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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.