Question

In the normal adjustment of an astronomical telescope, the objective and eyepiece are 32 cm apart. If the magnifying power of the telescope is 7, find the focal lengths of the objective and eyepiece.

• A. \( f_o = 7 \, \text{cm} \) and \( f_e = 28 \, \text{cm} \)
• B. \( f_o = 28 \, \text{cm} \) and \( f_e = 7 \, \text{cm} \)
• C. \( f_e = 28 \, \text{cm} \) and \( f_o = 4 \, \text{cm} \)
• D. \( f_o = 28 \, \text{cm} \) and \( f_e = 4 \, \text{cm} \)

Hint:

In an astronomical telescope, the magnifying power is given by the ratio of the focal length of the objective to the focal length of the eyepiece. The total length of the telescope is the sum of these two focal lengths.

Answer 1

The Correct Option is D

The magnifying power \( M \) of an astronomical telescope in normal adjustment is given by: \[ M = \frac{f_o}{f_e}, \] where: - \( f_o \) is the focal length of the objective, - \( f_e \) is the focal length of the eyepiece. We are given that the magnifying power \( M = 7 \), so: \[ 7 = \frac{f_o}{f_e}. \] Thus, we can express the focal length of the objective in terms of the focal length of the eyepiece: \[ f_o = 7 f_e. \]
Step 1: Use the total length of the telescope
In normal adjustment, the total length of the telescope is the sum of the focal lengths of the objective and the eyepiece: \[ f_o + f_e = 32 \, \text{cm}. \] Substitute \( f_o = 7 f_e \) into this equation: \[ 7 f_e + f_e = 32, \] \[ 8 f_e = 32, \] \[ f_e = 4 \, \text{cm}. \]
Step 2: Find \( f_o \)
Now that we know \( f_e = 4 \, \text{cm} \), we can calculate \( f_o \) using \( f_o = 7 f_e \): \[ f_o = 7 \times 4 = 28 \, \text{cm}. \] Thus, the focal lengths of the objective and the eyepiece are: \[ f_o = 28 \, \text{cm}, \quad f_e = 4 \, \text{cm}. \] Therefore, the correct answer is option (D).