Lens 1F System - Lens Fourirer Transforms

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In this video, I explain how lenses perform Fourier transforms by canceling the phase due to free space propagation. I use both the concepts of angular spectrum and Fresnel propagation, and the awesome result is that a lens performs a fourier transform if it is placed a distance of 1f away from an object, and the observation plane is 1f away from the lens.
This is part of my graduate series on optoelectronics / photonics, and is based primarily on Coldren's book on Lasers as well as graduate-level coursework I have taken in the EECS department at UC Berkeley.
Hope you found this video helpful, please post in the comments below anything I can do to improve future videos, or suggestions you have for future videos.

Опубликовано:

26 июн 2019

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Комментарии : 22

@fohker8296 3 года назад

Hey thank you so much for this short course, it's a blessing being able to watch this for free !

@DeepakKumar-lo9lk 2 года назад

Best video on RU-vid for Fourier Optics

@JoelRTLCosta 7 месяцев назад

Thank you for the classes!

@eunsjun6582 2 года назад

You saved my life.

@shihansajeed2110 3 года назад

that's an amazing video. thank you

@vladi21k 2 года назад

Thank you for your video, it was very helpful and it inspired me with new ideas how to teach my graduate class. However please note that by performing the Fourier transform, it doesn't mean that lens do something fundamentally different to imaging. In fact, it can be shown with Fourier optics that the lens does imaging by converting frequency information (sum of plane waves incoming from different directions) to spatial information, proportional to the amplitude of the plane waves (imaging).

@gamebm Год назад

Many thanks for the excellent classes. I have a question, is there any way to show that at the high-frequency limit (Eikonal limit), the Fourier optics also leads to the Lens maker's equation?

@thejll 3 года назад

Nice! Might you do the 4f correlator one day? How a Lyot stop works?

@abharadwaj6814 4 года назад

Thanks Jordan! I've a question. Suppose the object is placed at a distance greater than F from the lens, and the screen is placed at the imaging distance (as found out by lens equation) does g_out(x_image) = fourier transform of g_in(x_obj) ?

@JordanEdmundsEECS 4 года назад

No, imaging is not the same (unfortunately) as the Fourier-Transforming property of a lens. For a FT you need to place both object and plane exactly 1f from the lens. For imaging, this would be 2f.

@DrGJ420 2 года назад

Phase transformation by the lens between g- and g+ . Where can I study such transformation by optical components like convex and concave lens? Thanks

@vintech88 5 лет назад

Thank you for the video Can you explain the spatial frequency in detail and little intuitive, my mind somehow not accepting these terms

@JordanEdmundsEECS 5 лет назад

Yeah, it’s a very challenging concept, I’ll probably make a few videos on it. If you understand frequency in time you can think of it in exactly the same way:

@MM-vw1ck 2 года назад

So what you're saying is, one can cure Gout just by simpy computing a fourier transforkm?

@JordanEdmundsEECS 2 года назад

#bestBusinessIdeaOf2021 #cureEverythingWithTheFourierTransform

@poutineausyropderable7108 2 года назад

Isn't there a problem with the propagator. Isn't it. e^(ikf -ikx^2/2k) You're off by a facor of e^ikf. Which is a constant. Got it

@marlonsaveri Год назад

Suddenly a Ks appeared. What is Ks?

@brianma875 Год назад

What if I cascade 2 lenses with a distance in between? I can't cancel the term which represents the propagation between 2 lenses. I'm doing something wrong or it just don't work for 2 lenses apart from each other? If that's the case, the objectives of a microscope can't be treated as a Fourier lens.

@nohaivce2614 Месяц назад

@brianma875 Hi, if I understood your question correctly, you might want to look into a 4f-system which describes this problem. tldr it's a reverse fourier transform by the second lens

@brianma875 Месяц назад

@@nohaivce2614 Oh I mean the second lens is placed at a distance shorter than the focal length of the first lens. I'm doing this because I wish to get an adjustable joint focal length of those lenses by changing the distance between them.

@nohaivce2614 Месяц назад

@@brianma875 Hi, something that comes up for me as an idea when you say 'joint focal length' is using 2 lenses still in the focus of each other, but with different focus lengths each (so effectively it turns from a 4f into a 2*f_1 and 2*f_2 system. This would have the same effect as 4f where at the end-screen after the second lens you get a inverse fourier of the Fourier after the first plane, but with different size of the image. You can use ABCD law to get the magnification and other properties of this system.

@nohaivce2614 Месяц назад

@@brianma875 However, of course if you want to place the second one out of focus you can. I came across it being briefly discussed in J. Peatross' and M. Ware's Physics of Light and Optics in Section 11.1. As far as I udnerstood you can use Fresnel Prop. from aperture to the first lens, from the first lens to second and then get the Fraunhofer field of that in the focus of lens 2.