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Optics problem

  1. Mar 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Using the results of Problems 4.70, that is EQs. (4.98) and (4.99), show that

    Rparallel + Tparalllel = 1


    2. Relevant equations

    Rparallel = ( tan^2 ( thetai - thetat) ) / (tan^2 (thetai + thetat) )

    Tparallel = (sin (2*thetai) * sin (2*thetat))/ sin^2 (thetai + thetat)

    3. The attempt at a solution

    After getting this far (shown below) I took it to the math help center at my university and they couldn't solve it any further than what I had done:

    First put both in the same denominator

    sin^2 (thetai - thetat)) / cos^2(thetai - thetat) * cos^2(thetai + thetat/sin^2(thetai + thetat which gives a common denominator of cos^2(thetai-thetat)* sin^2(thetai + thetat)

    For brevity I will call thetai = i and thetat = t

    Now we have sin^2(i-t)*cos^2(i+t) + sin (2*i)*sin(2*t)/ cos^2(i-t)*sin^2(i+t)

    I tried (1 - cos^2(i-t)*(1-sin^2(i+t) + sin(2*i)*sin(2*t)/ cos^2(i-t)*sin^2(i+t)

    which puts the minus on cos and plus angle on sin which matches the denominator but that is as far as I got which was further than the help desk at my university.

    Can someone give me a hint as to which identities I should use to work this out?

    You have my undying gratitude and about a million photons of positive energy sent to you for your help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 14, 2014 #2
    Here's two identities that might help:

    ##\sin^2(x-y) = \sin^2(x+y) - \sin(2x)\sin(2y)##
    ##\cos^2(x-y) = \cos^2(x+y) + \sin(2x)\sin(2y)##
     
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