question archive Q1) For a human eye with a pupil diameter (aperture diameter D) of 0
Q1) For a human eye with a pupil diameter (aperture diameter D) of 0
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Q1) For a human eye with a pupil diameter (aperture diameter D) of 0.005 meters, and a focal length of 0.022 meters, there will be a resolution limit due to diffraction at 550 nm wavelength. If we space light-sensitive cells on the retina at this spacing, how many retina cells per square meter (of retina) would there need to be and why? You may assume that light-sensitive cells are spaced in a grid.
Q2) Now suppose there are giants which are exactly twice the size of humans in each dimension. So giant eyes have pupils of 0.01 meters, and a focal length of 0.044 meters. Does the spacing of light-sensitive cells change and if so, by how much?
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Q2
And now for giant, the aperture diameter is 0.01 m and focal length is 0.044 m, so the f - number is given by
N = f / D { here, f = 0.044 m, D = 0.01 m }
N = 0.044 / 0.01
N = 4.4 ........................(2)
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