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Ray Deflection of a Prism

  1. Aug 25, 2012 #1
    1. The problem statement, all variables and given/known data

    The problem I am having troubles with is proving the deflection of a ray by a prism can be represented by a specific equation : θd = θ - α + arcsin(sin(α)√(n2 - sin2(α)) - sin(α)cos(α)). I have derived another form of the deflection equation for a prism and I am attempting to move it into the form requested but I am getting stuck on some trigonometry.

    2. Relevant equations

    α is the apex angle of the prism and θ is the angle of incidence on the prism.

    3. The attempt at a solution

    See attachments.
     

    Attached Files:

    Last edited: Aug 25, 2012
  2. jcsd
  3. Aug 25, 2012 #2

    TSny

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    Hello, Axis001. Could you please define the symbols, especially the meaning of θ and α?
     
  4. Aug 25, 2012 #3
    My apologies not sure how I overlooked defining my variables. α is the apex angle of the prism and θ is the angle of incidence on the prism.
     
  5. Aug 25, 2012 #4

    TSny

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    Is your attachment showing your own work? Are you trying to find a mistake in the attachment?

    I'm guessing that θ1 and θ'1 are the angles of incidence and refraction, respectively, at the first refraction; whereas, θ'2 and θ2 are the angles of incidence and refraction, respectively, at the second refraction. (That is, the primes denote angles inside the prism.)
     
  6. Aug 25, 2012 #5
    The work in the attachment is mine and your assumption of origin of angles of refraction is correct. I have simplified the equation for the displacement angle I originally found θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ))) into terms of simply θ, n, and α. The last step shown in the attachment is where I am currently stuck. I currently have the displacement of the prism at θd = θ - α + arcsin(nsin[sin(α)√(n2 - sin2(α)) - sin(θ)cos(α))] and I simple cannot figure out how to reduce it to the required form of θd = θ - α + arcsin(sin(α)√(n2 - sin2(α)) - sin(θ)cos(α)). I have tried every single trigonometric identity I can think of and none of them give the reduced equation I am looking for.
     
    Last edited: Aug 25, 2012
  7. Aug 25, 2012 #6

    TSny

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    In the attachment, the third to last equation looks ok. However, in going to the final expression there appears to be two mistakes:
    (1) the cosine factor is not quite simplified correctly (going from the next to last line to the last line)
    (2) there’s a place where you took no/n1 to be n instead of 1/n (going from the third to last line to the next to last line)
     
  8. Aug 25, 2012 #7
    Firstly thank you for your reply I greatly appreciate it.

    (2) there’s a place where you took no/n1 to be n instead of 1/n (going from the third to last line to the next to last line)

    Ah yes I'm not sure how I missed.

    (1) the cosine factor is not quite simplified correctly (going from the next to last line to the last line)

    With the correction to the incorrectly stated index of refraction the correct reduction from the second to last line to the last line should be:

    θd = θ - α + arcsin(nsin[sin(α)√(n2 - sin2(α)) - (1/n)sin(θ)cos(α)])

    Which leaves me with the problem of the nsin(...) term in the equation. The required equation θd = θ - α + arcsin(sin(α)√(n2 - sin2(α)) - sin(θ)cos(α)) which is just the equation I currently have with the nsin(...) term eliminated from it. I have tried using sum property for sines again to see if it generates the required equation but it just provides a useless and incredibly complicated new equation.
     
  9. Aug 25, 2012 #8

    TSny

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    Note that cos[sin-1(β)] = √(1- sin2[sin-1(β)]) = √(1-β2)
     
  10. Aug 25, 2012 #9
    Ok from "start" to current step with the corrections you helped me find this is what I have.

    θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ)))
    θd = θ - α + arcsin(nsin[sin(α)cos[arcsin((1/n)sin(θ))] - cos(α)sin[arcsin((1/n)sin(θ))])
    θd = θ - α + arcsin(nsin[sin(α)sqrt[1 - (sin2(θ))/n2] - (1/n)cos(α)sin(θ))
    θd = θ - α + arcsin(nsin[(1/n)sin(α)sqrt[n2 - sin2(θ)] - (1/n)cos(α)sin(θ))
    θd = θ - α + arcsin[nsin[(1/n)[sin(α)sqrt[n2 - sin2(θ)] - cos(α)sin(θ)]]]
     
  11. Aug 25, 2012 #10

    TSny

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    Oh, I overlooked another mistake which I highlighted in red above. Should those sin's be there?
     
  12. Aug 25, 2012 #11
    Yes the nsin(...) terms should be there and it is what is giving me such a head ache since I just simply cannot get it out of the equation. A form of the deflection equation that I know is θd = θ - α + arcsin(nsin(α - arcsin((1/n)sin(θ))). It is easily derivable from applying Snell's Law at both "sides" of the prism and the expression of θd = θ1 + θ2 - α. Solving for θ2 into a expression of θ1 and α will result in the deflection equation I started this with but it results with a nsin(...) term I just can't seem to get out of my equations.
     
  13. Aug 26, 2012 #12

    TSny

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    Hmm. Which is correct:

    sin(α - β) = sinαcosβ-cosαsinβ

    or

    sin(α - β) = sin[sinαcosβ-cosαsinβ]
     
  14. Aug 26, 2012 #13
    :eek: I need to go to sleep. Thank you so much I really appreciate it.
     
  15. Aug 26, 2012 #14

    TSny

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    Me too. :zzz:
     
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