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 is 3.84 \times 10^{8} \mathrm{~m} from Earth.)
Strategy After substituting into Equation 25.10
\theta_{\mathrm{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_{\text {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 arc length of a circle is justified — the circular arc is very nearly a straight line. The Hubble Space Telescope has produced several gigabytes of data every day for over 20 years.