review: solid angle
in recent years I have been surprised to find that many students taking this course are not comfortable with the concept of solid angle, so here is a brief review
angle transverse distance at a distance solid angle transverse area at a distance
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luminance (as I use the term)
the light energy (usually already integrated over the spectral range of interest) per unit area (of the source) per unit solid angle (in the direction of interest) per unit time leaving the source surface watt/(m2 steradian) if not integrated over color then watt/(m2 steradian Hz) fundamentally for self-luminous sources to keep terminology simple*, I will also use this word for the light leaving an illuminated surface
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illumination (as I use the term)
the light energy (usually already integrated over the spectral range of interest) per unit area (of a target) per unit solid angle (in the direction of the source) per unit time reaching the target surface watt/(m2 steradian) [or watt/(m2 steradian Hz)] fundamentally of interest for illuminated targets to keep terminology simple, I will also use this word for the light reaching a sensor
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how it falls off with distance
crucial to know who means what by it! consider illumination from an idealized point source of light the energy per unit area (normal to the direction of the point source) per unit time falling on a target (or a sensor) falls off as 1/distance2 but a real source is an area, never a point if very close, it doesnt fall off at all with distance if line-like and not too close it falls off as distance-1
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from scene to lens
consider a small area AS of a scene emitting pS watt/(m2 steradian) in the spectral range of interest in the direction of the lens the power collected by lens PL is then PL = AS pS rL2 cosSL/(dSL/cosSL)2 this power is delivered by the lens to the sensor but to what area of the sensor? (of course, this treatment ignores all the actual losses to scattering, absorption, etc)
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ps
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from lens to sensor (detector D)
lens equation: 1/dSL + 1/dLD = 1/f for simplicity, assume dLD << dSL so dLD f image area Ai corresponding to scene area AS is then given by ratios Ai/f 2 = AS/dSL2 so the power per unit area on the sensor is pD=(ASpSrL2cos3SL/dSL2)/((AS/cosSL)f2/dSL2) = pS cos4SL (rL/f )2 = ( /4) pS cos4LD / f-number2 so scene-to-camera distance doesnt matter! but it says image gets dimmer as cos4
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dsl
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so what will the ultimate signal be?
we found pD ~ pS / f-number2 what does that tell us about the signal we can expect to see from the sensor? it depends on the sensor!
typical sensor output (CCD signal voltage, photographic film blackness) is proportional to pd times exposure time (independent of pixel area for CCD, but not for film!) others might deliver, e.g., output current proportional to pd times pixel area (but independent of exposure time!)
sensing is the preceding fundamentals sensors are these still-open details
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but it says image gets dimmer as cos4LD
and if you look at OLD photographs it does!
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assignment
5) For a scene illuminated by typical Pittsburgh sunlight (how many watts/m ?) estimate the
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