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

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The discussion focuses on calculating the focal length of a GRIN lens with infinite radii of curvature, where the refractive index varies with height. The participant is unsure how to mathematically express the refractive index n(y) given specific values at different heights. It is noted that despite the lens being plane, rays of light bend toward regions of higher refractive index, indicating the material still has a focal length. The conversation references the thin lens equation and compares the situation to optical phenomena like mirages. Ultimately, a suggested formula for n(y) is provided, emphasizing the complexity of the lens's behavior despite its geometric simplicity.
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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