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There is an option to filter only one color (R,G,B,Y) in my camera. What I want to know is can this effect be achieved in photoshop? And what is this called? I tried googling monochrome and came up with nothing. My camera calls it part color.

image image2

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So no matter what the picture is, it should be B&W except the the objects havibg the selected color. – Sid Dec 29 '12 at 5:58
Once you figure out how to do this, I would recommend using it in limited fashion. It can be effective in certain cases, but far too often it is overused. Just a friendly tip :) – dpollitt Dec 29 '12 at 22:48
@dpollitt Thanks. :) – Sid Dec 30 '12 at 2:51
up vote 9 down vote accepted

This technique is called Selective color.

Sometimes, you select a point (in this case, somewhere on the CD-R case), and the region around that point that is close enough to the same color retains its color, while the rest of the picture becomes black and white.

Other times, as you mention, you can select a color and a tolerance, or a range of colors, and the software will allow anything within that range to remain colored.

On the example on the Wikipedia page, it appears that the saturated region was hand-selected or masked.

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I think this is not that. What I want is to retain all the objects in Red to be as-is and the rest of the image to be B&W. See the new image – Sid Dec 29 '12 at 5:43
@Sid, it's still selective color, just more work is involved since you have more object parts to mask back in. Basically, you do whole image to select for the red, for example, then mask in (or alter the selection) the remaining parts of the mostly red objects that are not red. – John Cavan Dec 29 '12 at 14:03
@JohnCavan Er.. Could you direct me to a good tutorial please? Google only has ones doing invert selection and change to monochrome. – Sid Dec 29 '12 at 14:09 covers the basics, but you can use selection (e.g. color selection, magic wand, etc.) and inversion as a way to get most of your work done and then you just mask in the remaining bits as needed. – John Cavan Dec 29 '12 at 14:36
@JohnCavan Thanks :) – Sid Dec 30 '12 at 2:52

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