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Sigle slit

  1. Apr 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A laser 632.8nm is sent through a single slit of widith 0.30 mm. Wats the width of the central maximum on a screen 1.0m in the back of the slit?

    2. Relevant equations

    asin(theta) = m lamda

    3. The attempt at a solution
    lamda: 632.8nm

    0.30 mm= 3.0 x 10^-4 m

    sin(theta)= m lamda/ a
    sin(theta)= 1 (632.8x10^-9)/3.0 x 10^-4 m

    = 2.10x10^-3 m
     
  2. jcsd
  3. Apr 27, 2009 #2

    Ok you're on the right track, you found sintheta. Now use your inverse sin function and find the angle theta. Once you have that, you know the distance to the screen. Can you find a basic trig property to help you find the unknown "height" if you will of the screen? (remember, this height you find will only be half the width of the central maximum)
     
  4. Apr 27, 2009 #3
    do i inverse sin 2.10 E -3? i'm nto quick sure what to do?
     
  5. Apr 27, 2009 #4
    You don't know how to turn SinTheta = x into an angle?

    What Theta corresponds to SinTheta = 1/sqrt2 for example?
     
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