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Find angle of incidence given deviation angle

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data

    A ray of light passes from air into benzene. For its deviation angle δ = |θ1 - θ2| to be 9.5°, what must be its angle of incidence? The refractive index of benzene is 1.501.

    2. Relevant equations

    Snell's law

    3. The attempt at a solution

    From Snell's law we have sin θ1 = 1.501 sin θ2. Here θ1 is the angle of incidence. We can rewrite θ2 = θ1 - 9.5 and substitute that into the equation to get

    sin θ1 = 1.501 sin (θ1 - 9.5)

    But how is it possible to solve for θ1 from this equation? :confused:
     
  2. jcsd
  3. Aug 11, 2011 #2
    sin θ = (e-e-iθ)/(2i)

    (e-e-iθ)/(2i) = 1.501 (ei(θ-9.5)-e-i(θ-9.5))/(2i)
    e-e-iθ = 1.501 (ei(θ-9.5)-e-i(θ-9.5))
    e2iθ-e0 = 1.501 (ei(2θ-9.5)-e-i(-9.5))
    e2iθ-1 = 1.501 (e2iθe-9.5i-e9.5i)
    e2iθ-1.501 e-9.5ie2iθ = 1-1.501 e9.5i
    e2iθ (1-1.501 e-9.5i) = 1-1.501 e9.5i
    e2iθ = (1-1.501 e9.5i) /(1-1.501 e-9.5i)

    e2iθ = (1-1.501 e9.5i) /(1-1.501 cos(9.5) + i 1.501 sin(9.5))
    e2iθ = (1-1.501 e9.5i)(1-1.501 cos(9.5) - i 1.501 sin(9.5)) /|1-1.501 cos(9.5) + i 1.501 sin(9.5)|2
    e2iθ = (1-1.501 e9.5i)(1-1.501 e9.5i) /|1-1.501 cos(9.5) + i 1.501 sin(9.5)|2

    e2iθ = (1-1.501 cos(9.5) - i 1.501 sin(9.5))2 /[(1-1.501 cos 9.5)2+(1.501 sin 9.5)2]
    e2iθ = [(1-1.501 cos 9.5)²-(1.501 sin 9.5)² - 2i (1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]

    2iθ = ln|[(1-1.501 cos 9.5)²-(1.501 sin 9.5)² - 2i (1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]|
    + i arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

    2iθ = ln|[(1-1.501 cos 9.5)²-(1.501 sin 9.5)² - 2i (1-1.501 cos 9.5)(1.501 sin 9.5)]| - ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]
    + i arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

    2iθ = ln√{ [(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]²+[2(1-1.501 cos 9.5)(1.501 sin 9.5)]²} - ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]
    + i arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

    θ = -i/2 ln√{ [(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]²+[2(1-1.501 cos 9.5)(1.501 sin 9.5)]²} +i/2 ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]
    + 1/2 arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

    θ = i/2 [ln[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²] - ln√{ [(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]²+[2(1-1.501 cos 9.5)(1.501 sin 9.5)]²}]
    + 1/2 arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

    θ = i/2 ln {[(1-1.501 cos 9.5)²+(1.501 sin 9.5)²]/√[((1-1.501 cos 9.5)²-(1.501 sin 9.5)²)²+(2(1-1.501 cos 9.5)(1.501 sin 9.5))²]}
    + 1/2 arctan [2(1-1.501 cos 9.5)(1.501 sin 9.5)] /[(1-1.501 cos 9.5)²-(1.501 sin 9.5)²]

    Please excuse the terrible font. :)
     
    Last edited: Aug 11, 2011
  4. Aug 11, 2011 #3

    PeterO

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    Homework Helper

    You could use the expression sin(A-B) = sinA.cosB - cosA.sinB
     
  5. Aug 11, 2011 #4
    Aren't there an easier and less massy way to solve this question? This was just a 1 mark question in an old exam... I don't think it needs all the tedious trigonometry. Maybe there is a simple method?
     
  6. Aug 11, 2011 #5

    PeterO

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    Homework Helper

    If the paper is old enough, you would have been using a table of sines rather than a calculator, so you would be able to look at multiple sine values at the same time.

    The sin(A-B) method quickly reduces to a Tan function.
     
  7. Aug 11, 2011 #6

    PeterO

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    Homework Helper

    Strangely I received the message about your calculation, but don't find it in this thread????

    In your solution you omitted a bracket, so forgot to multiply sin9.5 by 1.501.

    I am pretty sure that if you do that you will get the expected answer.

    Note: I hope you were only rounding off values as you typed them in your post, and not before you had finished calculating with them.
     
  8. Aug 11, 2011 #7
    Thanks Peter. I had forgotten to multiply everything by 1.501 in that line of my working. Thanks a lot for your help. I got the answer right :)
     
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