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I have been studying interference of light waves for a while now and I have a doubt.

I have never seen any sources (books, internet , teachers) ever take into consideration about the "actual" physical angle between Electric fields of interfering waves (or equivalently that between the Magnetic fields of interfering waves) other than the phase difference.

  1. So do we (books , teachers , Physicists) take into account this angle, in general?
  2. In case they (books ,teachers , Physicists) do ignore this angle, what are the corresponding assumptions?

P.S. Any source dealing with the examples on the same would be highly appreciated.

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  • $\begingroup$ I’m not sure I understand...light always has E and B perpendicular. Even if it interferes. $\endgroup$
    – joseph h
    Commented Nov 18, 2020 at 5:49
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    $\begingroup$ @Drjh I meant angle between electric fields of two different waves. $\endgroup$
    – Tony Stark
    Commented Nov 18, 2020 at 5:51
  • $\begingroup$ Do you mean two beams polarised at different angles? Or do you means the angle between the two rays? $\endgroup$ Commented Nov 18, 2020 at 6:43
  • $\begingroup$ @JohnRennie Probably angle between two rays. $\endgroup$
    – Tony Stark
    Commented Nov 18, 2020 at 6:52
  • $\begingroup$ Please consider specifying who "we" or "they" are. "Do we take ..." cannot be answered without knowing who "we" are. $\endgroup$
    – my2cts
    Commented Nov 18, 2020 at 9:39

2 Answers 2

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The angle between the light rays is irrelevant. To calculate the interference pattern we are just calculating the electric field vector at a point in space.

We consider some point $(x,y,z)$ in space, and at this point the electric fields of the two rays will be $\mathbf E_1(x,y,z) \sin(\omega t + \phi_1)$ and $\mathbf E_2(x,y,z) \sin(\omega t + \phi_2)$. We simply vector add these to find the resultant field at this point. We do not need to know where the rays have come from, or where they are go to. All we need to know is their value at the point.

Note that we tend to assume the two rays are polarised in the same plane so the vectors $\mathbf E_1$ and $\mathbf E_2$ are parallel, or that the rays are unpolarised so they have equal intensities at all polarisation angles. If we were dealing with rays polarised at different angles this would affect the intensity since the magnitude of the vector sum depends on the angle between the vectors.

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  • $\begingroup$ "we tend to assume the two rays are polarised in the same plane" -- is this some sort of mathematical approximation (such as small angles approximation sin x = x) or we simply make an "ad havoc" assumption to simplify things? $\endgroup$
    – Tony Stark
    Commented Nov 18, 2020 at 8:21
  • $\begingroup$ We simply make an "ad hoc" assumption to simplify things. Did you really mean ad havoc? I feel as though I have seen students making ad havoc assumptions on many occasions :-) $\endgroup$ Commented Nov 18, 2020 at 8:46
  • $\begingroup$ I mean "ad havoc" covers a broad spectrum of my thoughts. $\endgroup$
    – Tony Stark
    Commented Nov 18, 2020 at 11:32
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This is a good question. In the Fraunhofer limit polarisation does not matter but when non parallel interfere polarisation will degrade contrast. It is caused by the polarisation direction parallel to the plane (TM) of incidence, formed by the two k vectors, as opposed to polarisation perpendicular to the plane (TE). Consider two waves propagating at an angle of 90 degrees. For TE the vector sum of E (or B) varies with phase between $E_1 \pm E_2$. For TM the sum is always $\sqrt{E_1^2+E_2^2}$, thus there is no interference at all. For the extreme case of 180 degrees difference both TE and TM show full interference but the two TM waves due to this angle are effectively 180 degrees out of phase and TE interference minima coincide with TM maxima and vice versa. For unpolarised light this will wash out the image contrast. In general there is contrast loss if the light is not purely TE polarised.

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