# Insights The Diffraction Limited Spot Size with Perfect Focusing - Comments

Tags:
1. Mar 13, 2017

2. Mar 14, 2017

One additional input is the stars are far enough away that the light from them is collimated and thereby the focused spot size that occurs with a perfectly paraboidal primary telescope mirror (neglecting atmospheric turbulence) is diffraction limited in size. The stars are far enough away that they are essentially point sources at infinity, but the image size in the focal plane of the primary mirror of the telescope is larger than a point and is determined by the diffraction limited spot size.

3. Apr 26, 2017

### Mech_Engineer

This is a cool article!

Some additional info which might be useful:
An important topic in diffraction-limited optics is the Airy Disk, which is defined as "the best focused spot of light that a perfect lens with a circular aperture can make, limited by the diffraction of light." The diameter of the Airy disk's first rings drive requirements in many optical imaging systems like cameras, telescopes, infrared imagers, etc.

Airy disk, intensity profile:

A useful equation for use in optical and imaging system engineering is the "Airy Disc Diameter," also described as the diameter to the first "intensity zero," and is useful for estimating the smallest resolvable feature in an image.

Airy Disc Diameter:
$$D=2.44*λ*f_{number}$$
Where:
D = Diameter to first intensity zero (microns)
λ = Working wavelength (microns)
f_number = Working f-number of the optical system

This equation can be simplified further with some approximations for visual imaging systems. If we assume the image is primarily green (~546 nm, 0.546 microns), the equation simplifies to:
$$D_{546nm}=1.33*f_{number}$$

Given this simplified version, we can see that imaging performance of a visible camera for example is driven by f/#, where a "faster" f/# (a.k.a. numerically lower) optical system like a well-designed SLR camera lens will give better diffraction-limited resolution. Additionally, as a general rule a digital camera's sensor will need pixels which are on the order of size of the airy disc diameter for the lens being used. Many modern cameras (especially cell phone cameras) are limited in resolving power due to the lens's f/#, rather than the pixel count on the image sensor.

Last edited: May 23, 2017
4. Apr 26, 2017

@Mech_Engineer I'm glad you liked the article. The Airy disc I believe is what you get for the shape and intensity of the focused diffraction spot if you assume a circular aperture. The result that I have in the article above results from assuming a square aperture. It is a somewhat specialized topic so that it hasn't received a tremendous number of views, but I'm so glad you found it interesting. :) :)

5. May 21, 2017

### fizzy

"&#x0394;x=&#x03BB;fb" role="presentation">Δx=λfb where b" role="presentation">b is the diameter of the beam and/or the diameter of the focusing mirror (the smaller of the two). Meanwhile, the focused spot area Af=(&#x0394;x)2=&#x03BB;2f2A" role="presentation">Af=(Δx)2=λ2f2A, where A" role="presentation">A is the area of the beam."

don't know whether that copy/paste will post correctly but ...

The main thing is that you start with diameter and move to area and lose a factor pi along the way. A>d^2 !!

So which is it ??

6. May 21, 2017

My argument is meant to show the conservation of energy in a qualitative sense and also to show that the peak intensity is proportional to the square of the area of the aperture. Very precise energy conservation could be shown in both cylindrical and x-y geometry. The calculations I did, (and it was quite a number of years ago=I simply summarized the result here), shows an energy conservation in x-y geometry where you have a product rule $I(x,y)=I_o i(x) i(y)$ with a rectangular or square aperture.

7. May 21, 2017

### fizzy

Energy conservation is not "qualitative" is QUANTITATIVE. You do not have a conservation law based on "quality".

"... the diameter of the beam and/or the diameter of the focusing mirror ...." clearly you were not discussing a rectangular aperture.

Last edited: May 21, 2017
8. May 21, 2017

Quantitatively, it worked out correctly in x-y coordinates. If you want to compute the diffraction integrals in cylindrical coordinates, I'm sure you would get complete energy conservation, and you would find my on-axis intensity value to be correct. Feel free to post the calculation in cylindrical coordinates, but I did it using x and y independently and I was satisfied with the result.

