Question 25.5: HUBBLE POWER GOAL Understand magnification in telescopes. PR...
HUBBLE POWER
GOAL Understand magnification in telescopes.
PROBLEM The Hubble Space Telescope is 13.2 \mathrm{~m} long, but has a secondary mirror that increases its effective focal length to 57.8 \mathrm{~m}. (See Fig. 25.12.) The telescope doesn’t have an eyepiece because various instruments, not a human eye, record the collected light. It can, however, produce images several thousand times larger than they would appear with the unaided human eye. What focal-length eyepiece used with the Hubble mirror system would produce a magnification of 8.00 \times 10^{3} ?
STRATEGY Equation 25.8
m = \frac{f_0}{f_e} [25.8]
for telescope magnification can be solved for the eyepiece focal length. The equation for finding the angular magnification of a reflector is the same as that for a refractor.
Solve for f_{e} in Equation 25.8 and substitute values:
m=\frac{f_{o}}{f_{e}} \quad \rightarrow \quad f_{e}=\frac{f_{o}}{m}=\frac{57.8 \mathrm{~m}}{8.00 \times 10^{3}}=7.23 \times 10^{-3} \mathrm{~m}
REMARKS The light-gathering power of a telescope and the length of the baseline over which light is gathered are in fact more important than a telescope’s magnification, because these two factors contribute to the resolution of the image. A high-resolution image can always be magnified so its details can be examined. A low resolution image, however, is often fuzzy when magnified. (See Section 25.6.)