Why shoot to the right

Introduction

The purpose of this article is to highlight and explain why it is important in digital photography to expose your picture so that the histogram is pushed as much as possible to the right (towards the white) without, however, clipping it.

Film Photography v/s Digital Photography

It is very important, to start with, to understand the difference between how the film reacts to light and how our digital camera’s senso reacts to it.
Film: Film emulation reaction to light is a highly non linear. This is why people who are used to shoot film will often tell you to slightly underexpose your picture to get more saturated colors.
Digital Cameras Sensor: Digital Cameras Sensors are, in the contrary, linear devices.

Result of the Digital camera’s sensor linearity

So the sensor of the Digital Camera is linear, what does it mean?

Well it simply means that if our digital camera’s sensor can capture 12 bits (that’s 2^12 = 4098) of data in RAW and assuming a Dynamic Range of 5 stops.

* The 1st stop will hold 1/2 of that i.e. 2048 – yes 1/2 the data recorded is in the highest stop
* The 2nd stop will hold 1024
* The 3rd stop will hold 512
* The 4th stop will hold 256
* The 5th stop will hold 128

To keep things simple, every time we push our histogram to the left (toward the dark area) we are giving our digital camera’s less data to record (because of the linearity explained above) and, since a stop is still a stop, this can only be compensated by equally spreading this available data over the stop range which explains why, in digital photography, blacks are always noisier.

How to expose

So to get the maximum data in our final picture, it is important to keep the maximum amount of the scene’s data in the highest stops of the Dynamic Range and that’s what is commonly known as “shooting to the right” of the histogram.

Shooting to the right it not really a technique, since it highly depends on the scene we are shooting. The easy way to do it is to dial some Exposure Compensation+1/3 … +1 … or even +2 depending on the scene but the only correct way to do it that I am aware of is to first take a “test shot” check the histogram then, according to the histogram, dial some EC and retake the shot one more time and repeat this process until we have a histogram that looks like pushed to the right but with no clipping.

More about RAW shooting

If you are a RAW shooterAnd my advise is, become one you should also know that the in digital camera histogram is not of the actual RAW data but rather of the embedded JPG so, almost always, the in camera histogram will be more to the right than the RAW data histogram you will see in you RAW converter. The actual relation between the two (JPG and RAW histogram) is camera dependent and you can only learn it by extensively testing your camera. But generally speaking, if your camera histogram shows no, or slight, clipping it means you are safe.

Another reason to shoot RAW over JPG

To add to all the benefits of RAW shooting like, flexibility, white balance adjustment etc… Here is another reason to shoot RAW.
“JPEGs throw away highlights, and compress the ones it keeps with low contrast settings.”quoted from John Sheehy

Conclusion

Getting the right exposure is crucial in both film and digital photography, but in the latter, it simply means pushing the histogram to the right as much as possible without actually clipping your whites.

Hope this helps

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  • Joss Delage

    You say:
    “* The 1st stop will hold 1/2 of that i.e. 2048 – yes 1/2 the data recorded is in the highest stop
    * The 2nd stop will hold 1024
    * The 3rd stop will hold 512
    * The 4th stop will hold 256
    * The 5th stop will hold 128″

    That doesn’t add up to the total. Am I missing something?

    Thanks,

    JD

  • http://www.adidap.com Antoine Khater

    Well let me try to put it in other terms, the idea is that each stop will simply capt 1/2 of the available data.
    So we are starting with 4096 then first stop will take 1/2 of that
    we remain with 2048 sot the 2nd stop will take 1/2 of that too so 1024
    etc…
    so if we had taken in the example a 6 stops dynamic range, then the 6stop would have 64 (1/2 of the remaining 128)

    I hope it does explain a little bit

  • http://www.flickr.com/groups/nikon_d90/discuss/72157622598302761 Magnera

    Hi!
    Great little article! I learned something new today.

    Used some quotes from you here: http://www.flickr.com/groups/nikon_d90/discuss/72157622598302761, link back is of course provided. Hope this is okay by you.