Two point sources distant `0.1m` away viewed by a telescope. The objective is covered by a screen having a hole of 1mm width. If the wavelength fo the light used is `6550 A^(0)`, then the maximum distance at which the two sources are seen just resolved, will be nearly

Options

A. `125.0m`
B. `164m`
C. `131 m`
D. `163m`

Answer

To solve the problem of determining the maximum distance at which two point sources can be resolved using a telescope with a given aperture, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Distance between the two point sources, \( x = 0.1 \, \text{m} \)
- Width of the hole in the screen (aperture), \( D = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Wavelength of light used, \( \lambda = 6550 \, \text{Å} = 6550 \times 10^{-10} \, \text{m} \)

2. **Use the Resolution Formula:**
The limit of resolution for a circular aperture is given by the formula:
\[ \theta = \frac{1.22 \lambda}{D} \]
where \( \theta \) is the angular resolution in radians.

3. **Calculate the Angular Resolution:**
Substitute the known values into the formula:
\[ \theta = \frac{1.22 \times 6550 \times 10^{-10}}{1 \times 10^{-3}} \]
\[ \theta = \frac{1.22 \times 6550 \times 10^{-10}}{1 \times 10^{-3}} = 8.003 \times 10^{-7} \, \text{radians} \]

4. **Relate Angular Resolution to Distance:**
The relationship between the distance \( d \) at which the sources can be resolved and the angle \( \theta \) is given by:
\[ \tan(\theta) \approx \theta \approx \frac{x}{d} \]
Rearranging gives:
\[ d = \frac{x}{\theta} \]

5. **Substitute Values to Find \( d \):**
Substitute \( x = 0.1 \, \text{m} \) and \( \theta = 8.003 \times 10^{-7} \, \text{radians} \):
\[ d = \frac{0.1}{8.003 \times 10^{-7}} \approx 124.9 \, \text{m} \]

6. **Final Result:**
Rounding off, we find:
\[ d \approx 125 \, \text{m} \]

### Conclusion:
The maximum distance at which the two sources can be resolved is approximately **125 meters**.