Question

In normal adjustment, for a refracting telescope, the distance between the objective and eyepiece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:

• A. 20 cm
• B. 30 cm
• C. 10 cm
• D. 15 cm

Hint:

In an astronomical telescope, the focal length of the objective lens is usually larger than the eyepiece lens to provide higher magnification.

Answer 1

The Correct Option is A

Step 1: {Understanding the Normal Adjustment Condition}
In a refracting telescope under normal adjustment, the total length of the telescope is: \[ L = f_o + f_e \] where:
\( f_o \) is the focal length of the objective lens,
\( f_e \) is the focal length of the eyepiece lens.
Step 2: {Using the Given Values}
It is given that the total length of the telescope is: \[ f_o + f_e = 30 \] Also, the magnification of the telescope is given by: \[ M = \frac{f_o}{f_e} \] Since \( M = 2 \), we get: \[ \frac{f_o}{f_e} = 2 \] Step 3: {Solving for \( f_o \) and \( f_e \)}
Rewriting the equation: \[ f_o = 2 f_e \] Substituting into the length equation: \[ 2f_e + f_e = 30 \] \[ 3f_e = 30 \] \[ f_e = 10 { cm}, \quad f_o = 20 { cm} \] Thus, the correct answer is \( 20 \) cm.