I have the Nikon D600, at ISO 100 has more then 14 EV of dynamci range. Now if I shoot a photograph in raw file and open it in Lightroom, with the exposure slide I can go only +5EV and -5EV.

I am confused, does that mean the in Lightroom I cannot use the full dynamic range of my camera? How I have to interpret the +5 when I push the exposure slide all the way to the right for example?

Thank you.


1 Answer 1


You are confusing the dynamic range that can be contained in a RAW file with the adjustment that tells Lightroom what portion of that dynamic range to place in the center of the displayed picture's dynamic range.

A photo can have some parts that are so bright they are pure white and other parts so dark they are pure black. The distance between the white parts and the black parts, expressed as Exposure Values (EV) are the image's dynamic range.

Your camera can detect differences in brightness of about 13 stops (the other EV in the camera's DR is eaten up by the noise floor - the level of brightness that is indistinguishable from digital noise). But the best an 8-bit image, such as a JPEG, can do is somewhere between 6 or 7 stops. This means a typical monitor that is 8-bits per color channel is also limited to about 6-7 stops. So are most printer ink and paper combinations.

When you open a RAW image in Lightroom, what you see on the screen is an 8-bit conversion of the RAW data using your LR default settings. As you move the controls, including the exposure slider, LR recomputes that 8-bit conversion and displays the results. Once you are happy with your edits, you can export the image and the conversion you see on the screen is saved as an 8-bit JPEG.

Some of the adjustments you can make in LR squeeze the wider 14-bit range between the brightest and darkest areas of the scene into the narrower range of an 8-bit image. With the exposure adjustment, though, you are telling LR to shift the point at which everything brighter than a particular brightness value is rendered as white and everything darker than a particular brightness value is rendered as black without changing the distance between the white and black points.

See also When comparing sensor dynamic range, what are those numbers based on?
And What does a dynamic range difference of 2.7 EV really represent?

  • \$\begingroup\$ That's is interesting. Thank you. Let us suppose I open the raw file and do not touch the exposure slider. The image has blown out areas (pure white). Now I push the highlights to the left and magically I see details there where I could only see white areas. Those details are taken then from the not mapped portion of the dynamic range of the raw file to the right of the limit of the jpeg dynamic range upper limit? \$\endgroup\$
    – az1mov
    Commented Aug 23, 2014 at 20:35
  • \$\begingroup\$ Another question is: I understand that theory of tonal mapping raw->jpeg applies everywhere. So the jpeg I see as a preview of the raw file in the back of my camera and which is used to generate the histogram I see in preview mode as well, well how is that jpeg's tonal range mapped on the one of the RAW file? \$\endgroup\$
    – az1mov
    Commented Aug 23, 2014 at 20:37
  • \$\begingroup\$ In the case of the highlights adjustment you are extending the EV of the white threshold and compressing the additional information into the very top of the DR of the display without moving the mid tones or shadows. For the preview it depends on the in camera settings and how they affect the shape of the light curve. Increasing the contrast setting will spread the shape of the histogram wider and more of the highlights will be pure white and more of the shadows will be black. Decreasing contrast raises the white point and lowers the black point, so the histogram is squeezed in the middle. \$\endgroup\$
    – Michael C
    Commented Aug 24, 2014 at 4:19

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