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Is QE linear in relation to exposure time?


Demonperformer

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An object emits light in two wavelengths, x & y, of equal intensity (same number of photons per second).

I have a camera that has a QE of 70 in wavelength x and 35 in wavelength y.

I also have two filters, one for each wavelength (assume all other criteria for the filters are identical).

To get images of equal brightness, would I simply need to double the exposure time for the image taken with the y filter, or is there some more complicated relationship between exposure time and QE?

Thanks.

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Probably expressed myself badly in the title of the thread.

The point I am trying to get at is that, for a waveband with a lower QE (for a particular chip/camera) I would need to increase the exposure to end up with an image of similar brightness. If the QE is half does the exposure simply have to be doubled, or is it a more complicated formula (and if that is the case, what is it)?

On 19/11/2017 at 10:12, Demonperformer said:

To get images of equal brightness, would I simply need to double the exposure time for the image taken with the y filter, or is there some more complicated relationship between exposure time and QE?

 

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16 minutes ago, Demonperformer said:

If the QE is half does the exposure simply have to be doubled, or is it a more complicated formula (and if that is the case, what is it)?

 

No, you've pretty much nailed it.

And since you can diddle (technical term) with the colour balance in your image processing tool, it doesn't even have to be that accurate.

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Correct. But while you can compensate difference in brightness by stretching more, you don't want to be too far off, because the lower brightness frames also have more noise.

Difference in qe is the reason why zwo designed a set of rgb filters for the asi1600mm. The red filter has a wider passband than the green and blue, and the green has a narrower band than the blue filter, to "equalize fluxes". This way you can have the same exposure time for all filters and end up with equally bright images (if you image a white light source = ceiling).

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