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Reflection and Wavefronts

  1. Sep 4, 2011 #1
    Hi,

    A wave is changing from medium A to medium B and it travels slower in medium B than in medium A. If I draw a normal line from the boundary to the wavefront the wavefront in medium B will move away from the normal? I attached to make it more clear

    Furthermore, I got this question:

    The angle between the wavefronts and the interface in region A is 60. The refractive index anb is 1.4.

    What I did was: if 60 between wavefront and boundary, 30 degrees between ray and boundary: sin 30 / sin r = 1.4. I then got between the wavefront and the boundary 69.1 degrees but it is wrong, the mark scheme got 38 degrees. Can anyone help me?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Sep 4, 2011 #2

    Doc Al

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    It looks like they were specifying the angle with respect to the normal to the surface. (They probably mixed them up.)


    Edit: On closer inspection, there's nothing wrong with the diagram or problem statement.
     
    Last edited: Sep 4, 2011
  4. Sep 4, 2011 #3
    Um, sorry, I am still a bit confused in relation to how they did it. I attached the question since I didn't word it very clearly in my first post:
     

    Attached Files:

  5. Sep 4, 2011 #4

    Doc Al

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    Here's the problem:
    What you need in Snell's law is the angle between the ray and the normal.
     
  6. Sep 4, 2011 #5
    Oh, sorry, that was a typo I guess... In the diagram, the angle between the wavefront and the normal is 60 degrees. So I did 90-60 to get the angle between the ray and the normal. I did sin 30 / 1.4 while the mark scheme did sin 60 1.4. So the mark scheme is wrong? :confused:
     
  7. Sep 4, 2011 #6

    Doc Al

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    No, the diagram is quite clear. The angle between the wavefront and the boundary is 60 degrees.
     
  8. Sep 4, 2011 #7
    Ah, never mind, I got it now. Thanks! But if you wouldn't mind, could you help me with the second part of the question I linked? I got the first part but I am confused as to what the marking scheme says:

    position of any one minimum closer to centre / minima closer together;
    frequency increased so wavelength decreased / correct explanation in terms of double-slit equation;
     

    Attached Files:

  9. Sep 4, 2011 #8

    Doc Al

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    I knew you'd figure it out. :wink:
    Are you sure you linked it? (See next post--you were still linking when I wrote this.)
     
  10. Sep 4, 2011 #9

    Doc Al

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    That's the trick. Do you realize that since the speed of the wave is fixed, increasing the frequency must decrease the wavelength? And how will a smaller wavelength affect the pattern? What does the double-slit equation tell you?
     
  11. Sep 4, 2011 #10
    Sorry Doc Al, I updated the post a few seconds after I posted it. I am tired, I shouldn't be studying until now so I am making those frustrating stupid mistakes :redface: But this is the last question and then I am done!
     
  12. Sep 4, 2011 #11

    Doc Al

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    Look above!
     
  13. Sep 4, 2011 #12
    Ah, the distance between the first destructive interference (half a wavelength) to the center will decrease?
     
  14. Sep 4, 2011 #13

    Doc Al

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    Yes. Smaller wavelength means that a given maximum or minimum will be closer to the center.
     
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