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Interference Intensity

  1. Nov 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Monochromatic light of wavelength 463 nm from a distant source passes through a slit that is 0.0350 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum (θ = 0∘) is 9.20*10-5W/m2 .
    What is the intensity at a point on the screen that corresponds to θ = 1.20∘.
    2. Relevant equations
    I=I0*[sin(πa(sinθ)/λ) / (πasinθ/λ)]2

    3. The attempt at a solution
    I = 9.20*10-5 * [sin(π*0.035*10-3*(sin1.20)) / 463*10-9 / π*0.035*10-3(sin1.20)/463*10-9 ] 2
    = 2.802*10-8 W.m-2

    This is the answer that I have found but unfortunately it has come up as wrong - I have made sure that my calculator is in degrees and not radians, and have tried multiple times to see if I have made a mistake whilst inputting in the calculator.
    Any idea where I have made a mistake ?

    PS: As I am new could you also give me any insight as to how to write my calculations out more clearly on this website, because I am aware that it is very tedious to read the calculations, thanks !
     
  2. jcsd
  3. Nov 3, 2015 #2

    BvU

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    So what do you get for ##\ {\displaystyle{ \pi \, a \sin\theta\over \lambda}}\ ## ?

    You can use LaTeX by putting latex source between ## \#\# ## or between ## $$ ##

    A good way to learn is to use the right mouse button while over a typeset formula and select "Show Math As | TeX Commands"​

    [edit] your "I have made sure that my calculator is in degrees" makes me suspicious about ##\ \ \sin \left ( \displaystyle { \pi \, a \sin\theta\over \lambda} \ \right ) \ \ ## :wink: !! 0.087 is dead wrong !

    Oh and for such small angles ##\sin\theta \approx \theta ## and you can leave your calculator on radians :rolleyes:
     
    Last edited: Nov 3, 2015
  4. Nov 3, 2015 #3
    For ##{ \pi \, a \sin\theta\over \lambda} ##
    I have got :
    π*0.035*10-3*sin(1.2) / 463*10-9 ≈ 4.97

    which would therefore give me##
    \ \ \sin \left ( \displaystyle { \pi \, a \sin\theta\over \lambda} \ \right ) ≈ -0.97 \ \

    ##And if I apply the whole formula : I=I0*[sin(πa(sinθ)/λ) / (πasinθ/λ)]2

    Is that correct?

    Sorry, about the problems using LaTeX, I'm still getting used to it and am trying to edit them out!

    Edit : The answer is indeed correct, thank you ! I was just confused as to why I needed to use radians instead of degrees, I think I understand but an explanation would still be welcome, thank you <3
     
    Last edited: Nov 3, 2015
  5. Nov 3, 2015 #4

    BvU

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    If you are really a beginner, you are doing quite well with LaTeX.

    The natural unit for angles is radians. (even in a miscreant like Excel !).

    In the expression for ##I/I_0## the ##\displaystyle { a \sin\theta\over \lambda} \ ## is a phase difference in number of wavelengths - so in natural units, not in degrees. See also here (but this is a calculation for a minimum).
     
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