Calculating Resolution of Space Camera for Imaging Earth

AI Thread Summary
To determine the resolution of a space camera for imaging Earth, the diffraction limit of 4.1*10^-7 m corresponds to the image plane. To find the resolution on the Earth's surface, one must consider the camera's focal length and orbital height. The relationship involves using the magnification factor, which is dimensionless, to convert the airy disk size into a practical resolution in meters. The correct equation to apply is (object) x (magnification) = (image). This approach will yield a realistic resolution for distinguishing objects on Earth from the camera's altitude.
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If I have a camera orbiting in space way above the Earth, and it has a diffraction limited resolution of 4.1*10^-7 m (diameter of airy disc), then how can I find out the corresponding value if the lens is supposed to take images of the Earth?
 
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What do you mean by "corresponding value"?
Presumably by the 4.1*10^-7 m you mean the resolution limit at the image plane.
If you knowthe focal length and orbital height you can work out the resolutionsize on the Earth's surface
 
mgb_phys said:
What do you mean by "corresponding value"?
Presumably by the 4.1*10^-7 m you mean the resolution limit at the image plane.
If you knowthe focal length and orbital height you can work out the resolutionsize on the Earth's surface

Yeah, how do I figure out the res on the Earth? I know the Focal length and orbital height. I was thinking you could just multiply the airy disk size by 100000 for 100km
 
Note that multiplying (airy disc diameter) x (orbital height) gives incorrect units of m^2, when we're looking for an answer in meters.

Hint: what is the camera's magnification factor?
 
So I'm looking for a dimensionless number? Not sure what it could be, I'm only given lens diamater, focal length and wavelength.
 
Resolution at the Earth's surface will have units of meters.

Eg.: the camera can distinguish objects on Earth that are ___ meters apart.
 
yeah, I'm having trouble figuring out how I can get that very small airy disc size to correspond to resolution on the Earth's surface, 100km's away. I tried using this
ea350ccdc729cf3d0385636f2115ce35.png

where di is distance to the image from the lens, and do is distance to the object. But, then I get a very small number and if i multiply this number with the airy disc size I don't get a realistic resolution on the Earth's surface.
 
Last edited:
Which of these seems more reasonable to you:

(object) x (magnification) = (image)
or
(image) x (magnification) = (object)

?
 
Redbelly98 said:
Which of these seems more reasonable to you:

(object) x (magnification) = (image)
or
(image) x (magnification) = (object)

?

First one.
 
  • #10
Good.

Does the 4.1*10^-7 m Airy disk correspond to an object or image?
 
  • #11
Redbelly98 said:
Good.

Does the 4.1*10^-7 m Airy disk correspond to an object or image?

image
 
  • #12
Sounds like you can use the 1st equation in post #8 to solve the problem now.
 
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