Aperture of human eye is 0.2cm. The minimum magnifying power of a visal telescope, whose objective has diameter100cm is.

Options

A. 500
B. 0.002
C. 2
D. 100

Explanation

To solve the problem of finding the minimum magnifying power of a visual telescope given the aperture of the human eye and the diameter of the telescope's objective, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Aperture of the human eye (D_eye) = 0.2 cm
- Diameter of the telescope's objective (D_telescope) = 100 cm

2. **Understanding the Concept of Resolving Power**:
- The resolving power (angular resolution) of an optical system can be expressed as:
\[ d\theta = \frac{1.22 \lambda}{D} \]
- Where \( \lambda \) is the wavelength of light and \( D \) is the diameter of the aperture.

3. **Calculate the Resolving Power for the Telescope's Objective**:
- For the telescope's objective:
\[ d\theta_{telescope} = \frac{1.22 \lambda}{D_{telescope}} = \frac{1.22 \lambda}{100 \text{ cm}} \]

4. **Calculate the Resolving Power for the Human Eye**:
- For the human eye:
\[ d\theta_{eye} = \frac{1.22 \lambda}{D_{eye}} = \frac{1.22 \lambda}{0.2 \text{ cm}} \]

5. **Finding the Minimum Magnifying Power**:
- The minimum magnifying power (M) of the telescope can be calculated by taking the ratio of the resolving powers:
\[ M = \frac{d\theta_{eye}}{d\theta_{telescope}} = \frac{\frac{1.22 \lambda}{0.2}}{\frac{1.22 \lambda}{100}} \]
- The \( \lambda \) cancels out:
\[ M = \frac{100}{0.2} = 500 \]

6. **Conclusion**:
- Therefore, the minimum magnifying power of the visual telescope is **500**.