Optics Rotation : Project 4

" WHICH-WAY " INTERFEROMETRY

with Prof. Metcalf

LOSS AND RESTORATION OF INTERFERENCE

Figure 1. Setup

P1, P2, and P3 are linear polarizers.

LOSS

Let b+, d+ and b, d be creation and annihilation operators of two branches. Let H and V be horizontal and vertical polarization. So we choose to have

bHI>= z I>, bVI> = 0 ,
dHI> =0, dVI> = w I>,

Combining the beams gives

a+H = a1b+H + a2d+H

a+V = a1b+V + a2d+V

aH I> = a1zI>, aVI> = a2w I>,

So if combined light is observed we get

<a+HaH + a+VaV> = Ia1z I2 + Ia2wI2.

There is no interference!

If we look at Quantum mechanics we see that we got right answer because we can distinguish two beams (paths) by their polarization.

RESTORATION

But if we now put the light in the combined beam through a plane polarizing filter at 45 deg between horizontal and vertical so that we have

a+P a+Ha+V.

The intensity of the observed light is proportional to

<a+PaP> = 1/2Ia1z + a2wI2.

Interference is back!

Again we have no problem with Quantum mechanics because we cannot distinguish light from the two branches using polarization.

MEASUREMENTS

I set everything as shown on Fig.1. and for each degree of polarization I made picture and later I plotted Intensity vs Position across the beam. I used IDL to plot profiles.

Table 1. Pictures of the screen and plots for two cases together

 A) two beams have perpendicular polarization without P3 B)  two beams have perpendicular polarization  with P3 at 0 deg

Plots show intensity vs. position (in pixels) along the line shown in the second row.

From these graphs I measured visibility

1. P1 was set at 45 deg, and P3 was removed, so what is measured is visibility vs. polarization angle of P1. Intensity of light was to strong to see fringes so I attenuated beam with two more polarizers.

2. Now I set P2 so that the visibility of fringes was minimal. What was measured is visibility vs. polarization of P3 with attenuated beam - now the intensity was very low but I left the beam attenuated just for comparison with case above.

3.  For this measurement I removed attenuation. Again P1 and P2 were at 45 deg and I measured visibility vs. polarization angle of P3.

Conclusion:

The two perpendicular polarizations are tags on the light going through the two paths. When you observe all the light, you can tell which way by looking at its polarization. Then you see no interference. The relative phase between two parts is there but it has no affect on the total intensity. On the other hand, it does effect light polarized at 45 deg from both paths polarizations angles. When you look at that, you see interference. But than you cannot tell which polarization light had before, so you cannot tell which path it took.