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

Telescope Magnification Formula Application:

- For a refracting telescope in normal adjustment, the total length \( L \) is calculated as:

\[ L = f_o + f_e \]

where:
\( f_o \) = objective lens focal length
\( f_e \) = eyepiece lens focal length
\( L = 1.00 \) m = total telescope length.

- The magnifying power \( M \) is determined by:

\[ M = \frac{f_o}{f_e} \]

- Given \( M = 19 \), the equation becomes:

\[ 19 = \frac{f_o}{f_e} \]

\[ f_o = 19 f_e \]

- Substituting this into \( L = f_o + f_e \):

\[ 1.00 = 19 f_e + f_e \]

\[ 1.00 = 20 f_e \]

\[ f_e = \frac{1.00}{20} = 0.05 \text{ m} = 5 \text{ cm} \]

\[ f_o = 19 \times 0.05 = 0.95 \text{ m} = 95 \text{ cm} \]

Consequently, the objective lens has a focal length of 95 cm, and the eyepiece lens has a focal length of 5 cm.