Question

A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. Calculate its magnifying power in normal adjustment and the distance of the image formed by the objective.

Hint:

The magnifying power of a telescope in normal adjustment is simply the ratio of the focal lengths of the objective and eyepiece. For an object at infinity, the image formed by the objective is at its focal point.

Answer 1

Step 1: The magnifying power \( M \) of the telescope in normal adjustment is given by the formula: \[ M = \frac{f_{\text{objective}}}{f_{\text{eyepiece}}} \] where \( f_{\text{objective}} \) is the focal length of the objective lens and \( f_{\text{eyepiece}} \) is the focal length of the eyepiece. Step 2: Substituting the given values: \[ M = \frac{150 \, \text{cm}}{5 \, \text{cm}} = 30 \] Thus, the magnifying power of the telescope in normal adjustment is 30. Step 3: To find the distance of the image formed by the objective, we use the lens formula: \[ \frac{1}{f_{\text{objective}}} = \frac{1}{v_{\text{objective}}} - \frac{1}{u_{\text{objective}}} \] where \( f_{\text{objective}} \) is the focal length of the objective lens, \( v_{\text{objective}} \) is the image distance, and \( u_{\text{objective}} \) is the object distance. Step 4: In normal adjustment, the object is at infinity, so \( u_{\text{objective}} \to \infty \). Therefore, the lens formula simplifies to: \[ \frac{1}{f_{\text{objective}}} = \frac{1}{v_{\text{objective}}} \] \[ v_{\text{objective}} = f_{\text{objective}} = 150 \, \text{cm} \] Thus, the distance of the image formed by the objective is 150 cm.