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Homework Help: How far is the screen from the slit? Diffraction problem.

  1. Nov 6, 2008 #1
    1. The problem statement, all variables and given/known data

    The distance d from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 centimeters. The wavelength of the red laser is 633 nanometers and the slit is 0.15 millimeters wide. How far is the screen from the slit?

    2. Relevant equations

    [tex]sin\theta = \frac{m\lambda}{a} [/tex]

    3. The attempt at a solution

    m = 3
    [tex]\lambda = 633 \times 10^{-9} m [/tex]
    [tex]a = 0.15 \times 10^{-3} m [/tex]

    [tex]sin\theta = \frac{y}{4.05 \times 10^{-2} m}[/tex]

    [tex]\frac{y}{4.05 \times 10^{-2} m} = \frac{m\lambda}{a}[/tex]

    [tex]y = \frac{4.05 \times 10^{-2} m (3)(633 \times 10^{-9}m)}{0.15 \times 10^{-3} m}[/tex]

    y = 0.00051273 m

    [tex]sin\theta = \frac{0.00051273 m}{4.05 \times 10^-2 m}[/tex]
    [tex]\theta = 0.725[/tex]
    [tex]cos 0.725 = \frac{r}{4.05 \times 10^{-2}}[/tex]
    [tex]r = 4.05 \times 10^{-2} \times cos 0.725[/tex]
    [tex]r = 4.05 \times 10^{-2}[/tex]

    This is not the correct answer. Where did I go wrong?
    Last edited: Nov 6, 2008
  2. jcsd
  3. Nov 6, 2008 #2
    Re: Diffraction

    The third maximum corresponds to m=2.
    First one is at theta=0 (m=0)
  4. Nov 6, 2008 #3
    Re: Diffraction

    It's not looking for the maximum it's looking for the minimum.
  5. Nov 6, 2008 #4
    Re: Diffraction

    Sorry, I did not look properly.

    You are calculating y but y is given (y is 4.05 cm).
    Tan(theta) = y/D
    D is the unknown here.
    I got approx 3 m for D.
  6. Nov 6, 2008 #5
    Re: Diffraction

    The correct answer is 3.20 m. How do you get that answer?
  7. Nov 7, 2008 #6
    Re: Diffraction

    I've told you already, write the formula as
    y/D = m*lambda/z

    Solve for D. y is given.
  8. Nov 8, 2008 #7
    Re: Diffraction

    So I guess in this case, since theta is small, sin(theta) is about equal to tan(theta), and setting them equal to each other is the key to solving this problem.
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