# Simulating off-axis beams: calculating the shift¶

Due to its use of the paraxial approximation, simulating an off-axis beam in Finesse requires the inclusion of higher order modes, see e.g. Section 9.2 in . In order to study such a misaligned beam in Finesse we can use a telescopic setup such as depicted in Fig. 11 with two mirrors (m) or beamsplitters (bs) and use the xbeta parameter to add a small mismatch angle $$\beta$$ to the angle of incidence of both beamsplitters (or mirrors).

We will derive here the shift in the position of the beam as a function of (among others) this angle $$\beta$$. In addition we will also derive the effect of this parameter on the phase of the beam, resulting from the changed path length between the laser and the detector. Both formulae will be verified using Finesse simulations in example Aligning and translating a beam with steering mirrors.

In the next section we will discuss the required number of higher order modes to be included and look at the effects of changing the beam parameters such as waist size and waist position.

## Introduction¶

In Fig. 12 we have drawn in detail the two different situations: with and without the tilt $$\beta$$.

The dark red part is the “untilted beam”, which has an incident angle $$\alpha$$ on both of the untilted (dark blue) beamsplitters and traverses a distance $$s$$ between the beamsplitters. After rotating both beamsplitters over the small angle $$\beta$$ (light blue) the beam follows the light red path with corresponding incident angle $$\alpha + \beta$$ and traverses now a distance $$\tilde{s}$$ between the beamsplitters. The tilted beam hits the second beamsplitter at a distance $$d$$ away from where the untilted beam hits it, which results in the relative vertical shift $$\Delta$$ of the position of the outgoing beam for which we derive Eq. (10) below

$\Delta = s \cdot \sin 2\beta$

The total difference in length between the light and dark red paths translates into a relative phaseshift $$\delta\varphi$$ when detecting the beam at the right of the telescope, which satisfies Eq. (18) below

$\delta \varphi = \frac{2 s \cdot \sin^2 \beta}{\lambda} \cdot 360^\circ$

## Calculation of shift in position $$\Delta$$¶

Using the sine rule for the triangle enclosed by $$d$$, $$s$$ and $$\tilde{s}$$

(6)$\frac{d}{\sin 2\beta} = \frac{s}{\sin \gamma} = \frac{\tilde{s}}{\sin(90^\circ+\alpha-\beta)},$

we can express the length $$d$$ as

(7)$d = s \cdot \frac{\sin 2\beta}{\sin \gamma}.$

Furthermore, from the total angle of that same triangle we can obtain from

$(90^\circ-\alpha-\beta+2\alpha) + 2\beta + \gamma = 180^\circ,$

the angle $$\gamma$$

(8)$\gamma = 90^\circ - \alpha - \beta.$

We need to calculate $$\Delta$$:

(9)$\Delta = d \cdot \sin (90^\circ-\alpha-\beta) = d \cdot \sin \gamma,$

where we used eq. (8). Inserting Eq. (7) we then obtain the shift

(10)$\Delta = s \cdot \sin 2\beta$

We see that all $$\alpha$$ dependence cancels out. Furthermore, we see that the shift increases linearly with the distance between the two beamsplitters. Since for larger values of $$\beta$$ the paraxial approximation requires more and more higher modes to be included in a simulation, increasing $$s$$ instead of $$\beta$$ could help simulating the same shift $$\Delta$$. This will be discussed further below.

## Calculation of difference in path length and $$\delta\varphi$$¶

Note that there are two parts contributing to the difference in path length:

1. $$p_1$$: the difference between $$\tilde{s}$$ and $$s$$

2. $$p_2$$: a horizontal part since we hit the beamsplitter further to the right

$$p_1$$ will have a positive contribution, while $$p_2$$ will have a negative contribution to the change in path length.

### calculation p1¶

Noting that

(11)$\sin \gamma = \sin(90^\circ - \alpha - \beta) = \cos(\alpha + \beta)$

we can obtain $$\tilde{s}$$ from the sine rule Eq. (6) to get

(12)$\tilde{s} = s \cdot \frac{\sin(90^\circ + \alpha - \beta)}{\sin \gamma} = s \cdot \frac{\cos(\alpha - \beta)}{\cos(\alpha + \beta)}$

so

(13)$p_1 = \tilde{s} - s = s \cdot \frac{\cos(\alpha - \beta) - \cos(\alpha + \beta)}{\cos(\alpha + \beta)}$

### calculation p2¶

Since part $$p_2$$ shortens the total path, we add an overall minus sign:

(14)$p_2 = - d \cdot \cos(90^\circ-\alpha-\beta) = -d \cdot \sin(\alpha + \beta)$

Using eqs. (7) and (11) this becomes

(15)$p_2 = - s \cdot \frac{\sin (\alpha + \beta) \sin 2\beta}{\cos(\alpha + \beta)}$

### combining the two¶

Combining eqs. (13) and (15) we get

(16)$s \cdot \frac{\cos (\alpha - \beta) - \cos (\alpha + \beta) - \sin (\alpha + \beta) \sin 2\beta}{\cos (\alpha + \beta)}$

We work out the numerator:

\begin{aligned} \text{numerator} &= \cos (\alpha - \beta) - \cos (\alpha + \beta) - \sin (\alpha + \beta) \sin 2\beta \\ &= \cos\alpha \cos\beta + \sin\alpha \sin\beta - \cos\alpha \cos\beta + \sin\alpha \sin\beta - \\ & \qquad \qquad (\sin\alpha \cos\beta + \cos\alpha \sin\beta) 2 \sin\beta \cos\beta \\ &= 2 \sin\alpha \sin\beta -2 (\sin\alpha \sin\beta \cos^2\beta + \cos\alpha \cos\beta \sin^2\beta) \\ &= 2 \sin\alpha \sin\beta ( 1 - \cos^2\beta) - 2 \cos\alpha \cos\beta \sin^2\beta \\ &= 2 (\sin\alpha \sin\beta - \cos\alpha \cos\beta) \sin^2\beta \\ &= -2 \cos(\alpha + \beta) \sin^2 \beta \end{aligned}

Putting this into eq. (16) we see that the $$\cos(\alpha+\beta)$$ factor cancels out and we find the total path difference

(17)$p_1 + p_2 = -2 s \cdot \sin^2 \beta$

Note that again the $$\alpha$$ dependence cancels out. The relative phase shift can be calculated by dividing the path difference Eq. (17) by the wavelength $$\lambda$$ and multiplying by $$360^\circ$$. We also pick up an extra overall minus sign, cancelling the one from Eq. (17), since we define a plane wave as $$\cos(\omega t - k x + \varphi)$$, see for example Eq. (1.4) in . Putting it together we get

(18)$\delta \varphi = \frac{2 s \cdot \sin^2 \beta}{\lambda} \cdot 360^\circ$

## Summary¶

• Both the shift of the beam eq. (10) and the change in phase eq. (18) are independent of the angle $$\alpha$$

• Given the square in eq. (17) the path length difference is typically small but not per se compared to the wavelength $$\lambda$$, leading to a significant phase shift.

## Effect on a finite size beam¶

The above calculation was done for an infinitesimally narrow beam hitting the beamsplitter exactly in the midpoint (its rotation point). We show here that the result is the same for a beam hitting the beamsplitter off-axis.

• this does not affect the shift $$\Delta$$ of the beam on the beam detector.
• this does not affect the total path length difference and hence $$\delta\varphi$$ since the dark red beam gains a bit on the left but loses the same amount on the right.