The trajectory of the light

Due to the angle $\gamma$ between the two mirrors, the light will change the angle it makes with the centerline after each reflection. Since we couple the laser in at the wide side of the collimator (the back side), the angle the light makes with the centerline of the collimator will decrease after every reflection with one of the mirrors. Coupling in at the wide side of the collimator has the advantage that the total number of reflections is larger than when the light is coupled in at the narrow side of the collimator (the front side). The incoming angle is $\alpha_0$, the angle at point $z$ in the collimator is given by:

$$\alpha(z) = \alpha_0 - n(z)\gamma$$ (1.9)

with $n(z)$ the number of times the laser beam has been reflected at point $z$ in the collimator. The distance $z$ can be expressed as a function of $n(z)$ and $\alpha_0$:

$$z = D\tan(\alpha_0) + D\tan(\alpha_0 - \gamma) + D\tan(\alpha_0 - 2\gamma) + ... + D\tan(\alpha_0 - n(z)\gamma)$$ (1.10)

with D the distance between the two mirrors. Since the angle $\alpha_0$ is small, we can take $\tan(\alpha_0)$ to be $\alpha_0$:

$$z = D\alpha_0 + D(\alpha_0 - \gamma) + D(\alpha_0 - 2\gamma) + ... + D(\alpha_0 - n(z)\gamma)$$ (1.11)

Rearranging this formula gives:

$$n(z) = \frac {\alpha_0 +1/2 \gamma}{\gamma} + \frac{1}{\gamma}\sqrt{1/4 \gamma^2 - \gamma \alpha_0 + \alpha_0^2 - 2 \gamma \frac {z}{D}}$$ (1.12)

Since $\gamma$ is smaller than $\alpha_0$ by a factor of 100, this can be simplified to:

$$n(z) = \frac{\alpha_0}{\gamma} + \frac{1}{\gamma}\sqrt{\alpha_0^2 - 2\gamma\frac{z}{D}}$$ (1.13)

If we replace $z$ by the length $L$ of the collimator, we get the total number of reflections $N_{\rm total}$ in the collimator. Substituting Eq. 1.13 in Eq. 1.11 gives:

$$\alpha(z) = \sqrt{\alpha_0^2 - 2 \gamma\frac{z}{D}}$$ (1.14)

This formula also shows that there is a minimum for $\alpha_0$ for which the laser light will reach the end of the collimator. If the angle is smaller than this value, the laser light turns around in the collimator, and will come out at the same side it was coupled in. This minimum value of $\alpha_0$ is:

$$\alpha_{0_{min}} = \sqrt{2\gamma\frac{L}{D}}$$ (1.15)

The angle $\alpha_0$ should be bigger than this minimum value, since it is not wanted that the laser light turns around in the collimator.

Another restriction for the angle $\alpha_0$ is the fact that the laser light, with a width of 0.5 cm, has to fall on the coating of the mirror without being blocked by the first mirror. Since the two mirrors are separated by 6 cm, and the coating stops at 3 mm from the edge of one of the mirrors, the minimum angle needed is $\alpha_0 = 0.133$ rad.