2
\$\begingroup\$

I took two pictures of two different flowers with portrait setting. One came perfectly exposed with dark background, the other seems over exposed with brighter background. I tried it again and the same result! Does the color of the object make the camera select wrong exposure? If yes, what is correct technical term for it. How can I fix such photos. Apparently problem is with non white colors only.

enter image description here

Portrait setting Nikon D5100 (the background is nice dark. In reality it was not dark at all)

enter image description here

Portrait setting Nikon D5100 (background is bright, more correct to what really it was but the flower seems to be over exposed. The focus was on flower)

\$\endgroup\$
2
  • \$\begingroup\$ Btw I made similar experiment with other plants and came across the same issue with non white plants. The portrait mode over exposed the flower color (well the color was lighter than what it should have been for a lightly violet flower), I fixed that in manual mode but then the surrounding was a little darker than it actually was. \$\endgroup\$
    – TheTechGuy
    Commented Sep 9, 2012 at 22:09
  • \$\begingroup\$ If anything, the red flower seems correctly exposed (although quite highly saturated) and the white flower is a little underexposed in the background \$\endgroup\$
    – Dreamager
    Commented Sep 10, 2012 at 9:38

2 Answers 2

5
\$\begingroup\$

When your camera adjusts your settings (aperture, shutter speed, and/or ISO) to get a certain exposure value it is basing those decisions on the tone of the image (different from color).

It could be any color, but the tone should average out to be 18% grey (thus the use of grey cards that are 18% grey). So if you had a bright white vs a dark white you would experience the same issues as you would bright green vs dark green.

If you have an image that is supposed to be predominately bright then the camera will darken it to reach 18% grey so you need to adjust your exposure to be brighter to compensate. Conversely, if your image is predominately dark the camera will brighten your image to get to 18% grey and so you need to lower your target exposure to fool the camera into getting the correct exposure.

This is why people use manual. If you are shooting a light skinned person with blonde hair and a white shirt against a dark background and that person takes up most of the frame the camera will darken the image and you can set your target exposure (Av and Tv modes) to get the correct brightness. However, if you move forward, backwards, or side to side and change how much of the frame your model takes up in the picture then it can change how your camera calculates the exposure. Using manual this won't happen (the meter will bounce around as it predicts changes, but the meter in manual mode is just a guide and doesn't actually do anything).

In your case the white flower is bright so the camera darkens the whole frame to reach 18% grey, thus the darker than normal background. In the case of the red flower, the flower was darker so it made the whole frame brighter which raises the background to be overexposed as well.

\$\endgroup\$
1
  • \$\begingroup\$ This answer is basically correct for most older cameras with monochromatic light meters. Many newer cameras now use RGBir meters that measure red, blue, green, and even near infrared light independently. \$\endgroup\$
    – Michael C
    Commented Mar 20, 2017 at 13:52
2
\$\begingroup\$

Color does figure into it on some cameras. Nikon SLRs use a small CCD with red, green and blue filters mounted in the pentaprism housing to do exposure metering, auto white balance and a few other things. This article discusses how metering works on the F5 film body, and almost all of what's there forms the basis for the metering in Nikon's digital bodies. I'm sure Canon and others do something similar, but I haven't explored it.

The lighting conditions in your photos vary enough that one isn't really comparable to the other. The white flower is brightly lit from the side, and given the tendency of whites to blow out and lose detail, the camera could well dial back the exposure to avoid that. The red flower and its background appear to be more uniformly lit, which is why it's less shadowy.

If you were to repeat the same experiment under identical lighting conditions, you'd probably find that the color really doesn't matter so much as the difference in light levels between the parts of the picture. Put the red flower in the same sidelit situation and the camera would back off to keep from blowing out the red channel.

In any case, some of the background in the first picture could be brought out using the fill light feature available in a number of post-processing programs.

\$\endgroup\$
4
  • \$\begingroup\$ Lighting was exactly the same in the above two cases (they were side by side). I took other pictures also using different frame and same result. \$\endgroup\$
    – TheTechGuy
    Commented Sep 9, 2012 at 23:08
  • 1
    \$\begingroup\$ @enthusiast: I just took a look at the histograms for several square areas that consisted of just the flower petals. The red one is pretty evenly exposed all the way around, but there's a definite shift in brightness between the left and right sides of the white one. That difference will have an effect on what the metering system sees. \$\endgroup\$
    – Blrfl
    Commented Sep 10, 2012 at 1:13
  • \$\begingroup\$ Isn't the histogram based on this picture and not on the actual lighting/condition itself? So if I took this picture severely under exposed, the histogram would show all black even though the actual lighting would be much better. \$\endgroup\$
    – TheTechGuy
    Commented Sep 10, 2012 at 10:34
  • \$\begingroup\$ @enthusiast: The histogram is just a plot of levels of brightness on the X axis and how many pixels are at that level on the Y axis. Usually this is done for the color channels plus one of the overall level. What I looked at was multiple histograms for small crops within the petal areas of each flower. The red picture's crops showed pretty consistent amounts of light all the way around where the white's showed less light on its right side than on the left. I did that as a way to quantitatively determine if one side was brighter than the other in case my eyeballs were off. \$\endgroup\$
    – Blrfl
    Commented Sep 10, 2012 at 15:12

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.