Hele fijne structure, snel, zonder muziek etc bedankt.

Thank you! Very useful explanation!

Very cleanl, love it

I like this lecture but this is a bit more of an advanced introduction. For example the Helmholtz equation and the transition from scalar to vectorial function. This is a good introduction for someone with advanced math and already in an optics class.

@ 0:00 Here, he leaves out the QUANTUM MODEL of light. In video 9. he corrects this.

Beautiful! My personal experience of light physics is a certain awkwardness around the subject. It took me ages to understand many things do not have a description and usually, tutors avoid saying it. They're also so many quantities that are derived by different models and used trivially in other light descriptions. Sure there is a link but going through so many approximations that you are lost if they are still valid or not. Further colleagues avoid relinquishing their light description of conversations since they will be exposed? Creating a babel. or in short, a light description is so confusing and is mainly because it is taboo to express you are lost.

@jacobvandijk6525

3 жыл бұрын

That's not a pleasant situation you describe. I'd say, organize a meeting with your colleagues and discuss your problems openly. I can imagine this could 'lighten up' the place considerably ;-) Good luck!

6:00. Plz can you clarify what exactly is causing this Fourier transform? e.g. if light is propagating from a point source, a lens can cause F.T. to make them parallel and vice-versa. Here delta function is transformed to constant and vice versa. What would be the function in this case and how is it transforming(I mean what is the imaging system here)? I don't quite understand. And great lecture series btw.

@SanderKonijnenberg

2 жыл бұрын

Thanks for your comment. I explain the Fourier transforming property of a thin lens in more detail in another video (see kzread.info/dash/bejne/i2l4w8mGkrOrd84.html at 22:34). The function that is being Fourier transformed is the complex-valued field (we assume fully coherent monochromatic light) in the front focal plane. The resulting Fourier transform corresponds to the complex-valued field in the back focal plane (if we neglect aberrations and the finite aperture of the thin lens).

you lost me within a minute, sigh :).

@waadfrelle

2 жыл бұрын

@@jacobvandijk6525 Wat aardig. Ik heb verder gezocht naar een meer niet-academische uitleg. Succes verder!