I am only 5 minutes into this video and have already learned more than in my entire advanced optics module.... brilliant video, this is the future of learning
I'm still learning a lot of fourier optics concepts...going over LARGE, demoralizing textbooks😑 and you just managed to compress so much into one video. Incredible!
Last year I took a class at photonics west on Fourier Optics. It was 8h long and $600. In 25 min I learn more with your video. Thank you so much for your work.
@ 0:55 Here, in H's Principle, the source of light is called a field too. One could call this starting field U(0,0). @ 1:40 If you also put in U(0,0), then you have an expression for "the propagator" from QFT. Here, U(0,0) = 1.
Hi! Thanks a lot for the extremely helpful videos! Could you please clarify how and why, at minute 21:16, you went from having an Object in the (xo,yo) object plane to the (x/M,y/M) plane? why is it the lens plane that gets demagnified? shouldn't it be (xo/M,yo/M)?
My apologies for the confusion: in this slide, (x,y) denotes the image plane coordinates (not the lens plane). If we ignore the effects of the PSF (i.e. assume that it's a delta peak), then the image I(x,y) should be a magnified copy of |O(x_o,y_o)|^2. That is, I(x,y)=|O(x/M,y/M)|^2, so that the object field at object coordinates (x_o,y_o) is mapped to image coordinates (x,y)=(M*x_o,M*y_o).
Generally, NA is defined NA=n*sin(theta), where n is the refractive index of the immersion medium. So NA>1 is typically achieved by immersing the sample in a medium with high refractive index, e.g. water or immersion oil. If the refractive index of a medium is n, then the wavelength of light in that medium is lambda/n (where lambda is the wavelength in vacuum). By decreasing the wavelength we increase the resolution. Therefore, the resolution is still larger than lambda/2 if we define lambda to be the wavelength in the immersion medium, but it can be smaller than lambda/2 if we define lambda to be the wavelength in vacuum.
Thanks for your comment. 'Global phase factor' means in this case a phase factor that does not depend on (x,y), and therefore does not affect the way that the field propagates.
That's a good question, I never thought about it before. Section 3.7 of Joseph Goodman's 'Introduction to Fourier Optics' (can be found online) says: "The Huygens-Fresnel principle, as predicted by the first Rayleigh-Sommerfeld solution (see Eq. (3-40)), can be expressed mathematically as follows: [...]". It appears that the Huygens-Fresnel integral is a heuristically derived principle, whereas the Rayleigh-Sommerfeld solution is more rigorously derived from the wave equation, and from which the Huygens-Fresnel principle follows. But ultimately, it appears they describe the same thing.
@@SanderKonijnenberg Thanks for the reply. I'm actually trying to accurately determine how the field of a Gaussian beam is transformed after impinging on a reflective metasurface (i.e. is it still a Gaussian beam?). I applied The Huygen-Fresnel diffraction integral mainly because it appeared to be a simpler version of the RS integral. I guess I have to checkout Goodman.
Just a minor question, at 15:30, I thought that in F.T. shift theorem the shift in F.T. has the same sign as the exponent of the phase shift, i.e. F{f(x)e^(-2i pi ax)}(x',y') = F{f(x)}(x'-a, y'-a). I guess the later equations will still hold if we define M as |z_i|/|z_0| without the negative sign... Any thoughts?
If the Fourier transform is defined as F{f(x)} (x') = int f(x)e^(-2 pi i x x') dx then F{f(x) e^(- 2 pi i ax)} (x') = int f(x)e^(-2 pi i x (x'+a)) dx = F{f(x)} (x'+a). Perhaps what you're thinking of is the inverse Fourier transform? If you multiply F{f(x)} (x') with e^(-2 pi i a x'), then inverse Fourier transforming get you f(x-a). Physically, we also know that a real image generated by a single lens is inverted, so one way or another, M must turn out to be negative.
@@SanderKonijnenberg Yes you are definitely right. It's easier to think of (x'+a) as a new dummy variable then it's quite obvious. I did confused it with an e^(-2 pi i a x') shift in x' domain which will lead to f(x-a) in x domain. Thanks for clarifying!
at 2:23 , i think the complex form of the wave function i thinkl is : z/r2 * ei(kr−ωt), why do you say it is z/r2 * e ikr, and ommit the −ωt part ? Thanks a ot.
Yes, technically the field has a time-dependence, but the advantage of complex notation is that it allows you to omit the time-dependent complex exponential e^{-i\omega t}, see my video ruclips.net/video/31072jVfIUE/видео.html at 7:39 - 9:33 . For a monochromatic field, we typically don't care about how each point oscillates in time, but only about the phase difference between fields at different points in space.
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I am blown away by how quickly, clearly, and concisely you present this material!
I am only 5 minutes into this video and have already learned more than in my entire advanced optics module.... brilliant video, this is the future of learning
I'm still learning a lot of fourier optics concepts...going over LARGE, demoralizing textbooks😑 and you just managed to compress so much into one video. Incredible!
Last year I took a class at photonics west on Fourier Optics. It was 8h long and $600. In 25 min I learn more with your video. Thank you so much for your work.
Thanks a whole semesters lecture in one coherent video
Good material for reviewing wave optics after dropping it for 20 years.
@ 0:55 Here, in H's Principle, the source of light is called a field too. One could call this starting field U(0,0).
