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Minimal angle of deviation through the prism

  1. Mar 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Minimal angle of deviation δm of light after passing through prism is connected with index of refraction n and inner angle of prism Φ:


    If index of refraction and inner angle of prism are directly measured, determine minimal angle of deviation,
    n=(1.52 ± 0.04)
    Φ=(30 ± 0.02)°

    3. The attempt at a solution
    When having dependent variables, we use formulae x=F(a,b,c,...)

    x(medium) =F(a(medium), b(medium),....)
    plus forumulae for uncertainty.

    My question here is, how to express delta as a function of phi and n?

    Thank you
  2. jcsd
  3. Mar 15, 2015 #2

    rude man

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    Hint: use arc sin.
  4. Mar 15, 2015 #3
    well, n * sin ( phi ) = sin 1/2 ( phi + delta )
    arcsin (n* sin (phi) = 1/2 ( phi + delta )
    2 * arcsin ( n* sin(phi))= ( phi + delta )
    => delta = 2*arcsin (n*sin (phi )) - phi

    idem there other way?
    i assume this one is correct tho
  5. Mar 15, 2015 #4

    rude man

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    That's the way!
  6. Mar 15, 2015 #5
    one more question.
    i got result in a problem i.e.
    x= 43.84
    Mx = 0.01559

    now, when i round it to one significant is this correct;

    Mx= 0.02
    x= 43

    x = (43 +- 0.02)
  7. Mar 15, 2015 #6
    or x = 40
  8. Mar 15, 2015 #7

    rude man

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    Not sure I can answer that question. I would have said x = 43.84 +/- 0.02 but I don't know the background.

    OK, I didn't see the error bars on φ. and n. I thought you just wanted nominal δm.

    Realize that φ and n are independent variables. δ depends on the material's n and on the prism angle φ.
    So you need dδ as a function of dφ and dn. What would that be? (Hint: dδ is an "exact differential" of φ and n.)
  9. Mar 16, 2015 #8
    I appreciate your help, but i solved it, the rest, by myself. thanks. :)
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