*((1/r_2)-(1/r_1))

if both radii are infinite and 1/infinity = 0.

Never been much for optics, so i hope this helps

Yeah I know this is strange but geometrical optics fails to explain this fenomena. The "lens" is totally plane and perpendicular rays entering the "lens" should suffer no refraction according to Snell's law but in reality the rays bend toward the region of the lens with greater refractive index. All in all the material has a focal length, as strange as it may seem.
 
fluidistic said:
Yeah I know this is strange but geometrical optics fails to explain this phenomena. The "lens" is totally plane and perpendicular rays entering the "lens" should suffer no refraction according to Snell's law but in reality the rays bend toward the region of the lens with greater refractive index. All in all the material has a focal length, as strange as it may seem.
can lens aberrations be at fault for this phenomena?

* Piston
* Tilt
* Defocus
* Spherical aberration
* Coma
* Astigmatism
* Field curvature
* Image distortion

i've looked through my theoretical and optics textbooks and i can't find anything that would suggest a focal point for a plane of glass.

Your best bet is to consult your prof, i never got help for optics questions on here lol
 
Hmm now that I think of it. Mirage (see http://en.wikipedia.org/wiki/Mirage) is exactly the same thing. Cold air has a higher refractive index than hot air. If you fire a laser horizontally over the ground, according to geometrical optics the laser beam shouldn't bend at all since it would go in a straight line due to a constant refractive index horizontally. However you know that the ray would bend toward the denser (colder) air.
 
The index varies as:
n(y) = n(0) - f/t(sqrt(1+(y/f)^2)-1) where t is the thickness and f is the focal length
 
Last edited:

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