Question

In normal adjustment, for a refracting telescope, the distance between the objective and the eyepiece lens is 1.00 m. If the magnifying power of the telescope is 19, find the focal length of the objective and the eyepiece lens.

Hint:

For a refracting telescope, the objective lens has a large focal length, while the eyepiece lens has a small focal length to achieve high magnification.

Answer 1

Calculation of Focal Lengths in a Refracting Telescope

1: Understanding the Formula
In a refracting telescope, the magnifying power (\( M \)) in normal adjustment is given by: \[ M = \frac{f_o}{f_e} \] where:
- \( f_o \) = Focal length of the objective lens,
- \( f_e \) = Focal length of the eyepiece lens. The total length of the telescope in normal adjustment is: \[ L = f_o + f_e \] Given: - \( L = 1.00 \) m = 100 cm, - \( M = 19 \).
2: Expressing \( f_o \) and \( f_e \)
Rearranging the magnification equation: \[ f_o = M f_e \] Substituting in the length equation: \[ M f_e + f_e = L \] \[ 19 f_e + f_e = 100 \] \[ 20 f_e = 100 \] \[ f_e = \frac{100}{20} = 5 \text{ cm} \] \[ f_o = M f_e = 19 \times 5 = 95 \text{ cm} \]
3: Conclusion
- Focal length of the objective lens: 95 cm.
- Focal length of the eyepiece lens: 5 cm.