Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Determine the largest value of angle optics problem

  1. Dec 30, 2005 #1
    A ray of light is normally incident on the face PQ of a plastic prism with an index of refraction n=1.25 as shown in Figure 1. Determine the largest value of angle [tex]\theta[/tex] so that the ray is totally reflected at the face PR.

    I just want to know whether the sum of the incidence and reflection angle must equal to right angle or not (for this diagram), does the ray of reflection must be perpendicular with the plane QR?

    If that so, the maximum angle of [tex]\theta[/tex] is equal to the critical angle,right?

    Attached Files:

  2. jcsd
  3. Dec 31, 2005 #2
    I can't view the diagram. I believe theta is the angle of incidence on side PR.

    Do you know what is the condition for total internal reflection?

    For smaller values of theta, the ray will pass through side PR and enter the air (say angle of refraction is [itex]\alpha[/itex]). As you increase [itex]\theta[/itex], at one point the light will be reflected back in to the prism. This angle is known as critical angle. This is what you need to find.

    Write Snell's law for side PR. Now use the fact that

    [itex] |(sin \alpha)| \leq 1 [/itex]

    which should give you the largest angle [itex]\theta[/itex]

    Hope this helps.
    Last edited: Dec 31, 2005
  4. Dec 31, 2005 #3
    Nono, theta is actually the angle of the prism
  5. Dec 31, 2005 #4
    Still, since your angle of incidence on the first face (PQ) is 90 degrees, from geometry angle of incidence on the second face (PR) is [itex]\theta[/itex]. So the critical angle of incidence on face PR is same as the critical angle of the prism.
  6. Dec 31, 2005 #5
    Is the sum of the incidence angle and reflection angle is always equal to 90 degrees in the prism?
  7. Dec 31, 2005 #6
    This will be true for a right prism with it's sides being equal in length. (I am sure there is a name for this type of triangles). i.e. if side PQ = QR and angle [itex] \angle PQR = 90[/itex]. Draw a ray diagram on such a prism and see for your self.

    I was not able to see your diagram yesterday. Now I can see it and find that the
    [itex] \angle QRP = \theta[/itex]. This means

    if the angle of incidence on face PR = [itex] \theta_1[/itex] then,

    [itex] \theta = 90 - \theta_1[/itex]

    I believe you can find the maximum value of [itex] \theta_1[/itex] as I described earlier. This will give you a minimum value for the prism angle [itex] \theta [/itex].

    I know that your question ask for the largest value of theta. Could there be a mistake in your problem or diagram?
    Last edited: Dec 31, 2005
  8. Jan 1, 2006 #7
    There is no problem with the question and the diagram.
  9. Jan 1, 2006 #8

    Doc Al

    User Avatar

    Staff: Mentor

    No; that's only true if [tex]\theta[/tex] is 45 degrees.

    No, but the minimum angle of incidence to side PR will equal the critical angle. From that you can determine the maximum [tex]\theta[/tex].
  10. Jan 1, 2006 #9
    Why is the minimum angle of incidence is equal with the maximm of theta?
  11. Jan 1, 2006 #10

    Doc Al

    User Avatar

    Staff: Mentor

    If you draw the normal to side PR, you will see (with a little geometry) the simple relationship between theta and the angle of incidence.
  12. Jan 1, 2006 #11
    Yes, I know incidence , reflection angle and [tex]\theta[/tex] is all the same when the reflection ray is perpendicular with QR ( sum of incidence and reflection is 90) . I am just not sure whether the reflection ray will always perpendicular with QR or not,no matter what is the value of [tex]\theta[/tex].
  13. Jan 1, 2006 #12

    Doc Al

    User Avatar

    Staff: Mentor

    That's true, but not relevant here.
    No reason why it should be.

    Hint: Consider the angle that the incident ray makes with PR. How does that relate to [tex]\theta[/tex]?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook