To solve the problem of finding the minimum separation between objects on the surface of the Moon that can be just resolved by a telescope, we will use the Rayleigh criterion for resolution. The formula states that the minimum angular resolution θ (in radians) is given by:
\[
\theta = \frac{1.22 \lambda}{D}
\]
where:
- \( \lambda \) is the wavelength of light,
- \( D \) is the diameter of the telescope's aperture.
### Step 1: Convert the given values into appropriate units
- The diameter of the telescope \( D = 10 \, m \).
- The distance from the Earth to the Moon \( d = 4 \times 10^5 \, km = 4 \times 10^8 \, m \) (since \( 1 \, km = 1000 \, m \)).
- The wavelength of light \( \lambda = 5500 \, Å = 5500 \times 10^{-10} \, m \) (since \( 1 \, Å = 10^{-10} \, m \)).
### Step 2: Calculate the minimum angular resolution \( \theta \)
Using the formula for \( \theta \):
\[
\theta = \frac{1.22 \times \lambda}{D}
\]
Substituting the values:
\[
\theta = \frac{1.22 \times (5500 \times 10^{-10})}{10}
\]
Calculating this gives:
\[
\theta = \frac{1.22 \times 5500 \times 10^{-10}}{10} = 6.71 \times 10^{-7} \, radians
\]
### Step 3: Calculate the minimum separation \( s \)
The minimum separation \( s \) between two objects on the Moon can be calculated using the small angle approximation:
\[
s = d \cdot \theta
\]
Substituting the values:
\[
s = (4 \times 10^8) \cdot (6.71 \times 10^{-7})
\]
Calculating this gives:
\[
s \approx 268.4 \, m
\]
### Step 4: Final conversion to kilometers
To express this in kilometers:
\[
s \approx 0.2684 \, km \approx 0.27 \, km
\]
### Conclusion
Thus, the minimum separation between objects on the surface of the Moon, so that they are just resolved, is approximately:
\[
\text{Minimum separation} \approx 0.27 \, km
\]
