University level introductory optics course
Lecture notes: drive.google.com/drive/folders/1C19nI8QTyyVAysR-pDcoJ27p6VQyVcPM?usp=sharing
TYPO: at 51:11, the minus sign in e^{ik(x sin theta - z cos theta)} magically changes into a plus sign, which it shouldn't
TYPO: starting from 1:43:49, I wrote ExB/dt instead of d(ExB)/dt
0:00 Overview and structure of the course
6:24 Ray model
11:48 Ray transfer matrix
15:10 Magnification (linear/angular), magnifying glass, microscope, telescope
23:23 Waves
35:09 Diffraction gratings
41:05 Grating spectroscopy
43:44 Interferometry (Michelson, thin film, Fabry Perot)
57:16 Resolution limit
1:07:15 Fourier optics
1:16:28 Coherence
1:26:03 Polarization
1:34:00 Fresnel equations (reflection/transmission coefficients)
1:42:02 Radiation pressure, Poynting vector
Пікірлер: 9
Impressive work Sander, thank you so much for this lecture, it's been fun and interesting to follow it along. Keep up the great work!
Sander thank you so much for this course. I enjoyed your previous ASP related topics also. Please make more videos on wave optics and Fourier optics.
1:11:44 The "extreme" cases of this explanation confuse me a little bit. If we have a single emitter, then the spherical wave becomes a plane in the far field. This is fine and agrees with the Fourier transform interpretation. But if we have infinitely many emitters arranged in a line, then their sum is a plane wave. So, in the far field, we still have a plane wave, when actually its Fourier transform should be a single point. Is there an explanation for that which agrees with the Fourier transform interpretation, or does this interpretation really fails in this case? Btw thank you for the great content!
@SanderKonijnenberg
Жыл бұрын
One thing to keep in mind is that far-field propagation is an approximation that is only valid if the observation plane is 'sufficiently far'. But what is 'sufficiently far'? It means sufficiently far compared to the size of the source (see e.g. the Wikipedia article on the Fresnel Number or my video on Diffraction Integrals). Therefore, if your source is infinitely large, then indeed there is no observation plane where the far-field approximation is valid.
@StefanHoffmann84
Жыл бұрын
@@SanderKonijnenberg I see. Thank you!
11:18 Is the virtual image here not reduced instead of magnified? At least that is what the picture shows, but you are talking about magnification. Confuses me...
@SanderKonijnenberg
Жыл бұрын
You are correct that this illustration is not representative for a magnifying glass. I intended it to illustrate the general concept of a virtual image, which may or may not be magnified. In case of a magnifying glass, the virtual image would be much larger and much farther away, which was simply inconvenient to illustrate due to limited space on the slide.
@StefanHoffmann84
Жыл бұрын
@@SanderKonijnenberg Okay, I see. Thank you.