In an astronomical telescope, the focal length of the objective lens is 100 cm and of eye-piece is 2 cm . The magnifying power of the telescope for the normal eye is

Options

A. 50
B. 10
C. 100
D. `1/50`

Explanation

To find the magnifying power of an astronomical telescope, we can use the formula for magnification, which is given by:

\[ M = \frac{F_{objective}}{F_{eyepiece}} \]

where:
- \( M \) is the magnifying power,
- \( F_{objective} \) is the focal length of the objective lens,
- \( F_{eyepiece} \) is the focal length of the eyepiece.

Given:
- Focal length of the objective lens, \( F_{objective} = 100 \, \text{cm} \)
- Focal length of the eyepiece, \( F_{eyepiece} = 2 \, \text{cm} \)

Now, substituting the values into the formula:

1. Write the formula for magnifying power:
\[ M = \frac{F_{objective}}{F_{eyepiece}} \]

2. Substitute the given values:
\[ M = \frac{100 \, \text{cm}}{2 \, \text{cm}} \]

3. Calculate the magnifying power:
\[ M = \frac{100}{2} = 50 \]

4. Since the magnifying power is conventionally expressed as a positive value for telescopes, we take the absolute value:
\[ M = 50 \]

Thus, the magnifying power of the telescope for a normal eye is \( 50 \).