Intensity of waves

Discussion in 'Physics' started by boks, Nov 15, 2008.

  1. boks

    Thread Starter Active Member

    Oct 10, 2008
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    I have calculated the amplitude of three interfering waves as a function of \theta:

    \sqrt{\frac{P_{2}}{4 \pi}} \frac{1}{r}[1 + 4exp(i(\phi_{3} - \phi_{2} + kdsin \theta)) + \frac{1}{4}exp(i(\phi_{1} - \phi_{2} - kd sin \theta))]

    I now want to find the intensity, i.e. the square of the amplitude. Should I start by converting the exp terms into cos and sine? And is it only the real part that I should concentrate on?
     
    Last edited: Nov 15, 2008
  2. steveb

    Senior Member

    Jul 3, 2008
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    The simplest way to find the intensity, which is the square of the magnitude, is to multiply the wave function times it's complex conjugate. The complex conjugate of the wave function is the same as the original wave function with i replaced by -i. Once you do this, the answer is automatically real.

    If you allow me to use unconventional notation (i=-i) to signify substitution of -i in for i, ....
    |\Psi |^2 = \Psi \; \Psi^* = \Psi(i=i) \; \Psi(i=-i)

    I haven't worked out the problem, but your suggestion to break e into the sin and cos components is usually a good idea, if you don't see an obvious simplification.
     
  3. boks

    Thread Starter Active Member

    Oct 10, 2008
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    That worked very well. I found the correct answer. After substitution:

    0.8 \cdot 10^{-5}[\frac{273}{16} + \frac{17}{2}cos(kd sin \theta) + 2cos(2kd sin \theta)]

    Is it possible to find how many maxima and minima there are? And how do I calculate the directions of these?
     
  4. steveb

    Senior Member

    Jul 3, 2008
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    The usual technique to find maximum and minimum is to take the derivative and set it equal to zero. The solutions to that equation will be the local maxima and minima. There are different ways to find the direction. One simple way is to just plot the function and then determine by looking. There are more rigorous methods using the second derivative of the function, but since I'm an engineer, I never do that. - I just plot it.
     
  5. boks

    Thread Starter Active Member

    Oct 10, 2008
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    OK, thanks!
     
  6. boks

    Thread Starter Active Member

    Oct 10, 2008
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    OK, so I want \frac{dI}{d \theta} = 0 when I = 0.8 \cdot 10^{-5}[\frac{273}{16} + \frac{17}{2}cos(kd sin \theta) + 2cos(2kd sin \theta)]

    If I've calculated correctly, this means that -8.5sin(8.1sin \theta)8.1cos \theta - 32.4sin(16.2sin \theta)cos \theta = 0 I don't see how to factorise this.
     
  7. steveb

    Senior Member

    Jul 3, 2008
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    First, you can factor out the  cos(\theta) . You will then have the product of  cos(\theta) and another quantity which is a sum of two sin terms. So the product is zero if either quantity is zero. So you can have two types of solutions. Whenever,  cos(\theta) is zero and whenever the other quantity is zero. The first case is easy, but the second case is tricky. You can figure it out with a little logic. If you have trouble figuring it out, then plot the function and note where it equals zero. When you see the final answers in the graph, it will be easier to work back and understand.
     
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