Question

The magnifying power of a telescope is \( 9 \). When adjusted for parallel rays, the distance between the objective and eyepiece is 20 cm. The ratio of the focal length of the objective lens to the focal length of the eyepiece is:

• A. 8
• B. 7
• C. 9
• D. 12

Hint:

The magnification of an astronomical telescope is given by \( M = \frac{f_o}{f_e} \). The larger the objective lens focal length, the higher the magnification.

Answer 1

The Correct Option is C

Step 1: {Understanding Magnifying Power of a Telescope}
The magnification \( M \) of an astronomical telescope in normal adjustment is calculated as the ratio of the objective's focal length (\( f_o \)) to the eyepiece's focal length (\( f_e \)):\[M = \frac{f_o}{f_e}\]Step 2: {Using Given Information}
We are given that the magnification \( M = 9 \), leading to the equation:\[\frac{f_o}{f_e} = 9\]Additionally, the total distance between the objective and eyepiece is 20 cm:\[f_o + f_e = 20\]Step 3: {Solving for \( f_o \) and \( f_e \)}
Substituting the first equation into the second:\[9f_e + f_e = 20\]This simplifies to:\[10f_e = 20\]Solving for \( f_e \) and then \( f_o \):\[f_e = 2 { cm}, \quad f_o = 18 { cm}\]Verifying the ratio of focal lengths:\[\frac{f_o}{f_e} = 9\]The final answer, representing the ratio of focal lengths, is \( 9 \).