# What is the name of this equation? Seems to be from Optics

I was given this equation:

##\frac{1}{\lambda f} e^{ikr^2f}##,

where ##\lambda## is the wavelength of light and ##f## is the focal length of the lens. I was told that it is called the "lens phase", but I have no luck in finding it using Google.

I suppose this is multiplied when a beam of light travels through a lens.

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Here's something on a thin lens with a similar although not identical equation:

http://en.wikipedia.org/wiki/Thin_lens

and here's another with what looks like your equation:

http://www.optique-ingenieur.org/en/courses/OPI_ang_M02_C01/co/Contenu_14.html
Thanks! But, the Wikipedia entry seems to have come from nowhere, so I'm not sure about that one. However, the second one seems to be similar (minus the ##\frac{1}{\lambda f}## term). The value for ##r## is indeed ##\sqrt{x^2 + y^2}##, but with respect to what?

P. S. I was also told that the maximum value of ##r## is the radius of the lens.

jedishrfu
Mentor
I think the x and y are relative to the center of the lens.

Andy Resnick
I was given this equation:

##\frac{1}{\lambda f} e^{ikr^2f}##,

<snip>
Something is amiss: kr^2f is not unitless. kr^2/f would be ok, and as jedishrfu mentioned, corresponds to the phase delay of light passing through a lens. Similarly, the prefactor looks incorrect; from Fresnel diffraction it should be 1/(i*λ*z) or 1/iλf if you are interested in the field distribution at the focal plane (z=f).

• ecastro and jedishrfu
Something is amiss: kr^2f is not unitless. kr^2/f would be ok, and as jedishrfu mentioned, corresponds to the phase delay of light passing through a lens. Similarly, the prefactor looks incorrect; from Fresnel diffraction it should be 1/(i*λ*z) or 1/iλf if you are interested in the field distribution at the focal plane (z=f).
Thanks, I never saw that the exponent has a unit. I might heard the equation wrong, though (it was not given to me in written form, sorry). Actually, this is indeed related to Fresnel Diffraction (I might have missed the imaginary unit).

Edit: Is this somehow related to the Point-spread Function (PSF)?

blue_leaf77
Homework Helper
PSF is defined as the response of a system ##S## to a point shaped input in the input plane. For a thin lens, the the output field for a point illumination located in the focal plane in front of the lens will be a uniform field (collimated beam) whose direction of propagation depends on the exact location of the point source. In other words, the PSF for thin lens should be a function of, in addition to the coordinate in the output plane (##x, y##), the coordinate in the input plane, which is the coordinate corresponding to the location of the point source.

Mathematically the relation between the output field ##U(x,y,z)## to the input field ##U(\zeta, \eta, 0)## in a system ##S## is

##U(x,y,z) = \int \int U(\zeta, \eta, 0) h(x,y;\zeta, \eta) d\zeta d\eta ##

where ##h(x,y;\zeta, \eta) ## is what we call as PSF. On the other hand, the function you have there is just a multiplicative phase factor for a field after passing through a thin lens.

I see. So the equation I showed is not PSF. This clears a few things.

By the way, how can I know the shape of the PSF given the system has a thin lens? Does it depend on the where is my input and output planes located? That is, the PSF changes shape depending on these distance parameters.

Andy Resnick
The PSF can generally be computed easily if you know the field distribution at the exit pupil. The specific relations are slightly different for coherent and incoherent imaging- for coherent imaging, the impulse response is given by the Fourier Transform of the exit pupil, while for incoherent imaging the PSF is the modulus squared of the coherent PSF.

blue_leaf77
Homework Helper
Does it depend on the where is my input and output planes located? That is, the PSF changes shape depending on these distance parameters.
PSF is the response of a given system with respect to an impulse input, if you change the input and output planes, or in general any element of the system, then you changed the system.

The PSF can generally be computed easily if you know the field distribution at the exit pupil.
PSF is the response of a given system with respect to an impulse input, if you change the input and output planes, or in general any element of the system, then you changed the system.
Thank you. I tried simulating this using an input plane with a white pixel at the middle, but the simulation returns a diffraction pattern similar to a square aperture, which I guess is not a PSF. Might be the PSF is the smallest resolvable point in the image?

blue_leaf77
Homework Helper
Where did you put the input plane? If it's located at the front focal plane and the output at the back focal plane, the field at both plane is related through a simple FT assuming the lens aperture is very big.

I have put my input plane at 400 meters from the lens (a thin lens, with a focal length of 1.0 meters). The input then propagates until at the front of the lens and I multiplied it with a thickness function

##t_{l} = e^\left[-i \frac{k}{2f} \left(x^2 + y^2\right)\right]##

to know the diffraction pattern at the back of the lens, the pattern then propagates until the sensor which is approximately 1.0256 meters from the back of the lens.

I think the front and back focal planes are equal to the focal length if the lens is thin, right?

Andy Resnick
Thank you. I tried simulating this using an input plane with a white pixel at the middle, but the simulation returns a diffraction pattern similar to a square aperture, which I guess is not a PSF. Might be the PSF is the smallest resolvable point in the image?
Two possibilities come to mind:

1) what is the shape of your aperture? If it's rectangular or square, that would explain your result.
2) Is your object truly a point or could a 'pixel' be interpreted by the code as a square/rectangular object? That would also explain your result.

