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Small angle approx. optics

  1. Aug 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Had the same problem as this threadstarter:

    https://www.physicsforums.com/showthread.php?t=109059

    2. Relevant equations



    3. The attempt at a solution

    I managed to find a ratio of tangents for the two angles. From there, it seems you're supposed to go "well tan(x) ≈ x for small angles so let's magically assume this is a small angle and go grab a donut".

    Why is the small angle approximation appropriate for this problem and how do I avoid getting stuck on similar problems in the future?
     
  2. jcsd
  3. Aug 8, 2013 #2

    NascentOxygen

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    Staff: Mentor

    You could try a range of small angles and judge for yourself, e.g.,

    x=1°: x=..... radians, sin x=....., tan x=.....

    x=2°: x=..... radians, sin x=....., tan x=.....

    x=3°:


    By working this out for yourself, you'll be left with a better appreciation of the result. :smile:

    Remember, the trig approximations expect x to be in radians.
     
  4. Aug 8, 2013 #3
    Do what nascent oxygen recommends.....you will be surprised how 'BIG' the angle can be yet still be considered 'SMALL'
     
  5. Aug 8, 2013 #4

    Integral

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    Staff Emeritus
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    Gold Member

    Of course the small angle approximation only works if you use radians.
     
  6. Aug 8, 2013 #5
    As recommended !
     
  7. Aug 8, 2013 #6
    No, I get the small angle approximation, I just don't get how I am supposed to know that it is applicable here. I mean, we're not given any angles. We're supposed to figure it out through trig/geometry trickery.Theoretically, the angles could be pi/2 for all I know.
     
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