Setup 2: Theory

Consider the superposition of two beams of light with identical frequency but different polarization and the resulting signal being picked up by the detectors.

Assume the first beam, A, is linearly polarized and travels along the z-axis. Its electric field components along x- and y-direction can be represented by a pair of equations:

A_x=|A| Cos(wt+D)cos(R)

A_y=|A| Cos(wt+D)sin(R)

where w is the frequency of the light, R is the angle from the x-axis at which the light is polarized, and D is the initial phase of the light.

A second beam, B, is circularly polarized. Its electric field components along x- and y-direction can be represented by the following pair of equations:

B_x =|B| Cos(wt+Z)

B_y =|B| Sin(wt+Z)

where Z represents the initial phase of the circularly polarized beam and w its frequency.

Given that the absolute phase of the two light beams is unimportant, and that only the phase difference between the two matters, D, will be set to be 0.

S_x and S_y represent the sum of the electric fields of the two beams in the x-, respectively, y-direction:

S_x = A_x+ B_x

S_y = A_y+ B_y

The two detectors will each produce a voltage proportional to the intensity, I, of the electric field along the x- or the y-axis and it corresponds to the average of electric field squared, <E²>:

Vx µ I_x = <E_x²> = <S_x²>

V_y µ I_y = <E_y²> = <S_y²>

Specifically I = <E²> = (1/2 pi) Integral of E² with wt from 0 to 2 pi.

Combining and evaluating all the equation this yields:

Vx µ I_x = (1/2) Cos²(R) [ |A|² + |B|² +2 |A| |B| Cos(Z) ]

Vx µ I_y = (1/2) Sin²(R) [ |A|² + |B|² +2 |A| |B| Sin(Z) ]

Two important terms have been highlighted in red. They represent the parametrized form of an ellipse centered on the origin:

X=A Cos(t)

Y=B Sin(t)

If R, the rotation angle of the linearly polarized beam A remains constant, then plotting V_y against V_x with changing phase Z traces out an ellipse.

Special Case:

Assume the following:

a) the rotation angle R is fixed at 45 degrees,

b) the amplitude of beam A is identical to B: |A| = |B|

c) and finally that the phase angle Z changes at a constant rate of either v₀ t or -v₀ t which happens when the distance beam B travels increases (or decreases) at a constant rate.

In this case, the expressions above simplify to:

V_x µ I_x = (1/2) |A|² [ |1 + Cos(v₀ t) ]

V_y µ I_y = (1/2) |A|² [ 1 + Sin(v₀ t) ]

(A graph of V_x vs. V_y is shown at the very end of this page.)

If v₀ is positive, then V_x vs. V_y traces out a circle in the counter clockwise direction; if it is negative, it is in the clockwise direction.