1
$\begingroup$

This photo was published at stern magazine online. contaminated water leaks out of vessel

I wonder which information about the physical quantities could be reconstructed or computed given this photograph, the exposure time, and the detector size.

The white dots in the image correspond to particle traces on the imaging sensor. Thus by just counting the number of white dots, we know how many gamma photons interacted with the sensor while passing through. I will neglect the fact that the white pixels could also be caused by defect pixels aka "dead pixels". Let's assume the photo was taken with a non-stationary new camera.

So the only physical quantity I could come up is:

$$\frac{\text{# of white dots}}{12\pi \text{sr}\ \text{sensor_area}\cdot \text{exposure_time}}$$

Is there a way to find out more with some very conservative assumptions about the environment? What would these assumptions be?

$\endgroup$
  • 2
    $\begingroup$ The white dots may very well correspond to damaged pixels. You would need to compare to another frame from the same imaging device to know. We used to see this kind of damage on the experimental hall monitoring cameras at JLAB on a regular basis after they had been exposed to the conditions in the hall while the beam was on for a few weeks or months. Cameras had to be replaced every year or so as they became so bad as that they stopped serving as effective remote sensing devices for keeping tabs on what went on in the hall while we were locked down. $\endgroup$ – dmckee Nov 15 '13 at 15:59
  • $\begingroup$ dmckee: I totally forgot about the degradation of the sensor. Good point. $\endgroup$ – elcojon Nov 15 '13 at 16:01
  • $\begingroup$ It would be better to start from raw image data of the camera, to get the real pixel data - the bayer pattern image before demosaicing. $\endgroup$ – Volker Siegel Jul 19 '14 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.