The diameter of the objective of the telescope is `0.1` metre and wavelength of light is `6000 Å`. Its resolving power would be approximately

Options

A. `6xx10^(-5)rad`
B. `6xx10^(4)`rad
C. `6xx10^(-3)`rad
D. `6xx10^(-6)` rad

Explanation

To find the resolving power of the telescope, we can use the formula:

\[ R = \frac{D}{1.22 \lambda} \]

where:
- \( R \) is the resolving power,
- \( D \) is the diameter of the objective lens,
- \( \lambda \) is the wavelength of light.

### Step 1: Identify the given values
- Diameter of the objective, \( D = 0.1 \) m
- Wavelength of light, \( \lambda = 6000 \) Å (angstroms)

### Step 2: Convert the wavelength from angstroms to meters
1 angstrom = \( 10^{-10} \) m, so:
\[ \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

### Step 3: Substitute the values into the resolving power formula
Now we can substitute \( D \) and \( \lambda \) into the formula:
\[ R = \frac{0.1}{1.22 \times 6 \times 10^{-7}} \]

### Step 4: Calculate the denominator
First, calculate \( 1.22 \times 6 \times 10^{-7} \):
\[ 1.22 \times 6 = 7.32 \]
Thus,
\[ 1.22 \times 6 \times 10^{-7} = 7.32 \times 10^{-7} \]

### Step 5: Calculate the resolving power
Now substitute this back into the equation for \( R \):
\[ R = \frac{0.1}{7.32 \times 10^{-7}} \]

Calculating this gives:
\[ R \approx \frac{0.1}{7.32 \times 10^{-7}} \approx 1.36 \times 10^{5} \]

### Step 6: Final result
Thus, the resolving power of the telescope is approximately:
\[ R \approx 1.36 \times 10^{5} \, \text{radians} \]