Technical name (radiometric/photometric/ray-optic) of the physical field that is measured with a camera A camera (that is FPA and objective) is a device that measures [intensity][1] or radiant intensity vs. two angles (horizontal & vertical). So it generates a representation of the physical quantity
$I(\phi, \theta)$
How is this "field" actually named in physics? I am looking for a radiometric, photometric, ray-optics or similar term.
A few options that come to my mind:

*

*Intensity field

*Radiant intensity field

*Light field (I think this is something else)

Examples: A voltmeter is a device that measures voltage, a clock is a device that measures time. An impedance analyzer is a device that measures an impedance spectrum (i.e. V/A vs. frequency). But what does a camera measure? (camera including objective, not just the sensor array)
[1]: For simplicity let ignore here color and also non-linear response of typical consumer cameras.
 A: The unit that best represents what a camera measures directly is radiance/luminance, not radiant/luminous intensity.  I will stick with the photometric versions of these units (luminance, luminous intensity) moving forward.
Luminance is defined as "flux per solid angle per projected area".  Projected is the key concept here.  (Note that flux per solid angle is also intensity; luminance is therefore intensity per area.)
The projected area concept is what gives the unit its most desired property: its value does not change with distance from the source.  This property allows the unit to represent the human perception of 'brightness': a source seems the same 'brightness' to us whether we are up close to it, or far away.
From wikipedia:

The 'S' here is some random surface, but you can imagine a sensor placed there.  A pixel in this sensor (of size dS) represents a projected area from the source.
The 'double solid angle' graphs, where one solid angle is projected from the image back onto the object, is what makes luminance unique.  A camera's pixel doesn't measure intensity directly, because it doesn't know what solid angle its pixel is imaging - you'd have to know the full geometry of the scene to calculate intensity.  But that pixel is getting something proportional to luminance, whether it really knows the geometry or not.
A camera's pixel is truly 'projected' onto the original source, via its optics.  I hope the picture helps you visualize this.
