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What is the angular width of the central maximum?

  1. Apr 23, 2006 #1
    A single slit of width 1.50 um is illuminated with light of wavelenth 500.0 nm. What is the angualar width of the central maximum?
     
  2. jcsd
  3. Apr 23, 2006 #2

    Hootenanny

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    Please show some work or initial thoughts.

    ~H
     
  4. Apr 23, 2006 #3
    the answer in the book is 38.9 degrees but how is this right?
     
  5. Apr 23, 2006 #4
    I tried the formula &= wavelenth / b and came up with (5 * 10^-9) / (1.5 *10^-6) and got 3.33 * 10^-3 then multiplied this by 2. is this right?
     
  6. Apr 23, 2006 #5
    anymore ideas anyone ???
     
  7. Apr 23, 2006 #6

    Hootenanny

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    The answer in the book is correct. You are simply a factor of [itex]10^{2}[/itex] out with respect to the [itex]\sin\theta[/itex]. Your error lies here:

    [tex]500nm \neq 5\times 10^{-9}m[/tex]

    Recheck your calculation. Also you need to find the angle (ArcSin)

    ~H
     
    Last edited: Apr 23, 2006
  8. Apr 23, 2006 #7

    nrqed

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    first, 500 nm = 500 times 10^-9.

    The second thing is that what you found is sin(theta). To find the *angle* you must do the inverse sin of that. Then multiply the result (in degrees) by two
     
  9. Apr 23, 2006 #8
    thanks guys it works, just made a silly error!
     
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