The Diffraction Limited Spot Size with Perfect Focusing - Comments

In summary, Charles Link has submitted a new PF Insights post titled "The Diffraction Limited Spot Size with Perfect Focusing." The post discusses the diffraction limited spot size that occurs with a perfectly paraboidal primary telescope mirror, and how it is determined by the size of the Airy disk. The post also includes additional information on the Airy disk and its importance in diffraction-limited optics, as well as an equation for estimating the smallest resolvable feature in an image. The author also mentions the conservation of energy and how the peak intensity is proportional to the square of the aperture's area. However, there is some discrepancy in the calculations and whether they apply to cylindrical or x-y geometry.
  • #1
Charles Link
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Charles Link submitted a new PF Insights post

The Diffraction Limited Spot Size with Perfect Focusing
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Continue reading the Original PF Insights Post.
 
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  • #2
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.
 
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  • #3
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:
airy_disk.GIF


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:
[tex]D=2.44*λ*f_{number}[/tex]
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:
[tex]D_{546nm}=1.33*f_{number}[/tex]

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.
 
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  • #4
Mech_Engineer said:
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:
View attachment 197155

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:
[tex]D=2.44*λ*f_{number}[/tex]
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:
[tex]D_{546nm}=1.33*f_{number}[/tex]

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. 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.
@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
"Δx=λfb" 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=(Δx)2=λ2f2A" 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
fizzy said:
"Δx=λfb" 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=(Δx)2=λ2f2A" 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 ??
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
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.

You got it wrong. Please correct your article.
 
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  • #8
fizzy said:
Energy conservations is not "qualitative" is QUANTITATIVE. You do not have a conservation law based on "quality". You got it wrong. Please correct your article.
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
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 ?!
 
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  • #10
fizzy said:
You wrote about diameters and now try to claim you were doing a rectangular aperture. What is the the diameter of a rectangle ?!
.
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".
 
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  • #11
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.
 
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  • #12
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
fizzy said:
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.
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. ## \\ ## Edit: 12-31-18 Here is a homework thread that covers the problem of the Airy disc created by a circular aperture: https://www.physicsforums.com/threa...n-using-bessel-functions.916241/#post-5775639
 
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Related to The Diffraction Limited Spot Size with Perfect Focusing - Comments

1. What is the diffraction limited spot size?

The diffraction limited spot size is the smallest possible spot size that can be achieved when focusing a beam of light. It is determined by the wavelength of the light and the numerical aperture of the focusing lens.

2. How is the diffraction limited spot size calculated?

The diffraction limited spot size can be calculated using the formula: spot size = 0.61 * wavelength / numerical aperture. This formula takes into account the diffraction of light as it passes through the lens.

3. What is perfect focusing?

Perfect focusing refers to the ideal situation where all the light in a beam is focused to a single point without any loss or distortion. This is difficult to achieve in practice, but it is the goal when designing optical systems.

4. What factors affect the diffraction limited spot size?

The diffraction limited spot size is affected by the wavelength of the light, the numerical aperture of the lens, and any aberrations or imperfections in the lens. Additionally, the presence of any obstructions or contaminants in the light path can also impact the spot size.

5. Can the diffraction limited spot size be improved?

In theory, the diffraction limited spot size cannot be improved beyond the limit determined by the wavelength and numerical aperture. However, by using specialized lenses and techniques such as adaptive optics, it is possible to reduce the impact of aberrations and achieve a smaller spot size in practice.

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