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 (parallel rays) 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.
Step 2: {Using Given Information}
It is given that \( M = 9 \), so: \[ \frac{f_o}{f_e} = 9 \] Also, the total distance between the objective and eyepiece is: \[ f_o + f_e = 20 \] Step 3: {Solving for \( f_o \) and \( f_e \)}
Using the given equations: \[ 9f_e + f_e = 20 \] \[ 10f_e = 20 \] \[ f_e = 2 { cm}, \quad f_o = 18 { cm} \] Thus, the ratio of focal lengths is: \[ \frac{f_o}{f_e} = 9 \] Thus, the correct answer is \( 9 \).