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Refraction of light

  1. Feb 6, 2008 #1
    Hi! I want to ask you something. When the light travels through medium 1, there is reflection of light, so there is energy released, but it needs time the energy to travel through medium 2, right?
  2. jcsd
  3. Feb 6, 2008 #2


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    Sorry can you try and explain again what you are asking?
  4. Feb 6, 2008 #3
    Look at this http://www.walter-fendt.de/ph14e/huygenspr.htm" [Broken] There are 2 mediums? Lets say glass and water. When the light pass through glass it reflects and refracts in same time. It reflects from the glass and refracts in the water, right? Why there is time delay in medium 2?
    Last edited by a moderator: May 3, 2017
  5. Feb 6, 2008 #4


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    when light hits an interface = the boundary between two different refractive indexes some of the light it reflected back (as if from a mirror) and some travels into the second material. This doesn't take any time.
    Light does go at different speeds in different materials, the refractive index is (in simple terms) the fraction of the speed of light in vacuum that light travles in that medium.
    this doesn't mean any energy is lost.

    ps. The applet seems faulty - the wave never gets into the second medium when I run it.
  6. Feb 6, 2008 #5
    Press 2 times on next step, you'll see that there is time delay.
  7. Feb 6, 2008 #6


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    When the incoming wave hits one of the pink "atoms" on the boundary, a wave starts speading out immediately from that atom in both directions (upward and downward). The downward-going wave travels more slowly than the upward-going wave, but they both start at the same time.

    I think you're seeing what looks like a slight delay in the slower medium because the "atoms" in the diagram are rather large and the waves start from the centers of the atoms. It takes a short time before the wave becomes visible. Before that the pink circle of the "atom" hides the wave.
  8. Feb 7, 2008 #7
    But look at the angle of refraction, it is different than the angle of reflection. so there is time delay on 2-nd medium
  9. Feb 7, 2008 #8

    Andy Resnick

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    Light scattering can be very non-trivial. The scattering of femtosecond pulses (i.e. the reflection and transmission) is very different than the equilibrium picture we have in textbooks. One must not take cartoon diagrams too seriously.
  10. Feb 7, 2008 #9


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    Are you really asking, "why is the speed of the wave slower in the second medium than in the first medium?"

    I thought you were asking, "why does the (refracted) wave in the second medium start a bit later than the (reflected) wave in the first medium?"
    Last edited: Feb 7, 2008
  11. Feb 7, 2008 #10
    I want to know, is the energy transfer from the atoms of the 1-st medium to the atoms of 2nd medium, reason, of which there is time delay, and the wave of refraction moves slower than the wave of reflection?
  12. Feb 7, 2008 #11
    yes I could say that my question is: "why does the (refracted) wave in the second medium start a bit later than the (reflected) wave in the first medium?"
  13. Feb 7, 2008 #12
    Don't make that animation confuse you: actually reflected and refracted waves starts exactly in the same instant of time, but the lower speed of the refracted wave makes you think it started after a delay. It's not.
  14. Feb 7, 2008 #13
    Why the refracted wave have lower speed than the reflected wave?
  15. Feb 8, 2008 #14
    lightarrow, jtbell?
  16. Feb 8, 2008 #15
    You have an intuitive answer in the Physics Forums FAQ (one of the first 3 sticky threads of General Physics) with the title "Do Photons Move Slower in a Solid Medium?":

    However, solving for Maxwell's equations in a medium, characterized by an electric permittivity and a magnetic permeability


    you will find that the solution is a function of coordinates and time which propagates with the speed

    [tex]v = 1/\sqrt{\epsilon\mu }[/tex].

    Since, at least for a linear medium, you can write

    [tex]\epsilon\mu = n^2\epsilon_0\mu_0[/tex]

    where n is the refraction index > 1, you have:

    [tex]v = 1/n\sqrt{\epsilon_0\mu_0} = c/n < c.[/tex]
    Last edited: Feb 8, 2008
  17. Feb 8, 2008 #16
    Ok, and if I get bigger peace of glass (lets say) and smaller one. Will the index of refraction be same?
  18. Feb 8, 2008 #17
    Certainly. It depends only on the chemical composition and on the physical properties of the cristal (at least at non very high frequencies or intensities).
  19. Feb 8, 2008 #18
    If I get peace of glass and how can I measure its index of reflection? I read the FAQ of the sticky tread, and can't understand still what it want to say, and why it negates the theory of electron transfer of energy.
  20. Feb 8, 2008 #19
    I don't know your knowledge of physics, however that is standard physics at first year university and probably even at high school. In the traditional technic you shape the piece of glass as a prism and you measure the vertex angle a and the angle of "least deviation" d_m. Then the refraction index n is: n = [sin(a + d_m)/2]/sin(a/2).
    Sincerely I don't know how to explain you better than that way.
  21. Feb 8, 2008 #20
    Ok, I understand now. Thank you.
  22. Feb 8, 2008 #21
    And why the air have 1.003 refractive index?
  23. Feb 8, 2008 #22
    I think in this case you should look for a quantum description, but there are many people more aknowledged than me about that.
  24. Feb 10, 2008 #23
    And what happens, when there is total internal reflection? Like on the animation, there is only reflection, but what happens with the waves that are going inside of the medium?
  25. Feb 10, 2008 #24
    It happens that...you have total reflection! :smile:
    What do you mean with your question?
    Snell's law states that

    [tex]sin\theta_1/sin\theta_2 = n_2/n_1[/tex]


    [tex]\theta_1, \theta_2, n_1, n_2[/tex]

    are the angle of the light beam in medium1, the angle in the medium2, the refractive index of the first and the second medium, respectively. If n1 > n2 so that n1/n2 > 1 you have

    [tex]sin\theta_2 = sin\theta_1 *n_1/n_2 > sin\theta_1[/tex]

    so I can have

    [tex]sin\theta_2 = 1[/tex]


    [tex]sin\theta_1 < 1[/tex]

    that is when

    [tex]sin\theta_1 = 1*n_2/n_1[/tex]

    If medium1 = water and medium2 = air, then

    [tex]n_2/n_1 \cong 0.75 [/tex]


    [tex]\theta_1 \cong 48.6^\circ[/tex]

    so this means

    [tex]\theta_1 > 48.6^\circ\Rightarrow \theta_2 > 90^\circ[/tex]

    that is: the refracted beam is not in air but goes back in water = you have total reflection.
    Last edited: Feb 10, 2008
  26. Feb 10, 2008 #25
    So there is only reflection, but on the simulation, there are also waves that go inside of the medium 2.
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