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: {Normal Adjustment Condition}
For a refracting telescope in normal adjustment, the total length \( L \) is the sum of the focal lengths of the objective lens \( f_o \) and the eyepiece lens \( f_e \):\[L = f_o + f_e\]Step 2: {Given Values}
The total length is given as:\[f_o + f_e = 30\]The magnification \( M \) is defined as:\[M = \frac{f_o}{f_e}\]Given \( M = 2 \), we have:\[\frac{f_o}{f_e} = 2\]Step 3: {Solving for Focal Lengths}
From the magnification equation, we can express \( f_o \) in terms of \( f_e \):\[f_o = 2 f_e\]Substitute this into the length equation:\[2f_e + f_e = 30\]Simplifying the equation:\[3f_e = 30\]Solving for \( f_e \) and \( f_o \):\[f_e = 10 { cm}, \quad f_o = 20 { cm}\]The objective lens focal length is \( 20 \) cm.