1
$\begingroup$

To better explain my question, I will need to give a brief description of the configuration used in 2D MEMS switches.

So, the next figure shows a configuration of a 2D MEMS switch, a light beam arrives from a fiber to the input port, travels inside the switch until it meets a mirror that is in the standing state to reflect the beam to the output port.

2D MEMS Switch

The output beam from the fiber is wide, therefore, they collimate it using a collimator at the input. This collimator is picked such that the beam waist is minimum at the mirror of the worst case scenario (i.e., longest path).

This path is shown in the next figure.

Analysis of the beam Gaussian divergence

During the analysis of the power loss, there is a loss due to the imperfection in the reflection of the mirror, which ranges from 1% to 3%, and there is a loss due to the Gaussian beam divergence and the existence of the mirror 1d in the beam's path. According to the papers, "The optical signal loss due to Gaussian-beam divergence for a mirror of radius R at distance z is":

loss due to Gaussian beam divergence

Usually, they maintain a ratio R/W(z) of 1.5 to 2.

Note that the beam propagates in the free space from the input port to the mirror and from the mirror to the output port. This is because mirrors other than the mirror 1d are in the sleeping state (i.e., they are out of the beam's path)

My question is:

If we assume a case in which the beam propagates in a similar configuration, however, at the positions of the mirrors from 1a to 1c, there are switching elements with the same positions and dimensions made of glass such that the beam propagates through them.

What would be the impact of these elements from 1a to 1c on the loss due to the Gaussian beam divergence? Would it be the summation of the losses using the provided equation by at different distances, in other words

$L_{Gauss(at\ 1a)}+L_{Gauss(at\ 1b)}+L_{Gauss(at\ 1c)}+L_{Gauss(at\ 1 d)}+...$

If not, then why? and how can I calculate it?

Reference: CY Li et al., "Using 2x2 switching modules to build large 2-D MEMS optical switches"

$\endgroup$
0
$\begingroup$

Wow, this is a very detailed question. Thanks for your effort.

Lets ignore diffraction effects, which will scatter some small amount of extra power out of the laser beam. The loss at elements 1a to 1d will not simply sum up. This is because the power will only be lost at one of the mirrors, and will not be there to be lost at the other mirrors. So, to calculate the power lost by clipping in the whole path, you simply need to calculate the power lost at the mirror/glass where the beam is the largest relative to the mirror/glass. A more accurate calculation of the loss which includes diffraction effects would probably require a computer simulation.

There is another source of loss in the system though. If you use a bare (no optical coating) piece of glass at elements 1a - 1c, then you will lose about 3.5% per interface. With 6 interfaces (front and back of elements 1a - 1c) you will transmit $$ T=(1-0.035)^6=0.81\qquad\Rightarrow\qquad L=19 \%. $$ So you will lose almost 20% of the light to reflectance at the glass interfaces. If you pay for an anti-reflective optical coating you can get reflectivities as a low as 0.1% pretty cheaply in which case you will lose almost nothing to this source of loss.

$\endgroup$
  • $\begingroup$ " to calculate the power lost by clipping in the whole path, you simply need to calculate the power lost at the mirror/glass where the beam is the largest relative to the mirror/glass." Does that mean in this case I need to calculate it at the switching element 1a? $\endgroup$ – BHamza Apr 4 '14 at 18:31
  • $\begingroup$ @BHamza That's correct. $\endgroup$ – Chris Mueller Apr 4 '14 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.