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Angular Width Of Central Diffraction

  1. Jun 1, 2016 #1
    350nm falls on a single slit of width of 0.20mm. What is the angular width of the central diffraction peak?



    I think that the width should be equal to 2Θ, where Θ=arcsin(m*λ/d).... m=1 and we have λ=350*10^-9 and d= 0.20*10^-6.....but when i do the calculations I get 1.75 and the arcsin is a maths error...Could anyone help me please?
     
  2. jcsd
  3. Jun 1, 2016 #2

    blue_leaf77

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    Are those numbers in the same unit?
     
  4. Jun 1, 2016 #3
    λ=nm and d= mm
     
  5. Jun 1, 2016 #4

    TSny

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    Note 1 mm ≠ 10-6 m.
     
  6. Jun 1, 2016 #5
    Ok so I must have 10^-7 ?
     
  7. Jun 1, 2016 #6

    blue_leaf77

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    You should have written the unit as well, if you write it correctly it will be ##d= 0.20 \times 10^{-6}## km. If you use km for d then you must also use km for ##\lambda##.
     
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