Question

In normal adjustment, for a refracting telescope, the distance between the objective and 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:

In a refracting telescope, the focal length of the objective lens is much greater than that of the eyepiece, and their sum gives the total length of the telescope.

Answer 1

Applying the Telescope Magnification Formula:
- In a refracting telescope in normal adjustment, the total length of the telescope is given by: \[ L = f_o + f_e \] where, \( f_o \) = focal length of the objective lens, \( f_e \) = focal length of the eyepiece lens, \( L = 1.00 \) m = total length of the telescope.
- The magnifying power \( M \) of the telescope is given by: \[ M = \frac{f_o}{f_e} \] - Given \( M = 19 \), we get: \[ 19 = \frac{f_o}{f_e} \] \[ f_o = 19 f_e \] - Substituting this into the equation \( L = f_o + f_e \): \[ 1.00 = 19 f_e + f_e \] \[ 1.00 = 20 f_e \] \[ f_e = \frac{1.00}{20} = 0.05 \text{ m} = 5 \text{ cm} \] \[ f_o = 19 \times 0.05 = 0.95 \text{ m} = 95 \text{ cm} \] Thus, the focal length of the objective lens is 95 cm, and the focal length of the eyepiece lens is 5 cm.