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 \mathrm{~m}. (a) What is its limiting angle of resolution at a wavelength of 6.00 \times 10^{2} \mathrm{~nm} ? (b) What’s the smallest crater it could resolve on the Moon? (The Moon’s distance from Earth is 3.84 \times 10^{8} \mathrm{~m}.)
STRATEGY After substituting into Equation 25.10
\theta_{\min }=1.22 \frac{\lambda}{D} [25.10]
to find the limiting angle, use s=r \theta to compute the minimum size of crater that can be resolved.
(a) What is the limiting angle of resolution at a wavelength of 6.00 \times 10^{2} \mathrm{~nm} ?
Substitute D=2.40 \mathrm{~m} and \lambda=6.00 \times 10^{-7} \mathrm{~m} into Equation 25.10:
\begin{aligned}\theta_{\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 10^{-7} \mathrm{rad}\end{aligned}
(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=\left(3.84 \times 10^{8} \mathrm{~m}\right)\left(3.05 \times 10^{-7} \mathrm{rad}\right)=117 \mathrm{~m}
REMARKS The distance is so great and the angle so small that using the arclength 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.