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Splitting and combining EM waves & amplitude/intensity

  1. May 24, 2015 #1
    I am reading through my course guide and problems worked out and something bothers me.
    I think at this point we are working with classical waves, moving towards QM eventually.

    The idealization we work with is this. We have a light source creating a wave [itex] ψ(x,t)=Acos(ωt+kx)[/itex] .

    The light beam hits a beam splitter/partially reflective mirror. Half the intensity is reflected, half the intensity is transmitted. Say we do this 3 times to get 4 beams.

    As [itex]I = A^2[/itex] , so the amplitude of these waves is A/2 and we get four waves of [tex] ψ(x,t)=\frac{A}{2}cos(ωt+kx+Φ) [/tex]

    Now, for combining waves, one is supposed to be able to add them. If the Φ term is made exactly the same, they should be completely identical. One should get [tex] ψ(x,t)=4(\frac{A}{2}cos(ωt+kx+Φ)) [/tex] .

    So now we have 2 times the original intensity. This is asymmetrical and something must be wrong somewhere. I don't know if this mixes different idealizations, say QM light as a photon phenomena and classical EM waves phenomena.

    Is it somehow impossible to get 100% constructive interference after adding these waves? I know QM light isn't described by a simple sinusoidal wave. I always thought the issue of getting 4 times the intensity with interference is that you get 0 intensity at other spots, so energy is conserved. But there are no plane waves or spherical waves here.
    The solution to some problems given clearly show that it adds two beams of 1/4th intensity to get a beam with equal intensity and amplitude to the original beam.

    Shouldn't waves be recombined the same was as they are split? So either half the intensity and double the intensity, OR, half the amplitude and double the amplitude.
  2. jcsd
  3. May 24, 2015 #2


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    And how would you combine the four beams so they end up on the same axis ? Ah, with beam splitters!
    (to be honest, This answer I found with our colleagues).
    Nice treatise here. Comparable, but with a transition to QM here
  4. May 24, 2015 #3
    Wave equation for a beam mean polarization light.
  5. May 25, 2015 #4
    So the answer is that for a beam splitter light will leave each output port in exactly that ratio as to create or destroy no energy?

    And a mirror will just refuse to reflect all the sudden (as it is a splitter with one output blocked, ie anything that goes that way will become heat)?

    Is this a trick to solve the math or a genuine quantum effect described classically?
  6. May 25, 2015 #5
    Polarization light reflection have different reflection rate for different polarization angle. The power ration refers to a statistical combination of large number of waves with total zero mean polarization. I don't know how you can recombine all these waves.
  7. May 25, 2015 #6
    Polarized light isn't part of this course. It is a biophysics course so we ignore the magnetic field for simplicity.

    I don't know how your objection links in with this issue. Is it a mere inaccuracy? Does it resolve the issue completely. You being cryptic doesn't help. Is the question not clear? Is it an issue of bad idealizations? This isn't helping. In the end light is QM phenomena and this isn't the QM board, so I don't ask about real light. I ask about the limitations I am given by the course material. Are these very unusual? I know you can't guess the limitations I am working under, but like I said I don't know if they are usual or unusual or if they aren't clear.

    I don't know how to set up the limitations right, as I am too ignorant of those. All I need to do is follow 3 weeks of intro to QM for biophysics and ace the test, like I usually do. But this irks me.

    Let me rephrase. I am told I have to split waves with half intensity and add waves mathematically. Ignore that all this tries to model real light, because the model fails as Maxwell equations aren't even part of this course.

    The problem sets are set up in such a way that energy is conserved, somehow by luck. But we must follow the rule that waves are split with half intensity or amplitude over √2 and that we add the waves to superimpose them. This is not symmetrical.
    I feel I can't move on as long as this irks me. I will ask my teacher tomorrow.

    I don't have time to learn all of optics, all of waves and all of classical EM in a 3 week rushed intro course for biophysicists. The test is one day after the last class.
  8. May 25, 2015 #7
    I'm not cryptic. If I give you the solution, my post will deleted by admin. I said that this problem is not real. Is something to do for something reason. Your teacher can help you much better because he knows the assumptions for this problem.
  9. May 25, 2015 #8
    Yeah it was Saturday, Sunday and white Money. I wasted my time here, I guess.

    Ill ask him tomorrow.
  10. May 25, 2015 #9


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    Looks like you did. If you had checked out the links I gave you you would have found out easily.

    For the simpler case of one splitter a beam E1 into rE1 and tE1 and a recombination in a second beam splitter, one beam would have (r2 + tt')E1 and the other (rt + tr')E1.

    As shown in the link, the standard (ideal, theoretical) 50-50 beam splitter has $$ \begin{pmatrix} t' & r \\ r' & t \end {pmatrix} = {1\over \sqrt 2} \begin{pmatrix} 1& j\\ j& 1 \end {pmatrix}
    $$ leading to beam one having amplitude 0 E1 and beam 2 having amplitude ## {1\over 2} (2j)## E1
  11. May 25, 2015 #10
    Actually, I found those documents before I posted here. I don't see how they help. Maybe I am too stupid.
  12. May 26, 2015 #11


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    Temper, temper...
    The link taught (*) me what r, t, r' and t' are and why one of the outgoing beams has amplitude 0 (for proper optical paths)


    (the reflections from the simple mirrors and the optical paths are equal for both routes, so they introduce a common (complex) factor with norm 1)

    (*) yes, I'm learning this too, so I'm grateful for your thread...

    Last edited: May 26, 2015
  13. May 26, 2015 #12


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    (Temper temper was referring to the "Maybe I'm too ...". You have me convinced you are not. )
    I re-read posts and your post #4 is now easily adressed:
    No, a mirror simply has r = 1, t = 0 and we don't care about the r'and t', they can be taken to be 0. No cheating trick (and room for a phase jump as well: no problems as long as |r| remains 1.

    Key to this whole treatment is western michigan link eqn 4 through 9. Intensity conservation is pretty high in the pecking order, and the treatment makes sure it is. Same with probability conservation in QM (barring absorption, transitions etc; we are still talking ideal splitters/mirrors). So you can rest assured the analogue treatment in the rochester link is realistic. I like their (15) where you see that non-equal paths introduce a relative phase difference ##\phi## and the waves come out as described above for ##\phi = 0## and out of the other one for ##\phi = \pi##. For all ##\phi## you have that ##\sum E^2## is conserved. As it should be.

    I really like this (have done some interferometry a long time ago, but this way of dealing with it is also new to me). Thanks again for bringing it up !
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