1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Refracting angle of a prism

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Why the angle of a prism can be calculated by measuring θ , and then θ/2 ?

    2. Relevant equations



    3. The attempt at a solution

    Really no idea about it , please guide .Thank you.
     

    Attached Files:

  2. jcsd
  3. Feb 24, 2013 #2
    One more question , why the light ray from the collimator reflected not refracted?
    Thank you.
     
  4. Feb 24, 2013 #3

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Is the prism immersed in a spherical liquid of n > n of prism? Then you can get total internal reflection of the incoming beam back into the sphere.
     
  5. Feb 24, 2013 #4
    No, the outside medium is air. I read and checked books but I can't find one to prove θ/2=refracting angle of prism. Please guide . Thank you.
     
  6. Feb 24, 2013 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Not sure I understand the picture. It looks like a circular prism with a triangular hollow.
     
  7. Feb 24, 2013 #6

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Which could account for total internal reflection back into the circular prism.
     
  8. Feb 24, 2013 #7
    ABC is a transparent prism .
    A is the refracting angle of a prism.
    T is telescope to receive the reflected light.
    This is an experiment to find A.
    A lot of books did say this is an easy way to find A . But I cant find how . Please guide. Thank you.
     
  9. Feb 24, 2013 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Ok, so this nothing to do with refraction. It's simply a geometry problem. The prism might as well be a pair of mirrors at angle A.
    The incident light makes an angle A/2 to the surface, and is reflected at angle A/2 to the surface. Therefore each reflection is at angle A to its corresponding incident ray. Since the rays each side are turned through angle A in opposite directions, they now diverge at angle 2A = theta.
     
  10. Feb 24, 2013 #9
    How do we make the incident light at an angle A/2 to the surface of mirror?


    How to know reflection is at angle A to its incident ray ? or do you mean the incident light makes an angle A/2 to the normal?

    Thank you
     
  11. Feb 24, 2013 #10

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Simple geometry. Draw a line bisecting the angle A. It is parallel to the incident beam, so that line and the beam make the same angle to the surface.
    However, it is important that it does not need to be exactly A/2. Suppose you set it up a little bit askew, so the beam makes an angle A/2+x to one surface, and therefore an angle A/2-x to the other.
    A reflected beam makes the same angle to the surface as the incident beam, so these are also A/2+x, A/2-x respectively. One beam is therefore 'turned' through an angle of A+2x, the other through an angle of A-2x,. Adding these up still gives 2A as the divergence between them.
     
  12. Feb 25, 2013 #11
    Totally understand already . Really thank you . No wonder physics books don't show, it is just a math problem. Thank you so much
     
  13. Feb 25, 2013 #12
  14. Feb 25, 2013 #13

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It doesn't define the point R, but it looks like it is chosen as the point where the normals at Q and S intersect. So by definition the angles that AR subtends at Q and S are each right angles. That makes AQRS a cyclic quadrilateral, i.e. its vertices lie on a circle, and AR is a diameter of that circle.
     
  15. Feb 25, 2013 #14
    But do we have enough to prove Q and S are lie on the circumference?
    Use the centre of circle?
     
  16. Feb 25, 2013 #15

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes. You can define a circle uniquely by declaring that AR is diameter. Any right-angled triangle you then draw with AR as hypotenuse will have its third vertex on the circumference of the circle. This is pretty basic geometry.
     
  17. Feb 26, 2013 #16
    Yup, you are totally correct.
    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Refracting angle of a prism
  1. Prism Refractions (Replies: 2)

  2. Refraction in a prism (Replies: 0)

Loading...