I'm working on a CubeSat mission which uses two Raspberry Pi V2 cameras (without special filters) as sensors. I am trying to estimate if they could suffer thermal damage if they were to have the Sun in full or partial view. Whilst I have found calculations on how a magnifying glass concentrates the solar flux, I'm struggling to understand how this can or should be applied to the case of a camera, especially when it has a 220deg FOV fisheye.

Not sure if I can apply the simple lens and magnification equations and not sure how the distortion of the fisheye lens changes things. Is it the size of the image on the CMOS sensor that I need to calculate (in pixels or mm) and relate to the solar flux (1366W/m2)? Do I need to consider how that is distorted by the fisheye? Is the effect (if there is any) the same as long as the Sun is within that 220deg arc or does it matter where in the field of view it is?

  • $\begingroup$ The link you used does tell you how a fisheye lens concentrates the solar flux. If the sun is directly in front of the lens, the flux is bounded by a circular aperture. If the sun is off to the side, the flux is bounded by an ellipse. $\endgroup$
    – mmesser314
    Apr 28, 2021 at 14:17
  • $\begingroup$ @mmesser314 I don't think there's any mention of fisheye lenses in that post. What is unclear to me is if I need to consider the image of the sun as the source of the heat on the CMOS, potentially only damaging those pixels on which it appears, or is it the entire sensor. If it is the image of the sun, how do I calculate the size of it given a 220deg FOV fisheye lens. $\endgroup$ Apr 28, 2021 at 17:31

1 Answer 1


You can use the analysis from Heat production of magnifying glass. The fact that it is a fisheye lens just means that it is a short focal length lens. This kind of lens requires some oddly shaped elements to make it perform well, and it usually doesn't perform perfectly. For example there is typically noticeable distortion near the edge of the field. However, it is a lens. Given an object, it will form an image.

To see if the camera will be damaged by an image of the Sun, you only need to check the on axis case, where the image is in the center of the field of view. This is where the lens performs best. The focus is the sharpest. The image is not enlarged by distortion. In this orientation, the aperture captures the most light.

You can calculate the intensity as shown in your link. All that power will be concentrated onto a region on the sensor plane the size of the Sun's image.

I don't know how to decide if that is enough to damage the sensor. A reason for optimism is that the sensor should be designed to take an image of the Sun at its brightest on Earth. A reason for pessimism is that the spectrum in space will include more UV, and there is no air for cooling. The Raspberry Pi website FAQ says the computer is designed to take up to 85 C. You said no special filter. but if you are using the standard filter, it cuts off everything outside the visible.


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