Question

The resolution of a lens can be estimated by treating the lens as a circular aperture. The resolution is the smallest distance between two point sources that produce distinct images. This is similar to the resolution of a single slit, related to the distance from the middle of the central bright band to the firstorder dark band; however, the aperture is circular instead of a rectangular slit which introduces a scale factor. Suppose the Hubble Space Telescope, 2.4 m in diameter, is in orbit 90.4 km above Earth and is turned to look at Earth. If you ignore the effect of the atmosphere, what is the resolution of this telescope for light of wavelength 557 nm?

Answer 1

y = 2.56 10⁻² m

Step-by-step explanation:

The resolution of this telescope is given by the Rayleigh criterion, for the phenomenal diffraction the first minimum for a linear slit is in

a sin θ = λ

in general the angles are very small, so we approximate

sin θ = θ

we substitute

θ = λ / a

in the case of circular slits we must use polar coordinates, which introduces a numerical factor, leaving the equation

θ = 1.22
`(\lambda )/(D)`

where D is the diameter of the circular opening

In this case they indicate the lens diameter D = 2.4 m, the observation distance r = 90.4 km = 90.4 10³ m

how angles are measured in radians

θ = y / r

we substitute

y / r = 1.22\frac{\lambda }{D}

y = 1.22 \frac{\lambda r }{D}

let's calculate

y =
`1.22 ( 557 \ 10^(-9) \ \ 90.4 \ 10^(3) )/(2.4)`

y = 2.56 10⁻² m

this is the minimum distance that can differentiate two objects on Earth