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Question about refraction.

  1. Feb 6, 2009 #1
    Now, I've asked about this on some other forums: and it seems all people are able to to is repeat what happens at refraction, but not actually to answer the question:

    [PLAIN]http://ryansuchocki.co.uk/media/tmp.jpg [Broken]

    From the above image, why does the first sittuation occur rather than the second. (the white lines are the refracted waves.) I have been told that the waves which make up a perpendicular wavefront are not 'bonded', and are not inclined to remain 'in step' (making void the marching army or 'car off cliff' analogies) - so what is the scientific reason behind the connection between speed and direction, in an individual wave?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 6, 2009 #2

    Andy Resnick

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    The diagram 'wavefront 2' fails because the wavefront within the grey medium is not a physical wavefront- it is discontinuous, whereas prior to entering the medium, it was continuous.
  4. Feb 6, 2009 #3
    Why can't a wavefront break apart - when it is simply a row of perpendicular, in-step, waves?
  5. Feb 6, 2009 #4


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    The second case is exactly what is happening - just draw it with a large number of rays very close together.
  6. Feb 6, 2009 #5
    But then the entire wavefront would continue to move in the same direction, but rotated.
  7. Feb 6, 2009 #6
    As has already been said, once you keep drawing more and more rays into your diagram #2, the wave-front will look like in #1.
    You are only using half of the wave theory: your approach isn't sophisticated enough to calculate the intensity in each direction, so it is fairly natural to be unsure which direction the wave is moving. Nonetheless, you will later learn that the concepts of position and velocity are intertwined for waves.
  8. Feb 6, 2009 #7
    Thankyou very much,

    do you know of any websites/books etc from which i can learn this second half of the wave theory? half-understanding is frustrating!
  9. Feb 6, 2009 #8
    Are you sure? The second example looks like what would happen if you accounted for the slow down in velocity of the ray but failed to account for ray bending at the interface.

    To my knowedge, the green line does not have the same ray parameter as the white line and its energy would cancel out by Huygen's principle.
  10. Feb 6, 2009 #9


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    A ray (photon) doesn't bend at the surface it just slows down, it's the effect of rays slowing down that causes the wavefront to bend.
  11. Feb 6, 2009 #10
    Okay, but then surely something like this would happen:

    http://www.ryansuchocki.co.uk/media/tmp2.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  12. Feb 6, 2009 #11
    I've found the answer!!!!

    It all has to do with probability!

    Why didn't someone just tell me to look up Huygens-Fresnel principle!!!?

    Thank's for an interesting discussion every-one!
  13. Feb 6, 2009 #12


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    No the green lines are a wavefront, they are drawn between the ends of rays at the same time point.
    So draw the green wavy line between points on the green arrows at the same length and you get a bent beam
  14. Feb 7, 2009 #13
    Yes, bent, but still traveling in the same direction.
  15. Feb 7, 2009 #14
    I come from a background in seismology where rays are only (infinite frequency) approximations of the path along which energy travels from a source to a receiver.

    I am surprised that rays only travel in straight lines. I thought they travelled the shortest path between source and receiver, which would mean that when they reached a medium with higher/slower velocity they would bend to find the shortest path. How would you explain mirages without light bending?

    http://www.gps.caltech.edu/~carltape/research/pubs/Tape00_mirage.pdf [Broken]
    Last edited by a moderator: May 4, 2017
  16. Feb 7, 2009 #15
    To find the intensity at a point, you have to add up all the contributions for every point that the wave could have come from (e.g., by integrating along an earlier wave-front) and every possible path it could have taken in between (so the math is a little advanced), keeping track of how much the contribution is constructive or destructive. If you're not interested in an Optics textbook (such as Saleh and Teich) you might appreciate Feynman's little QED book. The simplified result is that the ray keeps its direction perpendicular to the wave-fronts.
  17. Feb 7, 2009 #16
    You are using the Huygens construction, whereby every point of the incident wave acts as a source of spherical waves. The physical interpretation ultimately comes from Maxwell's equations, and the derivation of Snell's Law from Maxwell's equations is found in any upper level E&M book.
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