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

enter image description here

Why is it that the maximum fringe contrast is observed when the signal mean is half the amplitude? Based on the equation

$$I_\text{FC} = \dfrac{I_{\text{max}} - I_{\text{min}}}{I_{\text{max}} + I_{\text{min}}} = \dfrac{I_{\text{amp}}}{I_{\text{mean}}},$$

it seems that we maximise the signal amplitude when we minimise the signal mean, and I don't see why the signal mean would have a lower bound of half the amplitude.

The Pointer
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  • You have sinusoids, so the means are halfway between max and min. Then, as per the second figure, the intensity ranges from max down to **zero**, giving maximum contrast. From the first figure, with the big DC offset, the intensity simply goes up and down about that constant offset. So brighter and dimmer about a constant useless level. The contract suffers. – Ed V Sep 11 '21 at 23:45
  • And, in your equation, note that I_FC = 1 if I_min = 0, but I_FC < 1 if I _min > 0, i.e., there is a DC offset. – Ed V Sep 11 '21 at 23:50
  • @EdV Oh, yes, because \$ \sin \$ is maximum at \$ \pi/2 \$? – The Pointer Sep 11 '21 at 23:54
  • The rest of what you said isn't quite clear to me. Can you please elaborate? – The Pointer Sep 11 '21 at 23:56
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    Not sure what you mean by that. But look at your first figure. All three sines have the same mean. But only the biggest sine goes all the way down to zero. It has maximum amplitude equal to the mean, so it swings plus and minus the mean. The other two sines have smaller swings, so their maximum amplitudes are less than the mean. So they cannot get as bright as the big sine and cannot get as dim as the big sine. Hence lower contrast. – Ed V Sep 11 '21 at 23:58
  • @EdV Ohh, ok, yes, I see what you mean. I think I understand now. Thanks for the clarification! – The Pointer Sep 12 '21 at 00:02

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