Question 25.7: Resolving Craters on the Moon GOAL Calculate the resolution ...
Resolving Craters on the Moon
GOAL Calculate the resolution of a telescope.
PROBLEM The Hubble Space Telescope has an aperture of diameter 2.40 m. (a) What is its limiting angle of resolution at a wavelength of 6.00 × 10² nm? (b) What’s the smallest crater it could resolve on the Moon? (The Moon’s distance from Earth is 3.84 × 10^8 m.)
STRATEGY After substituting into Equation 25.10 to find the limiting angle, use s = rθ to compute the minimum size of crater that can be resolved.
θ_{\text{min}}=1.22\frac{λ}{D} [25.10]
(a) What is the limiting angle of resolution at a wavelength of 6.00 × 10² nm?
Substitute D = 2.40 m and λ = 6.00 × 10^{-7} m into Equation 25.10:
\theta_{\mathrm{min}}=1.22\,{\frac{\lambda}{D}}=1.22\,{\left({\frac{6.00~\times~10^{-7}\,{\mathrm{m}}}{2.40\,{\mathrm{m}}}}\right)}
=\;3.05\times\mathrm{10}^{-7}\,\mathrm{rad}
(b) What’s the smallest lunar crater the Hubble Space Telescope can resolve?
The two opposite sides of the crater must subtend the minimum angle. Use the arc length formula:
s=r\theta=(3.84\times10^{8}\,{\mathrm{m}})(3.05\times10^{-7}\,{\mathrm{rad}})=~117 ~m
REMARKS The distance is so great and the angle so small that using the arc length of a circle is justified because the circular arc is very nearly a straight line. The Hubble Space Telescope has produced several gigabytes of data every day since it first began operation.