9. May 21, 2017

### fizzy

• Post has been edited to remove insults
You wrote about diameters and now try to claim you were doing a rectangular aperture. What is the the diameter of a rectangle ?!

Last edited by a moderator: May 22, 2017
10. May 21, 2017

When I get a chance I will demonstrate the precision of energy conservation of the diffraction pattern in one dimension using the sinc^2 function.(I'm busy at the moment.) (It integrates to be precisely the result that I used in the article. ) Meanwhile, I do have considerable experience working problems in diffraction theory, so that I'm not pretending to be someone that I am not. $\\$ Editing... $\\$ For a slit of width $b$ , $I(\theta)=I_o \frac{sin^2(\frac{\pi \, b \, sin(\theta)}{\lambda})}{(\frac{\pi \, b \, sin(\theta)}{\lambda})^2}$. It can readily be shown that with $sin(\theta)=\theta=\frac{x}{f}$, (small angle approximation with a focused image), and with a narrow pattern (wide slit) that $\int I(\theta) \, d \theta =I_o (\frac{\lambda}{b})$. (The calculation uses the substitution $u=(\frac{\pi b}{\lambda} )x$ and uses the result that $\int\limits_{-\infty}^{+\infty} (\frac{sin^2 u}{u^2}) \, du=\pi$. The limits on the integral are justified (instead of stopping at $\theta=\pi/2$ , etc.) because the beam pattern is narrow.) With this kind of precision in a one dimensional calculation, it would seem unnecessary to try to solve a complex diffraction integral in cylindrical coordinates. Notice that this makes the effective width of the pattern $\Delta \theta_{eff}=\frac{\lambda}{b}$ The result is exactly what is needed. It can be readily extended to two dimensions. $\\$ I think it is justifiable to call $\Delta x=f (\Delta \theta_{eff})$ the focused spot size diameter for the purposes of the article. Otherwise I could call it the "effective width of the focused spot".

Last edited by a moderator: May 22, 2017
11. May 22, 2017

Additional comment is that the mathematics shown in post #10 that assumes an image of the far field pattern occurs in the focal plane of the focusing gets the same result as the diffraction integral of Huygens sources over the surface of the paraboidal mirror using the equation $x^2+y^2=4fz$ for the paraboloid. For the more detailed diffraction integral, this results in a term in the diffraction integral that takes account of the phase change caused by the paraboidal mirror of $exp^{i k\frac{((x')^2+(y')^2)}{2f}}$. This author has previously performed the integral in this manner, but by simply assuming a focusing of the far field diffraction pattern in the focal plane at position $x= f \theta$ as a function of $\theta$, the same result for the intensity pattern of the focused spot is obtained. $\\$ Additional comment is that this focused spot has only an approximate width or diameter because it is the result of a diffraction pattern. A useful parameter for its size is the effective area $A_{eff}$ defined by $P_{total}=E_f A_{eff}$ where $E_f$ is the irradiance (watts/cm^2) at the center of the spot and $P$ is the total power.

Last edited: May 22, 2017
12. May 22, 2017

### fizzy

where b is the diameter of the beam .... where A is the area of the beam. You equate A and b^2 , you have a problem with what you wrote.

13. May 22, 2017

The assumption is made that the aperture of the incident beam onto the paraboidal focusing mirror is square. It will produce a focused spot that has effective area $A_{eff}=\frac{ \lambda^2 f^2}{b^2}$. Mathematically it works out just as it should. (In the article, for a beam diameter "b", I simply said the spot size was "approximately" $\Delta x=\frac{\lambda f}{b}$.) The computation can be done quite precisely, as I have done in posts #10 and #11. $\\$ Additional comment: I believe doing this same calculation using a cylindrical geometry would be quite difficult, but the calculation with the rectangular geometry is rather straightforward.