Calculating Focal Length of 1D Fresnel Lens

In summary: The peak will be at the point where the error function is zero.In summary, the focal length of a 1D Fresnel lens is given by the equation $$T(\xi)=\frac 1 2(1+\cos(\alpha \xi ^2)).$$ The transmittance function tells you how much light passes through the lens per unit time, and the peak intensity of light passing through the lens will be at the point where the error function is zero.
  • #1
skrat
748
8

Homework Statement


Calculate the focal length of 1D Fresnel lens, whose transmittance is given as $$T(\xi)=\frac 1 2(1+\cos(\alpha \xi ^2)).$$

Homework Equations


Anything you wish

The Attempt at a Solution


I have no idea. I tried to use the equation for diffraction image $$u_p=C\int _0 ^{R_p}T(\xi)e^{ik\frac{r^2}{2z}}rdr$$ which is simply a Fresnel diffraction multiplied with lens transmittance. I thought this would bring me to something useful, but I couldn't see how this would help.

Any ideas/hints on how to solve this problem would be highly appreciated!
 
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  • #2
Anything you wish
... nope, it's anything you choose ... part of the point of this section is to tell us how you are thinking about the problem. We already know how we are thinking of the problem.

I have no idea.
You could try looking at your notes - consider: what does the transmittance function tell you?
How would you use the transmittance function to work out how a plane wave would be affected?
I thought this would bring me to something useful, but I couldn't see how this would help.
... so you did have some idea - did you follow through on it?
Sometimes you won't be able to see how something will turn out until after you'd gone down the path for a bit.

Also see:
http://inst.eecs.berkeley.edu/~ee290f/fa04/Lenses_post.pdf
 
  • #3
The problem is here:
$$u_f=C\int T(\xi)e^{i\frac{k}{2R}\xi ^2}d\xi=C\int \frac 1 2(1+\cos(\alpha \xi^2))e^{i\frac{k}{2R}\xi ^2}d\xi=$$ $$=C\int \frac 1 2 (1+\frac 1 2(e^{i\alpha \xi ^2}+e^{-i\alpha \xi ^2}))e^{i\frac{k}{2R}\xi ^2}d\xi=$$ $$=C\int[\frac 1 2e^{i\frac{k}{2R}\xi ^2}+\frac 1 4 e^{i(\frac{k}{2R}-\alpha)\xi ^2}+\frac 1 4 e^{i(\frac{k}{2R}+\alpha)\xi ^2}]d\xi$$
In in the focal point the intensity (which is ##\propto |u_f|^2##) should have a huge peak. And from the equation above I should be able to already see that. Yet I don't. I am confused because of the imaginary argument in my exponents. How do I in this case find out where the peak is? :/
 
  • #4
skrat said:
I am confused because of the imaginary argument in my exponents.
You shouldn't be too concerned about this. These integrals are a variant of Gaussian integrals and they have the form:-
##\int_0^x e^{t^2}\ dt = -i\sqrt{\frac{\pi}{2}}erf(ix)##
where ##erf(ix)## is the error function.
 

1. What is a 1D Fresnel lens?

A 1D Fresnel lens is a type of lens that has a flat surface on one side and a series of concentric grooves on the other side, which bend light in a specific way to create a magnifying effect.

2. How is the focal length of a 1D Fresnel lens calculated?

The focal length of a 1D Fresnel lens can be calculated using the formula f = R/2, where R is the radius of the lens. This assumes that the lens is thin and the light rays are parallel to the lens axis.

3. What factors affect the focal length of a 1D Fresnel lens?

The primary factors that affect the focal length of a 1D Fresnel lens are the radius of the lens, the index of refraction of the material the lens is made of, and the angle of the incident light rays.

4. Can the focal length of a 1D Fresnel lens be changed?

Yes, the focal length of a 1D Fresnel lens can be changed by altering the radius of the lens or changing the angle of the incident light rays. This can be done by adjusting the position or orientation of the lens.

5. What are some common applications of 1D Fresnel lenses?

1D Fresnel lenses are commonly used in lighthouses, magnifying glasses, and projectors. They are also used in optical devices such as cameras, telescopes, and microscopes to focus and magnify light.

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