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Wave Angle Nature of Light

  1. Aug 29, 2007 #1
    Ok, the following problem I've tried several times to solve, although I suspect that somewhere along the line I'm just approaching what appears to be a simple problem in the wrong way:

    A two-point source operates at a frequency of 1.0 Hz to produce an interference patter in a ripple tank. The sources are 2.5 cm apart and the wavelength of the waves is 1.2 cm.

    Calculate the angles at which the nodal lines in the pattern are located far from the sources. (Assume the angles are measured from the central line of the pattern.)

    So here's what is given:

    d = 2.5 cm
    f = 1.0 Hz
    lambda = 1.2 cm

    Some formulas/expressions to work with:

    1. v = f(lambda)

    'v' is the speed of the wavelength

    2. (d)sin(theta) = (n-1/2)lambda

    Where 'theta' is the angle of the nth nodal line, 'lambda' is the wavelength, and 'd' is the distance from the sources.

    3. lambda = (d)sin(theta)/(n-1/2)

    This is just a variation of (1).

    4. theta + alpha = 90 degrees


    5. theta prime + alpha = 90 degrees

    theta prime = theta

    sin(theta prime) = x/L, sin(theta) = (n-1/2)(lambda/d)

    6. Since sin(theta prime) = sin(theta)

    x/L = (n-1/2)(lambda/d), so therefore

    lambda = (x/L)[d/(n-1/2)]

    'x' is the perpendicular distance from the right bisector to the point P on the nodal line
    'L' is the distance from the midpoint between the two sources to the point P

    Although I think it's safe to say that the 'x' and 'L' variables are irrelevant in this case because they aren't given, and are not in the formula used to solve for angles. I suppose that the same goes for the 'n' nodal number, since it's not given, I need to derive an equation which only involves the variables given, as well as sin(theta), so that I can solve for theta, and, in addition, alpha.

    If there any helpful suggestions on how to approach this problem, it would be much appreciated.
    Last edited: Aug 29, 2007
  2. jcsd
  3. Aug 29, 2007 #2


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    Homework Helper

    You have the equation: [tex]dsin\theta = (n-\frac{1}{2})\lambda[/tex]

    Just solve for [tex]\theta[/tex] and plug in your values for d and [tex]\lambda[/tex]
  4. Aug 30, 2007 #3

    I would need to have the nodal line value 'n' in order solve for [tex]\theta[/tex], but I don't. Is there some way to solve for 'n'? That's the part I'm lost at.
  5. Aug 30, 2007 #4


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    Just leave the n in there I think.

    so [tex]\theta = sin^{-1}(\frac{(n-\frac{1}{2})\lambda}{d})[/tex]

    Just plug in your values for d and [tex]\lambda[/tex]... and that'll be your final answer. It's asking for the angles... so I think they want the angle as a function of n.
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