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Homework Help: Light + prisma

  1. Mar 18, 2005 #1

    I'm trying to figure out what angel the light-wave will leave the glasprisma (it enters the glasprisma from the air).

    http://home.tiscali.se/21355861/bilder/prisma1.JPG [Broken]

    So I need to use Snell's law

    [tex]n_{1}sin\theta_1 = n_{2}sin\theta_2[/tex]

    n for the glasprisma equals 1.6

    So when it enters the prisma me should get

    [tex]\theta_2 = sin^{-1}{\left(\frac{1*sin0}{1.6}\right)} = 0[/tex]

    So when it enters the glasprisma the angle still is 0. When it's leaving the prisma we get this.

    http://home.tiscali.se/21355861/bilder/prisma2.JPG [Broken]

    Where v = 90 - 35 = 55degres.


    [tex]\theta_2 = sin^{-1}{\left(\frac{1.6*sin-55}{1}\right)} = Error![/tex]

    What am I doing wrong?

    Thank you.
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Mar 18, 2005 #2

    Doc Al

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    Staff: Mentor

    total internal reflection

    It just means that the angle of incidence on that surface exceeds the critical angle for total internal reflection. The light will reflect from that surface and exit from the bottom surface.
  4. Mar 18, 2005 #3
    So in this case it's total reflection? Then I've misscalculated.
  5. Mar 18, 2005 #4

    Doc Al

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    Staff: Mentor

    What makes you think you've miscalculated? You still can calculate the angle that the light leaves the prism.
  6. Mar 18, 2005 #5
    The change of the direction should be 57degrees. If it leaves the prism from the bottom surface the change of direction is probably more then 100degrees. Or maybe I've missunderstod you.

    Maybe it should be something like this.

    http://home.tiscali.se/21355861/bilder/prisma3.JPG [Broken]
    Last edited by a moderator: May 1, 2017
  7. Mar 18, 2005 #6

    Doc Al

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    Staff: Mentor

    Don't guess, calculate.

    Last edited by a moderator: May 1, 2017
  8. Mar 18, 2005 #7
    Thank you Doc Al. I'll calculate until I get the correct answer. :)
  9. Mar 18, 2005 #8
    Have you tried using; n= [sin( (A+D)/2)] / sin(A/2) where A is the angle of the prism and D the maximum angle of deviation. when you calculate the max angle of deviation you can use; 2i=A+D where i is the angle of incidence on the prism.

    hope this helps
  10. Mar 18, 2005 #9
    Hi zanazzi78.

    Sorry but you lost me there. Do you think you cold point out int the picture where A, D and i is? It would be really helpfull if you could attach the picture :smile:
  11. Mar 18, 2005 #10
    I`m new to this forum thing !!! how do i post pictures?
  12. Mar 18, 2005 #11
    Look at the attachment.

    http://home.tiscali.se/21355861/bilder/picture.PNG [Broken]
    Last edited by a moderator: May 1, 2017
  13. Mar 18, 2005 #12
    Oops here you go
  14. Mar 18, 2005 #13
    zanazzi78 it looks like it doesn't work. I got the same problem but I thought it only wasn't working from me. You can upload the picture here

    Choose "Bilder".
    Last edited by a moderator: Apr 21, 2017
  15. Mar 18, 2005 #14
    I can't get the pic small enough to post!

    The angle A is the interal angle between the sides of the prism.

    The angle D is the change in angle, between, where the ray would have gone(if not refracted), and the actulla path that the ray leaves the prism!

    The angle i is the angle the ray makes with the prim.
  16. Mar 18, 2005 #15

    Sorry for the delay My pc crashed (Damn technology!)
  17. Mar 18, 2005 #16
    Thank you zanazzi78. Now I understand :)
  18. Mar 18, 2005 #17

    Doc Al

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    Staff: Mentor

    That's the right attitude. But you're almost done. Figure out (from geometry) the angle of incidence that the light makes with the bottom surface, then apply Snell's law one last time.
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