# Adjust the intensity of images with nonuniform light

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Steven on 4 Oct 2015
Commented: Image Analyst on 5 Oct 2015
My laser light is blinking (I don't know why?!), so the images I have is something like the followings. The intensity is different for every frame. How can I fix this? I mean how can I have uniform intensity throughout?
I thought of normalizing by a reference image, but there could not be a unique reference frame for each image.
Thank you!
Steven

Image Analyst on 4 Oct 2015
I'd contact the manufacturer of your laser.
If you're stuck with it, then the best you can do is to set the camera exposure so that no frame will be overexposed, then adjust all the frames by multiplying the image by the ratio of the desired mean to their current mean. That assumes you have a linear system, not one with a non-linear gamma being imposed by the camera.

Image Analyst on 5 Oct 2015
Yes it will, if you have a linear intensity response. Let's say that your brightest image has a mean of 200. Let's define that to be the reference, and you want all images to have that same mean. So if you have an image where the mean is 150, you need to multiply that image by 200/150 to bring it's intensities up to where you want them to be -- at 200.
Steven on 5 Oct 2015
Yes. But Then the mean of all image frames is 200, I agree, but this does not mean all images would have similar intensity distribution as the reference one right? I mean even after this multiplication,the original problem (sudden change of intensity for each frame) still exists, only the values are different. Right? Thanks
Image Analyst on 5 Oct 2015
Well the means would be the same. The distributions would be the same as what they originally were, rather than the same as the reference image. So each corrected image will keeps its unique shape of histogram, it would just be scaled so that it's mean is at the same location (200) as the reference. The flickering of the laser still occurs of course, but you've corrected the images so the flickering seems to be mitigated and is less noticeable in the images than it used to be.