Angular Width Of Central Diffraction

  • Thread starter Chris18
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  • #1
Chris18
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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?
 

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  • #2
blue_leaf77
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λ=350*10^-9 and d= 0.20*10^-6
Are those numbers in the same unit?
 
  • #3
Chris18
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λ=nm and d= mm
 
  • #4
TSny
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Note 1 mm ≠ 10-6 m.
 
  • #5
Chris18
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Ok so I must have 10^-7 ?
 
  • #6
blue_leaf77
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d= 0.20*10^-6
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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