1
\$\begingroup\$

I have a friend who took some macro photos of snowflakes with the WB preset to "Cloudy", but lighted from an 11-watt flourescent bulb. Snowflakes look more pleasing with a blue tint, rather than "warm", so to correct for it, the WB slider had to be slid all the way to the left to make it look good. The question I have is, would such a drastic change in WB in any way contribute to the "noisiness" of the photo? I would have been inclined to say no, but I found an article at Red.com which seems to indicate otherwise. To quote:

Ultimately, any white balance change amplifies some colors compared to others. Extreme deviations from daylight can therefore affect image noise in a similar way to higher ISO speeds. This is particularly true with neutrally balanced candle or tungsten lighting, since white balancing amplifies cooler colors compared to warmer ones. This disadvantage can be offset by using a warming or cooling filter directly in front of the lens, or gels with studio lighting. However, these often also end up reducing available light—and progressively more so depending on the strength of the gel or filter.

(emphasis mine)

What got me to thinking about it, was the "known" fact that WB changes to a RAW file are completely non-destructive. So if changing the WB does indeed affect noise, then shooting a scene with "extreme deviations from daylight" will likely affect the the noise/grain in the photo; whether the camera sets the WB, or I change it manually in Adobe Lightroom or similar software.

Does this sound reasonable? Or is there something simple that I'm missing?

Here's a related question, but it doesn't really answer either. Does the camera white balance setting affect the raw image at all?

\$\endgroup\$
0

2 Answers 2

5
\$\begingroup\$

Just because using different WB settings to demosaic the raw data is non-destructive to the raw data doesn't mean one WB setting will be more or less noisy than another. It's not so much that you will get equal quality regardless of how you choose to interpret the raw data, but rather that regardless of whatever interpretation you choose to use for the raw data, that interpretation does not change the raw data contained in the file - it only changes the way an image created from that data looks.

Beyond that, a raw file has no white balance. The raw data is a set of monochrome luminance values. When the raw data is demosaiced a WB profile is applied to the raw data, but there is no WB information contained in the actual raw data that was read out from the sensor. Whatever WB was selected in the camera at the time the photo was taken has absolutely no effect on the readout from the sensor data, it is only used to apply a WB to the raw data to produce the thumbnail/preview jpeg image attached to the raw file. It is also included in the file's metadata, but the application you use to open the raw file with may or may not apply the in camera WB setting. Many raw converters ignore the in-camera WB setting and either apply their own default value, apply a value included in a preset created by the user, or apply their own version of Auto WB.

\$\endgroup\$
1
  • \$\begingroup\$ Thanks a lot, I guess that pretty much answers my question! So that's why it doesn't make any difference whether the WB is set in camera or in post. (For RAW files that is...) \$\endgroup\$
    – ezmopho
    Commented May 11, 2016 at 1:16
0
\$\begingroup\$

Extreme deviations <= Emphasis mine. Any extreme manipulation will.

A "light" manipulation on a raw file won't. That is why you are working with a raw file in the first place, to have room to play with, including the white balance.


was the "known" fact that WB changes to a RAW file are completely non-destructive.

Yes, all the "changes" in a RAW file, with the proper software are non destructive.

The RAW file stays untoched. You simply prepare instructions to generate another file. A 16 bit psd for example or a JPG file.

An extreme situation could be using sodium lamps for a product shoot. That would be not a good light to do that.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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