An astronomical telescope in normal adjustment has objective focal length $f_o = 100$ cm and eyepiece focal length $f_e = 5$ cm. Its magnifying power is:

**Astronomical telescope, normal adjustment** (final image at infinity): $M = \dfrac{f_o}{f_e}$. $M = \dfrac{100}{5} = 20$. **Compare with the compound microscope.** For a microscope $M = \dfrac{L \cdot D}{f_o \cdot f_e}$ — focal lengths are in the *denominator* (both should be small). For a telescope $f_o$ is in the *numerator*, so it should be **large**, while $f_e$ stays small. The opposite role of $f_o$ is the load-bearing difference between the two instruments.