The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the obje

Question

The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The focal lengths of lenses are:

Accepted Answer

1.  **Magnification Formula:** For an astronomical telescope in 'normal adjustment' (where the final image is at infinity and rays are parallel), the magnifying power ($M$) is defined as the ratio of the focal length of the objective ($f_o$) to the focal length of the eyepiece ($f_e$).
    $M = \frac{f_o}{f_e} = 9 \implies f_o = 9f_e$.
2.  **Tube Length Formula:** In normal adjustment, the distance between the objective lens and the eyepiece lens is the sum of their focal lengths.
    $L = f_o + f_e = 20	ext{ cm}$.
3.  **Calculation:** Substitute the value of $f_o$ from the magnification equation into the length equation:
    $9f_e + f_e = 20$
    $10f_e = 20$
    $f_e = 2	ext{ cm}$.
    Now, calculate $f_o$:
    $f_o = 9 	imes 2 = 18	ext{ cm}$.
4.  **Conclusion:** The focal lengths of the objective and eyepiece are $18	ext{ cm}$ and $2	ext{ cm}$, respectively.

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