University level introductory optics course

แชร์
ฝัง
  • เผยแพร่เมื่อ 5 ส.ค. 2024
  • Lecture notes: drive.google.com/drive/folder...
    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

ความคิดเห็น • 10

  •  2 ปีที่แล้ว

    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!

  • @optiondrone5468
    @optiondrone5468 3 ปีที่แล้ว

    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.

  • @heyjianjing
    @heyjianjing 16 วันที่ผ่านมา

    As I am revisiting these material, sometimes I am bit frustrated as to what details can be omitted, and what details are critical.
    For example, the Haidinger fringes, if two extended sources are separated by delta_z, then, using the same lens, they should not converge to the same point on the screen, but slightly offset due to delta_z, correct? is that because the extended source is very large compared to delta_z, so the offset is negligible?
    Then, when you show theta \approx r/f, isn't here r needs to be small compared to f for the approximation to be valid? I find it difficult to see how this approximation is making sense here, as the ring pattern is observed at a scale much larger than f.
    Then, at thin film interference, isn't that the derivation using wave formula assume the screen/lens to be horizontal, as the comparison is done at two refraction points rather than the two points on a line perpendicular to two rays, like the derivation using geometry?
    In addition, only the phase accumulation from z direction is counted for the ray travels in medium, but not accumulation in x direction, when the ray travels 2dtan(theta_2)?
    I tried to find answer to these questions in books and on internet, but no one seems to bother to mention.
    Hopefully someone can answer these questions, I fully understand that my understanding may be limited and my questions may be silly.

  • @StefanHoffmann84
    @StefanHoffmann84 ปีที่แล้ว

    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
      @SanderKonijnenberg  ปีที่แล้ว +1

      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
      @StefanHoffmann84 ปีที่แล้ว

      @@SanderKonijnenberg I see. Thank you!

  • @StefanHoffmann84
    @StefanHoffmann84 ปีที่แล้ว

    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
      @SanderKonijnenberg  ปีที่แล้ว +2

      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
      @StefanHoffmann84 ปีที่แล้ว

      @@SanderKonijnenberg Okay, I see. Thank you.