Two possibilities come to mind:

1) what is the shape of your aperture? If it's rectangular or square, that would explain your result.
2) Is your object truly a point or could a 'pixel' be interpreted by the code as a square/rectangular object? That would also explain your result.
The pixel does look like a mini-square aperture, so I guess it does explain why I have square diffraction pattern. But then, how would I acquire the PSF if this is the case?

blue_leaf77
Homework Helper
What do you want to achieve in this work? Do you want simulate what PSF is or merely want to get the mathematical expression of the PSF?
If you want the math, you can simply propagate the wavefront of a point source up to the lens. Then multiply with the lens transmission (phase) function, beyond this point it should be easy to deduce the output field at any output plane behind the lens. Furthermore if this is an imaging system, you will note that the distances of the source, image, and focal point is related through the usual equation for lens imaging.
Speaking of your simulation, when you observe the diffraction pattern reminiscent of that of rect aperture, did you only look at the square amplitude of it? It makes sense that you got such result due to the pixel shape influence of the computer, but if you plot the real part (or imaginary part) of the output field I think you should observe a ripple-like pattern, which is caused by the quadratic phase at the output plane. Taking into account this phase will make your simulation closer to reality as one normally would observe a spherical wave out of a lens.

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What do you want to achieve in this work? Do you want simulate what PSF is or merely want to get the mathematical expression of the PSF?
If you want the math, you can simply propagate the wavefront of a point source up to the lens. Then multiply with the lens transmission (phase) function, beyond this point it should be easy to deduce the output field at any output plane behind the lens. Furthermore if this is an imaging system, you will note that the distances of the source, image, and focal point is related through the usual equation for lens imaging.
I want to simulate the PSF of my system, and use the PSF to de-convolve the blurred image acquired by the output. Here are the results of my simulation by using a single pixel as my point source:  The first image is the diffraction pattern at the back part of the lens (the diffraction pattern at the front is similar, only brighter); and the second image is at the output plane a certain distance from the back of the lens.

Speaking of your simulation, when you observe the diffraction pattern reminiscent of that of rect aperture, did you only look at the square amplitude of it? It makes sense that you got such result due to the pixel shape influence of the computer, but if you plot the real part (or imaginary part) of the output field I think you should observe a ripple-like pattern, which is caused by the quadratic phase at the output plane. Taking into account this phase will make your simulation closer to reality as one normally would observe a spherical wave out of a lens.
Is this what you're talking about? This image is the intensity pattern at the output plane. I also tried enlarging the point pixel by making 3 x 3, rather than 1 x 1. Here is the resulting image at the output plane: Is this the PSF? Also, I found this thread: https://www.mathworks.com/matlabcentral/newsreader/view_thread/307631. There is a suggestion here of using the 'deconvreg' of MatLab to acquire the PSF.

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Andy Resnick
The pixel does look like a mini-square aperture, so I guess it does explain why I have square diffraction pattern. But then, how would I acquire the PSF if this is the case?
Typically, the PSF is larger than 1 pixel- that is, the system should be sensor limited, not lens limited. Here's an example of a 'measured' PSF (scaled x10 so you can see it better) Computationally, this can be simulated by applying a blur filter (gaussian convolution, for example) to your single-pixel PSF. Try a 3x3 blur or 5x5 blur, and you may want to start with a 2x2 or 3x3 unblurred PSF instead of a single pixel.

Then, are all these points possible candidates for the PSF? They're somehow similar to your PSF. Although this image is produced when the input plane and output plane are 1.0 meter at the front and back of the lens, respectively (with a thin lens having a focal length of 1.0 meter). Then, are all these points possible candidates for the PSF? They're somehow similar to your PSF. Although this image is produced when the input plane and output plane are 1.0 meter at the front and back of the lens, respectively (with a thin lens having a focal length of 1.0 meter).
View attachment 84701

Andy Resnick
I can't make out what your image is- an array of points? If I look closer, it seems that each bright point is surrounded by an airy-like pattern, with interference between neighboring patterns... yes?

I guess I should mention the PSF image I posted is actually |PSF|, since it's intensity and not field.

Andy Resnick
Then, are all these points possible candidates for the PSF? They're somehow similar to your PSF. Although this image is produced when the input plane and output plane are 1.0 meter at the front and back of the lens, respectively (with a thin lens having a focal length of 1.0 meter).
View attachment 84701
Ah- now I can zoom in. Try adjusting the object and image distances so that your image is a true image of the object and see what it looks like.

I can't make out what your image is- an array of points? If I look closer, it seems that each bright point is surrounded by an airy-like pattern, with interference between neighboring patterns... yes?

I guess I should mention the PSF image I posted is actually |PSF|, since it's intensity and not field.
Yes, it is an array of points with interference-like patterns in between points. The intensity of the image is ##I = |U|^2##, where ##U## is the diffraction pattern of the electric field (I don't know if this is correct. The object propagates by Fresnel Diffraction with the Angular Spectrum Method, and Fourier Transform is necessary).

Ah- now I can zoom in. Try adjusting the object and image distances so that your image is a true image of the object and see what it looks like.
I did try, here is the image: The object is located 5 meters from the front face of the lens, and the image is 1.25 meters from the back face of the lens. I used the Thin Lens Formula to calculate the location of the image.

blue_leaf77 