# Will two coherent sources of orthogonal polarization angle form the known interference pattern or there will be difference?

if i have a laser beam then i split it into two beams with ratio 1 : 1 then i take one of them to change its polarization angle (90) degree by passing it through an optically active substance then we make an interference between them both, does the interference pattern change what i really want to understand in the above problem is.. when the two beams are orthogonally polarized when they meet, the two electric fields (of the light beams) will be perpendicular to each other and so the magnetic fields, how can the electric field interfere with the magnetic field?

There will be an interference pattern, but not a visible interference pattern.

This is easily demonstrated: place a polarizing filter in the region where the two beams overlap, and a visible fringe pattern will appear downstream within that region.

If a polarization-sensitive recording material is placed in the region of overlap, a hologram will be recorded in the form of planes of equal birefringence -- instead of the usual form which is planes of equal refractive index or of equal absorption.

In a typical double slit experiment set up, you will probably not see an interference pattern, the intensity you see on the screen will be roughly the sum of the intensities due to individual beams.

The details of the intensity pattern on a screen will depend on your setup: how far away you are from the slits, their spacing, even what the slits are made of if your screen is very close to the slits.

The intensity of on the screen is proportional to the component of the average Poynting vector normal to the screen. For harmonic waves, the Poynting vector is

$$\vec{S} =\frac{1}{2} \vec{E} \times \vec{H}^* = \frac{1}{2} (\vec{E}_1 + \vec{E}_2) \times (\vec{H}^*_1 + \vec{H}^*_2) = \frac{1}{2} \vec{E}_1 \times \vec{H}^*_1 + \frac{1}{2} \vec{E}_2 \times \vec{H}^*_2 + \frac{1}{2} \vec{E}_1 \times \vec{H}^*_2 + \frac{1}{2} \vec{E}_2 \times \vec{H}^*_1$$ $$= \vec{S}_1 + \vec{S}_2 + \frac{1}{2} \vec{E}_1 \times \vec{H}^*_2 + \frac{1}{2} \vec{E}_2 \times \vec{H}^*_1$$

where $$\vec{S}_1$$ and $$\vec{S}_2$$ are the Poynting vectors due to individual slits. The cross terms are the tricky part.

If the screen is far away from the slits, then the waves are very well approximated by TEM waves, meaning $$\vec{E}$$ and $$\vec{H}$$ due to each source are orthogonal. For each source, we can write $$\vec{H}=\frac{1}{\eta}(\hat{n}\times\vec{E})$$ where $$\eta = \sqrt{\mu/\epsilon}$$ is known as the wave impedance of the propagation medium and $$\hat{n}$$ is the unit normal vector along the direction of propagation. Using vector identities, the first cross term is $$\frac{1}{2} \vec{E}_1 \times \vec{H}^*_2 = \frac{1}{2\eta^*} \vec{E}_1 \times (\hat{n}_2\times\vec{E}_2^*) = \frac{1}{2\eta^*} (\vec{E}_1·\vec{E}_2^*) \hat{n}_2 + \frac{1}{2\eta^*} (\vec{E}_1·\hat{n}_2) \vec{E}_2^*.$$

The first term is zero since the E-fields are orthogonal. Adding the second cross term, the total Poynting vector is $$\vec{S}=\vec{S}_1 + \vec{S}_2 + \frac{1}{2\eta^*} (\vec{E}_1·\hat{n}_2) \vec{E}_2^* + \frac{1}{2\eta^*} (\vec{E}_2·\hat{n}_1) \vec{E}_1^*$$

If $$\hat{N}$$ is the unit vector normal to the screen (pointing into the screen), the brightness on the screen is proportional to $$\textrm{Re}(\vec{S}·\hat{N}) = \textrm{Re}(\vec{S}_1·\hat{N}) + \textrm{Re}(\vec{S}_2·\hat{N}) + \textrm{Re}\left[\frac{1}{2\eta^*} (\vec{E}_1·\hat{n}_2) (\vec{E}_2·\hat{N})^* + \frac{1}{2\eta^*} (\vec{E}_2·\hat{n}_1) (\vec{E}_1·\hat{N})^*\right].$$

For a distant screen, near the center of the intensity pattern, the unit normals $$\hat{n}_1$$ and $$\hat{n}_2$$ are approximately the same, with the angle between them roughly equal to the angular separation of the two slits as seen from the screen. This means the factors $$\vec{E}_1·\hat{n}_2$$ and $$\vec{E}_2·\hat{n}_1$$ are close to $$\vec{E}_1·\hat{n}_1 = \vec{E}_2·\hat{n}_2 = 0$$, so the dot products contribute a factor no more than $$|\sin(\Delta\theta)|$$ where $$\Delta\theta$$ is this angular separation. Furthermore, the E-fields are roughly orthogonal to the screen, with the angle between the E-fields and $$\hat{N}$$ no more different from $$90^\circ$$ than $$90^\circ \pm \theta$$, where $$\theta$$ is the angular distance of the point on the screen from the center of the screen, as viewed from the slits, so these dot products contribute a factor no more than $$|\sin(\theta)|$$. Consequently, the intensity pattern can be expressed as

$$I=I_1 + I_2 + I_{cross},$$ with $$|I_{cross}| \le 2\sqrt{I_1 I_2} |\sin(\theta)| |\sin(\Delta\theta)|.$$

In most double slit setups, $$|\sin(\Delta\theta)\sin(\theta)|$$ will be so small that you can safely assume you will not see an interference pattern.

• Its not clear whether the question refers to double slit or possibly some kind of interferometer in her/his setup. We know that in the EM field, EM waves must superimpose as observations show beams do not "interfere" with each other when crossed for example. Jul 9 '19 at 15:16
• In the classical world we say there is "interference" when we see a dark (or bright spot) on a screen, but we also know that interference in the dark spots would be a violation of energy conservation. There are mounds and mounds of papers thru the early 1900s and education that tries to explain this classically, but it is not until we understand the wave function of light that we realize dark spots are where there is no energy or no photons as pathways here are not allowed. Jul 9 '19 at 15:18

No, in this situation there will be no interference pattern. Changing the relative phase between those two polarizations (which is what happens when you look at different points on the screen) will change the resulting polarization from diagonal to circular to antidiagonal and back to circular, but it won't affect the intensity.

In the typical 2 slit interference experiment the interference is old classical thinking. The modern explanation is not interference but that the pattern is formed by allowed paths ( n wavelength multiples) the dark areas have no photons present and the bright spots have many photons. Thus polarizing the beam has no effect.

• Classical electrodynamics is certainly capable of predicting/explaining the outcome of such an experiment, except possibly at extremely low intensities. The claim that changing the polarization of one of the beams has no effect on the outcome is also incorrect, whether you use quantum mechanics or not. Photons are "polarized": in this case the optical mode of interest that photons occupy is the one in the form of the superposition of two cross-polarized fields. This is a different problem than if these fields had the same polarization.
– Puk
Jul 8 '19 at 4:23
• The question itself is a little confusing as written, my interpretation is to take the simple double slit experiment and illuminate it with polarized light and then add a second polarized source at 90 degrees to the first of equal intensity. There will be no observed difference in the pattern accept that it is twice as bright. Jul 9 '19 at 15:05