A small telescope has an objective of focal length $140 ext{ cm}$ and an eyepiece of focal length $5.0 ext{

Question

A small telescope has an objective of focal length $140	ext{ cm}$ and an eyepiece of focal length $5.0	ext{ cm}$. The magnifying power of the telescope for viewing a distant object is:

Accepted Answer

1. **Magnifying Power Formula:** For an astronomical telescope in normal adjustment (viewing a distant object with the final image formed at infinity), the magnifying power ($m$) is given by the ratio of the focal length of the objective lens ($f_o$) to the focal length of the eyepiece ($f_e$).
    $m = \frac{f_o}{f_e}$
2. **Given Data:** 
    * Focal length of objective, $f_o = 140	ext{ cm}$
    * Focal length of eyepiece, $f_e = 5.0	ext{ cm}$
3. **Calculation:**
    $m = \frac{140}{5.0} = 28$
4. **Conclusion:** The magnifying power of the telescope is 28.

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