Question 25.5: Hubble Power Goal Understand magnification in telescopes. Pr...
Hubble Power
Goal Understand magnification in telescopes.
Problem The Hubble 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. 25.11.) The telescope doesn’t have an eyepiece, because various instruments, not a human eye, record the collected light. However, it can 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_{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}} \rightarrow 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 result of this magnification is an image with “good” resolution. However, the light-gathering power of a telescope largely determines the resolution of the image, and is far more important than magnification. A high-resolution image can always be magnified so its details can be examined. Such details are often blurred when a low-resolution image is magnified.