Week 6b: Fabry Perot Interferometers

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  • Опубліковано 27 лис 2024

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  • @sarkan4508
    @sarkan4508 Рік тому +3

    You're a very enthusiastic teacher, and can make yourself clear. I wish I was at least a half as good as you at my lectures in optics 🙂Greetings from the Czech Technical University in Prague

  • @jamesdennis6120
    @jamesdennis6120 2 роки тому

    I literally cannot express how much I appreciate him as a professor. He literally has done more for me than any other professor I've had. Not only does he understand my struggles with the level ADHD that I have to live with everyday (which is much higher than the average ADHD person), he also approaches things that help me conceptualize things in a way that eliminates my wandering mind and doesn't judge my understanding based on the faulty system of our education approach when it comes to someone struggling with ADHD. It's gunna suck when I no longer get to have him as a mentor. I really hope I am able to have him as a mentor for when I eventually work for my masters. And maybe one day have him as a partner for a phd and more.

  • @anandarunakumar6819
    @anandarunakumar6819 Рік тому

    Super clear, this is precisely what I wanted to learn about lasing cavity preparation.

  • @zelinma2403
    @zelinma2403 2 роки тому +2

    Learned much more than that in my laser class!

  • @sixfeetunder888
    @sixfeetunder888 2 роки тому +2

    thank you so much for these videos, you explanations are amazing and you're probably going to save me for my optics exam

  • @zelinma2403
    @zelinma2403 2 роки тому +2

    Thank you very much! Finally figured out FP cavity!

  • @robertmiller-ig1sk
    @robertmiller-ig1sk 4 місяці тому

    Amazing video, doing some experiments with these but haven’t taken any classes where I would have learned this yet. You taught this very well!
    You should make a similar video involve a confocal FPC

  • @juanarcila5173
    @juanarcila5173 2 роки тому +1

    You're extremely good at explaining this!

  • @PewrityLab
    @PewrityLab 10 місяців тому +2

    First of all, this is the best explanation of the FP on the entire internet, thank you for making it!
    27:47 I have one more question here. You mention that a single laser photon can't pass through the FP. But can't photons interfere with themselves? Each one can be thought as a wave packet of some length that suddenly collapses. For instance, sending one photon at a time through a double slit still produces the interference pattern. As unintuitive as that sounds.. shouldn't a laser photon pass straight through both mirrors? Thank you for any response!

  • @JJZB9
    @JJZB9 3 роки тому +2

    thank you for this, very helpful and insightful!

  • @PreangkaRoy
    @PreangkaRoy 3 місяці тому

    It's really a nice explanation, thank you so much for the lecture

  • @giorapeniakov3153
    @giorapeniakov3153 2 роки тому +1

    Excellent tutorial, thank you!

  • @jamesdennis6120
    @jamesdennis6120 2 роки тому

    Haha the part where he is like "Let's uh, put a 2 squared here. Just cause." made me bust out laughing hahahaha

  • @FloThePro1231
    @FloThePro1231 2 роки тому +2

    best explanation of fabry perot i found on the internet... thanks
    are these "lectures" part of a physics bachelor degree? or can you simply study optics in america?

    • @AndrewMacRae
      @AndrewMacRae  2 роки тому

      Glad it helped! These lectures are for a third year course in the physics program at the University of Victoria (Phys 325: Optics)
      Sadly, no Optics program here - just Physics or Astronomy. However some universities, such as Rochester do offer optics degrees specifically.

  • @tushargosain8638
    @tushargosain8638 6 місяців тому

    At 3.51 how come the transmitted wave acquire a phase exp(ikl)? What if the beam is incident at an angle theta, then what will be the phase acquired?
    I know i will be klcos(theta_t) but how does that come?

  • @ArioMoradi
    @ArioMoradi Рік тому +1

    Hello I have made a homemade Fabry-Perot-Interferometer and I want to knew who can I major gravitational waves with my Interferometer?
    I need it for my school project.

    • @AndrewMacRae
      @AndrewMacRae  Рік тому

      OK ... in principle, LIGO is a big interferometer (Michelson & FP) so in principle yes. But, you would essentially be rebuilding LIGO (a multi-decade effort) which would be on the challenging side. It is often fun to see what the real limitations are in a given instrument. Here there will be many.
      I believe there is a saying "In principle, there is no difference between practice and principle - but in practice there is"
      So I'd say in principle yes, in practice, no.

  • @rijoraju7133
    @rijoraju7133 7 місяців тому

    How is the resonance peaks measured experimentally?

  • @karls738
    @karls738 3 роки тому

    Great explanation, thank you very much. The funny thing is that you look very similar and your facial expressions are also very similar to mine. Hahah it looks like we are doppelgangers.