@ 1:40 If you also put in U(0,0), then you have an expression for "the propagator" from QFT. Here, U(0,0) = 1.
Thank you, this is really condensed and purified content
Great Video! Thank you!
Great video!
Hi! Thanks a lot for the extremely helpful videos!
Could you please clarify how and why, at minute 21:16, you went from having an Object in the (xo,yo) object plane to the (x/M,y/M) plane? why is it the lens plane that gets demagnified? shouldn't it be (xo/M,yo/M)?
My apologies for the confusion: in this slide, (x,y) denotes the image plane coordinates (not the lens plane). If we ignore the effects of the PSF (i.e. assume that it's a delta peak), then the image I(x,y) should be a magnified copy of |O(x_o,y_o)|^2. That is, I(x,y)=|O(x/M,y/M)|^2, so that the object field at object coordinates (x_o,y_o) is mapped to image coordinates (x,y)=(M*x_o,M*y_o).
@@SanderKonijnenberg thank you so much for the reply!
18:06 from angular spectrum method we have that the resolution is limited to lambda/2. But if we have a lens with a NA>1 we can have a resolution
Generally, NA is defined NA=n*sin(theta), where n is the refractive index of the immersion medium. So NA>1 is typically achieved by immersing the sample in a medium with high refractive index, e.g. water or immersion oil. If the refractive index of a medium is n, then the wavelength of light in that medium is lambda/n (where lambda is the wavelength in vacuum). By decreasing the wavelength we increase the resolution.
Therefore, the resolution is still larger than lambda/2 if we define lambda to be the wavelength in the immersion medium, but it can be smaller than lambda/2 if we define lambda to be the wavelength in vacuum.
3:58 why in approximation on complex exponential, we should change the sign?
Because exp(ik * lambda/2)=exp(i*pi)=-1 (recall k=2pi/lambda)
thanks, very clear, the minus come from exp(i*pi). I was misleading to expansion generating the minus sign.
What does global phase factor in 10:36 mean? Thank you for this wonderful video!
Thanks for your comment. 'Global phase factor' means in this case a phase factor that does not depend on (x,y), and therefore does not affect the way that the field propagates.
My 5am savior
Whats the difference between Rayleigh-Sommerfeld integral and Huygen-Fresnel integral?
That's a good question, I never thought about it before. Section 3.7 of Joseph Goodman's 'Introduction to Fourier Optics' (can be found online) says: "The Huygens-Fresnel principle, as predicted by the first Rayleigh-Sommerfeld solution (see Eq. (3-40)), can be expressed mathematically as follows: [...]". It appears that the Huygens-Fresnel integral is a heuristically derived principle, whereas the Rayleigh-Sommerfeld solution is more rigorously derived from the wave equation, and from which the Huygens-Fresnel principle follows. But ultimately, it appears they describe the same thing.
@@SanderKonijnenberg Thanks for the reply. I'm actually trying to accurately determine how the field of a Gaussian beam is transformed after impinging on a reflective metasurface (i.e. is it still a Gaussian beam?). I applied The Huygen-Fresnel diffraction integral mainly because it appeared to be a simpler version of the RS integral. I guess I have to checkout Goodman.
Just a minor question, at 15:30, I thought that in F.T. shift theorem the shift in F.T. has the same sign as the exponent of the phase shift, i.e. F{f(x)e^(-2i pi ax)}(x',y') = F{f(x)}(x'-a, y'-a). I guess the later equations will still hold if we define M as |z_i|/|z_0| without the negative sign... Any thoughts?
If the Fourier transform is defined as
F{f(x)} (x') = int f(x)e^(-2 pi i x x') dx
then
F{f(x) e^(- 2 pi i ax)} (x') = int f(x)e^(-2 pi i x (x'+a)) dx = F{f(x)} (x'+a).
Perhaps what you're thinking of is the inverse Fourier transform? If you multiply F{f(x)} (x') with e^(-2 pi i a x'), then inverse Fourier transforming get you f(x-a).
Physically, we also know that a real image generated by a single lens is inverted, so one way or another, M must turn out to be negative.
@@SanderKonijnenberg Yes you are definitely right. It's easier to think of (x'+a) as a new dummy variable then it's quite obvious. I did confused it with an e^(-2 pi i a x') shift in x' domain which will lead to f(x-a) in x domain. Thanks for clarifying!
very nice video
Can you share the MATLAB code?
The Google drive link in the video description also contains a zip-file with the Matlab codes. Hopefully those include what you're looking for.
any reference textbooks on this that can be recommended?
I'd go with J. Goodman's 'Introduction to Fourier Optics'
at 2:23 , i think the complex form of the wave function i thinkl is : z/r2 * ei(kr−ωt), why do you say it is z/r2 * e ikr, and ommit the −ωt part ? Thanks a ot.
Yes, technically the field has a time-dependence, but the advantage of complex notation is that it allows you to omit the time-dependent complex exponential e^{-i\omega t}, see my video ruclips.net/video/31072jVfIUE/видео.html at 7:39 - 9:33 . For a monochromatic field, we typically don't care about how each point oscillates in time, but only about the phase difference between fields at different points in space.
really helpful.
Thank you so much.@@SanderKonijnenberg
why are that many books so long and boring without clear explanation，disappointing until I found this