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Step-by-step Solution
Step 1: Identify the relevant formula
The minimum angular separation θ that a circular aperture (telescope’s objective) can resolve is given by the Rayleigh criterion:
$$\theta = 1.22 \frac{\lambda}{D}$$
where
- $ \lambda $ is the wavelength of light, and
- $ D $ is the diameter of the objective lens.
Step 2: Relate angular separation to linear separation
If the telescope is focused on objects at a distance L away, then the linear separation x between two point objects that can just be resolved is:
$$x = L \times \theta = L \times 1.22 \frac{\lambda}{D}.$$
Step 3: Substitute the given values
• Distance L = 1 km = $1000\,\text{m}$
• Diameter D = 10 cm = $0.10\,\text{m}$
• Wavelength $ \lambda = 5000\,\text{\AA} = 5000 \times 10^{-10}\,\text{m} = 5 \times 10^{-7}\,\text{m}$
• The factor 1.22 is used in the Rayleigh criterion.
Step 4: Calculate the resolvable distance
Plugging these into the formula:
$$
x = 1000\,\text{m} \times 1.22 \times \frac{5 \times 10^{-7}\,\text{m}}{0.10\,\text{m}}
= 1000 \times 1.22 \times \frac{5 \times 10^{-7}}{0.1}.
$$
Step 5: Simplify the expression
First simplify inside the fraction:
$$
\frac{5 \times 10^{-7}}{0.1} = 5 \times 10^{-6}.
$$
Now multiply by 1.22 and 1000:
$$
x = 1.22 \times 1000 \times 5 \times 10^{-6}
= 1.22 \times 5 \times 10^{-3}\,\text{m}
= 6.1 \times 10^{-3}\,\text{m}.
$$
Step 6: Convert to mm and conclude
In millimeters:
$$
6.1 \times 10^{-3}\,\text{m} = 6.1\,\text{mm}.
$$
This value is of the order of 5 mm. Hence, the minimum distance between the two objects that can be resolved by this telescope is about 5 mm.
