The diameter of an object of a telescope, which can just resolve two stars situated an angular displacement of `10^(-4)` degree, should be (`lambda`=5000 `A^(@)`)

Options

A. `35mm`
B. `35 cm`
C. `35 m`
D. `24 cm`

Answer

To solve the problem of finding the diameter of the telescope's objective that can just resolve two stars separated by an angular displacement of \(10^{-4}\) degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Angular Displacement to Radians**:
The angular displacement given is \( \Delta \theta = 10^{-4} \) degrees. To convert this to radians, we use the conversion factor \( \frac{\pi}{180} \):
\[ \Delta \theta = 10^{-4} \times \frac{\pi}{180} \approx 1.74 \times 10^{-6} \text{ radians} \]

2. **Convert Wavelength to Meters**:
The wavelength \( \lambda \) is given as \( 5000 \) Angstroms. We convert this to meters:
\[ \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

3. **Use the Resolving Power Formula**:
The formula for the minimum angular resolution of a telescope is given by:
\[ \Delta \theta = \frac{1.22 \lambda}{D} \]
where \( D \) is the diameter of the telescope's objective. We can rearrange this formula to solve for \( D \):
\[ D = \frac{1.22 \lambda}{\Delta \theta} \]

4. **Substitute the Values**:
Now we substitute the values of \( \lambda \) and \( \Delta \theta \) into the formula:
\[ D = \frac{1.22 \times (5 \times 10^{-7})}{1.74 \times 10^{-6}} \]

5. **Calculate the Diameter**:
Performing the calculation:
\[ D = \frac{1.22 \times 5 \times 10^{-7}}{1.74 \times 10^{-6}} \approx 0.35 \, \text{m} \]

6. **Convert to Centimeters**:
Since \( 1 \, \text{m} = 100 \, \text{cm} \):
\[ D \approx 0.35 \, \text{m} = 35 \, \text{cm} \]

### Final Answer:
The diameter of the telescope's objective should be \( 35 \, \text{cm} \).