1

I am currently reading the Fringe Contrast section of this document. This section says the following:

Be sure to use a DC-coupled detector when measuring fringe contrast. Adjust the interferometer alignment to maximize the fringe contrast in the optical signal. Then, use electronic filtering and amplification to maximize the signal for analog-to-digital conversion.

I want to see if I can measure fringe contrast in a basic interferometer setup. However, this explanation seems “high-level” and lacks details. What are the practical details of such a setup?
What specifically is the “dc-coupled detector”, and how does it fit into the setup?
How is the oscilloscope used as part of this setup, and what interferometry “signal” is it capturing?
I am a novice to electronics itself, so I would appreciate a more “basic” explanation.

The Pointer
  • 1,241
  • 1
  • 11
  • 24

1 Answers1

0

enter image description here

Figure 1. From Fundamentals of Light and Interference, page 12.

For whatever reason, to calculate IFC you need to know Imean. This means you need to allow the DC value of your signal through to the measurement circuit.

Also, why does calculating \$ I_{mean} \$ imply that we need to allow the DC value of our signal through the measurement circuit?

If there was no DC component to the signal then the waveform would oscillate about the zero value and \$ I_{max} = I_{amp} \$ and \$ I_{min} = -I_{amp} \$.

That would mean that \$ I_{mean} = \frac 1 2 ( I_{amp} + (-I_{amp})) = 0 \$ which gives you an error.

Transistor
  • 168,990
  • 12
  • 186
  • 385
  • Thanks for the answer. https://en.wikipedia.org/wiki/Interferometric_visibility#Visibility_in_optics shows the derivation of \${\displaystyle \nu ={\frac {I_{\max }-I_{\min }}{I_{\max }+I_{\min }}}}\$. Also, why does calculating \$I_{\text{mean}}\$ imply that we need to allow the DC value of our signal through the measurement circuit? – The Pointer Aug 07 '21 at 02:24
  • 1
    I missed that \$ I_{amp} \$ was only peak and not peak-peak. That explains that. I'll answer the other bit in the question shortly. – Transistor Aug 07 '21 at 02:28