  • @XinyanXiang
    @XinyanXiang 3 місяці тому

    amazing video

  • @yuwang-n3e
    @yuwang-n3e Рік тому

    In the experiment, how does the FPI interact with the incident laser, I mean how does it act as an optical filter, how does it determine the frequency of a scattered signal? Convolution or what? thanks for your answer

    • @AndrewMacRae
      @AndrewMacRae  Рік тому

      The picture that I like is that it really is nothing more than a pair of mirrors - it lets a small portion of the light in and reflects the rest. If the frequency corresponds to a resonant frequency of the FPI, then the field can become very large inside the cavity between the mirrors due to constructive interference.
      It isn't that it is "selecting" a particular frequency, but that for certain frequencies, the field inside the cavity is perfectly in phase with the incident light, leading to a large build-up. For other frequencies you get a sum of many waves with different phases - this sum turns out to be very small.

  • @karls738
    @karls738 3 роки тому +1

    What is actually measured on the interference pattern?
    The distance between the maxima?
    What does that mean? The wavelength or the bandwidth of the laser?

    • @AndrewMacRae
      @AndrewMacRae  3 роки тому +1

      Good question! It depends on what is being scanned. If the laser frequency (or wavelength) is being scanned, you are measuring the spacing of the cavity - you get one peak for every wavelength. This is a way to make very accurate spatial measurements. If you are scanning the mirror spacing, you are measuring the frequency content of the light source. In this configuration, you have what is known as an "optical spectrum analyzer". Typically the bandwidth of the laser is very large and there are better ways to measure the bandwidth. This is very useful though for seeing if there are several discrete frequencies (modes) within your laser, which appear as different peaks in the Fabry Perot Transmission.

    • @karls738
      @karls738 3 роки тому

      Thank you for the answer!
      So, for example, if you have a laser with a bandwidth smaller than the linewidth of the etalon's transmission peaks and the etalon is set at such an angle to the incident laser beam that the resonance condition is satisfied.
      How to determine the frequencies of the laser? Do you tilt the angle of the etalon and determine the transmitted power? (The configuration where the highest power is transmitted, the line widths of the laser and etalon match, and you know the line width of the laser is smaller, right? - but that wouldn't be very precise...) Or analyze the interference pattern in transmission? (What excatly measuring on the interference pattern, I assume a camera chip is illuminated)

    • @AndrewMacRae
      @AndrewMacRae  3 роки тому +2

      @@karls738 One confusing point is terminology here. There is bandwidth, and linewidth. The linewidth is the spread in frequencies surrounding a central frequency (or wavelength). The smaller the line width the better, typically and values of 100kHz or so are pretty standard. Bandwidth on the other hand refers to the range of central wavelengths that can selected, i.e. you are emitting a very narrow range about some central wavelength and the range of potential wavelengths are what you'd call the bandwidth.
      A Fabry Perot Interferometer is an excellent tool for discerning between closely spaced frequencies in a laser spectrum - as the cavity mirrors are tilted/separated, different frequencies appear as individual, narrow peaks in intensity (usually measured on a photodiode, which is like a (1-pixel) camera chip). This produces a periodic pattern with frequency spacing F = c/2L. From this, the separation between the peaks compared to c/2L tells you the frequency difference.
      On the other hand a FP cavity is a lousy device for measuring absolute wavelength - you only know that the frequencies are a multiple of wavelengths, not which multiple (15312 or 54312 wavelengths, say). If you want to measure the absolute frequency/wavelength, you're better off with a Michelson or Sagnac interferometer. This is how most commercial wave meters are built. The spacing between bright and dark spots in such an interferometer _is_ a function of absolute wavelength in these devices.
      Hope this helps.

  • @momensameh2032
    @momensameh2032 3 роки тому +1

    thank you very much

  • @Vlado93CG
    @Vlado93CG 2 роки тому

    Very neat. Thank you!

  • @karls738
    @karls738 3 роки тому

    Are these transmission peaks tunable by sending the laser at a certain angle to them?

    • @AndrewMacRae
      @AndrewMacRae  3 роки тому +2

      Yes, this is one way to tune the peaks, when flat mirrors are used. More often spherical mirrors are used and there is only a narrow range of input angles so that the primary way of tuning the peaks is changing the spacing.

  • @rosarioaldava-fisiket99
    @rosarioaldava-fisiket99 3 роки тому

    Great explanation!

  • @alejandrogomez2305
    @alejandrogomez2305 2 роки тому

    Does someone know why did he write t^2 = T = 1-R ? It's actually (n_t/n_i)t^{2} = T what is the reason for the approximation n_t/n_i = 1?

    • @AndrewMacRae
      @AndrewMacRae  2 роки тому +1

      Good question! It's a little subtle. There are actually two faces to a mirror (usually one has high reflective coating and the other has anti-reflection coating. The first face gives n_t/n_i and the second face gives a n_i/n_t so the ratio you mention cancels out overall. However, for the intensity _inside_ the glass you must include the n_t/n_i ratio to conserve energy.
      This subtlety was on the final exam for this course this year btw, so good observation :)

  • @harukabbh
    @harukabbh 2 роки тому

    thank you very much!

  • @stevehill6263
    @stevehill6263 2 роки тому

    Thank you

  • @dirpapa
    @dirpapa 3 роки тому +1

    drake equation (i looked it up)