Double-slit interference with phasors

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  • Опубликовано: 26 авг 2024
  • Using phasors to investigate the interference pattern I(θ) produced when a wave diffracts through two narrow slits separated by a distance d.
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    About me: I studied Physics at the University of Cambridge, then stayed on to get a PhD in Astronomy. During my PhD, I also spent four years teaching Physics undergraduates at the university. Now, I'm working as a private tutor, teaching Physics & Maths up to A Level standard.
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Комментарии • 4

  • @mingmiao364
    @mingmiao364 5 месяцев назад

    More optics content please!

  • @apq3710
    @apq3710 5 месяцев назад

    Is it possible to derive this result from the expression for the displacement produced by a single slit? I mean A is proportional to sin(½ka*sin(theta))/½ka*sin(theta) where a is a width of the slit
    I tried but got something really different. However, double slit is clearly just a superposition of two slits
    Btw great video!

    • @DrBenYelverton
      @DrBenYelverton  5 месяцев назад +1

      It is indeed, but note that the result you quoted applies to a slit of finite width while I'm assuming in this video that the slits are infinitesimally thin. The full analysis is mostly the same as in the video, except |A| is no longer a constant and should really be replaced by your expression involving the sinc function. So the phasor diagram I drew is still valid, but you have to imagine that the length of the two phasors we're adding changes as θ varies. That means the full solution is really the product of the expression in the video and the sinc function. In the limit a -> 0, the single-slit part of the expression tends to a constant independent of θ, because sinc(0) = 1, and you recover the result from the video.

    • @apq3710
      @apq3710 5 месяцев назад +2

      @@DrBenYelverton makes sense, thank you!