Question

A telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6·0 cm. The magnifying power and the length of telescope tube will be respectively :

• A. 24, 150 cm
• B. 42, 138 cm
• C. 24, 138 cm
• D. 42, 150 cm

Hint:

For an astronomical telescope under normal adjustment, the intermediate image is formed at the common focus of the objective and the eyepiece. Hence, the total mechanical length of the system must simply equal \(f_o + f_e\).

Answer 1

The Correct Option is A

Concept: An astronomical telescope in normal adjustment means that the final image is formed at infinity, allowing the observer's eye to view it in a completely relaxed state. Under normal adjustment conditions, the following standard design formulas apply:
• Magnifying Power (\(m\)): The angular magnification is equal to the ratio of the focal length of the objective lens (\(f_o\)) to the focal length of the eyepiece lens (\(f_e\)): \[ m = \frac{f_o}{f_e} \]
• Length of the Telescope Tube (\(L\)): The separation distance between the objective lens and the eyepiece lens is the sum of their respective focal lengths: \[ L = f_o + f_e \]

Step 1: Extracting given data and calculating magnifying power.
The problem provides the following focal lengths:
• Focal length of the objective lens, \(f_o = 144\text{ cm}\)
• Focal length of the eyepiece, \(f_e = 6.0\text{ cm}\) Using the magnifying power formula: \[ m = \frac{f_o}{f_e} = \frac{144}{6.0} \] Dividing 144 by 6: \[ m = 24 \] Hence, the magnifying power of the telescope is 24.

Step 2: Calculating the length of the telescope tube.
Under normal adjustment, the intermediate image is created exactly at the focal point of both lenses simultaneously. Thus, the total tube length \(L\) is given by: \[ L = f_o + f_e \] Substituting the known numerical parameters: \[ L = 144\text{ cm} + 6.0\text{ cm} = 150\text{ cm} \] Therefore, the magnifying power is 24 and the tube length is 150 cm, matching Option (A).