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 m long, but has a secondary mirror that increases its effective focal length to 57.8 m. (See Fig. 25.11.) 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 × 10³?
STRATEGY Equation 25.8
m={\frac{\theta}{\theta_{o}}}={\frac{h^{\prime}/f_{e}}{h^{\prime}/f_{o}}}={\frac{f_{o}}{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}}\qquad\rightarrow\qquad f_{e}=\frac{f_{o}}{m}=\frac{57.8\; m}{8.00~\times~10^{3}}=7.23\times10^{-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.)