a positive lens with a diameter of 30 cm forms an image of a satellite passing o

Question

a positive lens with a diameter of 30 cm forms an image of a satellite passing overhead. The satellite has two green lights (wavelength = 500 nm) spaced 1.0 m apart. If the lights can just be resolved according to the Rayleigh criterion, what is the altitude of the satellite? Most of this made sense, it's the last part of the calculation I can't follow. The correct answer is 492 Km.

thetam = 1.22 lamda/D = 1.22 (500 x 10^-9 m/ 0.3m) = 2.03 x 10^-6 rad. Up to here I'm good.

The last formula I can't find (so I'm assuming my teacher skipped some steps again). h=d/2.03 x 10^-6 rad = 4.92 x 10^5m = 492 km.   Thanks for the help with the last line.

Explanation / Answer

the Rayleigh criterion is that objects are resolvable if sin(theta) = 1.22 L/d where L is the wavelength, d is the diameter of the lens (or eye), and theta is the angle subtended by the object under observation here, L=5.5x10^-7m d=2x10^-3m so sin(theta) = 1.22(5.5x10^-7)/2x10^-3 = 3.35x10^-4; this is the whole angle, the half angle, or the angle between the line of sight and one car is half of this, or 1.67x10^-4 for very small angles, sin(theta) = tan(theta), and we can use this to estimate the height tan(theta)=1.45/h =1.67x10^-4m => h=8643m

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