Power from an image

Hi, I have a certain image with an intensity distribution.

How do I calculate the Power?

Is it correct to take the area * intensity?

How to I estimate the total intensity from the laser beam? sum(sum(image))= 623614876? im a bit confused becuase intensity is usually in unit W/m2 or W/sr? The value deosnt look correct.

My aim is to only take a part of this intensity distribution, and determine the total intensity of this part (a certain angle) by assuming the power is for example 4 W. How much intensity do I have within a certain angle distribution.

Image is gray. It shows the intensity from the laser distribution that has a certain rectangular shape. I don't know what the maximal pixelvalue is, but it doesn't seem saturated. The image is 2048 x 2048 pixels.

Thanks in advance!

Answer 1

[Walter Roberson]

https://www.mathworks.com/matlabcentral/answers/57838-calculate-signal-power-of-image#answer_70246

[Walter Roberson]

You are confused because you are hoping for absolute power units like watt, but you have not calibrated your intensity levels, and you have not calibrated that your sensors are linear.

Without calibration information you do not know whether 255 represents a strong laser with a sensor that is not very sensitive, or a strong laser with a more sensitive sensor and a good filter, or a weak laser with a very sensitive sensor.

[thestargazer]

How do I obtain the information about the calibration? I know which camera sensor that was used for the measurement. The maximal pixel doesnt seem to be 255 btw, it is much larger values per pixel.

[Walter Roberson]

Which sensor was used?

Also, could you show

filename = 'YourImageFileNameHere.tif';

info = imfinfo(filename);
if hasfield(info, 'DigitalCamera')
  disp(info.DigitalCamera)
else
  disp('No Exif, sorry')
end

[thestargazer]

A Camera IDS UI-3370CP-NIR-GL with lens was used.

When I ran that script, I got following message:

Undefined function 'hasfield' for input arguments of type 'struct'.

    Error in test (line 4)
    if hasfield(info, 'DigitalCamera')

The image is in format .png

[Walter Roberson]

Sorry should be isfield() instead of hasfield()

Unfortunately PNG files do not store metadata in the same way as JPG or TIF do, so it is more difficult to dig out any calibration attributes that might be stored with the image.

The camera manuals require registration :(

[thestargazer]

Result: No Exif, sorry

[Walter Roberson]

Slightly more on the png situation: https://www.mathworks.com/matlabcentral/answers/303156-how-to-convert-dicom-image-to-jpeg-png-with-out-loosing-it-s-metadata

The hope was that image metadata might have stored calibration information.

[thestargazer]

Wait, I checked the folder. There is a tif image, but your code produced: No Exif, sorry..

I guess that means it not useful. Sadly I didnt do these measurements I dont have the original images (if there are any).

[Guillaume]

Just knowing the specs of the camera is not enough to calculate the light source power. Since a lens was used, you need to know how much it transmit at the laser wavelength.

Presumably, to avoid damage to the camera, the laser wasn't aimed straight at it but reflected off a surface, in that case you also need to know the surface reflectance.

The best way to deal with all these parameters is to ignore them and calibrate the whole acquisition chain with a uniform light source with a known power.

[thestargazer]

Ok, so you can determine the total energy (J) of the lobe by suming the pixel values.

sum(imageArray(:))

This gives I = 623614876 J ? This doesnt make sense since these energies are usually in range of 200-100 nJ.

Ok, so I have a distribution of my laser lobe. Would the Power then be

P = 623614876 * area of the lobe?

If it makes any sense?

I do know 1 pixel is 5*5 um^2. I have the # pixles of the lobe and the firld distribuion. Does this make sense? This image has no noise in the background, its possible it has been pre-processed.

[Guillaume]

The sum of the pixels is a value proportional to the total energy (assuming the camera sensitivity is uniform). To know that proportionality constant, you need to know all sorts of things about the sensor (Quantum Efficiency, gain, etc.), the transmissivity of your optics and the reflectance of your screen. Without it, you cannot convert to Joules.

Since your sensor is 12-bit, the maximum pixel intensity would be 4095. If all the pixels were saturated, the total sum would 2048*2048*4095 = 17,175,674,880, much more than the 623,614,876 so that value is not unreasonable.

Once again, that value is not Joules. It is Joules * constant.

[thestargazer]

If you know Quantum Efficiency, gain etc how can you determine the proportionality constant? Is there an formula for it?

The camera is 16 bit.

[Guillaume]

"The camera is 16 bit". Not according to the manufacturer. The image will be uint16 since that the smallest type that can fits 12 bit per pixel but maximum intensity will be 2^12-1 = 4095.

"Is there an formula for it." Most likely. One you'll have to derive yourself though. It's going to be a product of the gains and losses of each optical and electrical component and will also involve the collection angle. As I said, the best way to know that constant is by calibrating your image system with a uniform light source of known power.

[Walter Roberson]

Better yet a whole series of known powers.

But you mention that you do not have the original images. Is the implication that you cannot run calibration tests?

[Image Analyst]

I agree with Walter and Guillaume: your best bet is to take pictures of known non-saturated patterns of known powers and basically build up a lookup table or function that gives you power out as a function of total integrated gray level in. That should work as long as you don't change the exposure time per frame (per video or still frame) after making this calibration function.