 Polarization-dependent visibility

ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2018/section-b/b2/polarization-dependent-visibility

Let the two arms deliver intensities

I_{1}, I_{2}

e_{1}, e_{2}

. For a perfectly coherent input, the recombined intensity under relative phase

ϕ

I (ϕ) = I_{1} + I_{2} + 2 I_{1} I_{2} Re [e_{1}^{†} e_{2} e^{i ϕ}] .

V = \frac{I _{m a x} - I _{m i n}}{I _{m a x} + I _{m i n}} = \frac{2 I _{1} I _{2}}{I _{1} + I _{2}} ∣ e_{1}^{†} e_{2} ∣.

For equal returned intensities, two linear polarizations at angle

θ

have overlap

cos θ

and therefore

V_{linear, linear} = ∣ cos θ ∣.

Choose one linear direction as

e_{L} = (1, 0)^{T}

e_{C} = (1, \pm i)^{T} / 2

. Their overlap has magnitude

1/ 2

V_{linear, circular} = \frac{1}{2} .

A non-polarizing beam splitter does not by itself guarantee equal arm losses; if

I_{1} \neq = I_{2}

, multiply either boxed polarization factor by

2 I_{1} I_{2} / (I_{1} + I_{2})

. Perfect temporal coherence does not compensate for polarization or intensity imbalance.

 New to topics? Read the docs here!