A telescope has an objective lens of diameter 5 inch. Calculate the resolving power of the telescope when light of wavelength `5,500 Å` is used.

Answer

To calculate the resolving power of the telescope, we will follow these steps:

### Step 1: Convert the diameter of the objective lens from inches to meters.
Given:
- Diameter (D) = 5 inches

To convert inches to meters, we use the conversion factor:
1 inch = 0.0254 meters.

So,
\[ D = 5 \text{ inches} \times 0.0254 \text{ m/inch} = 0.127 \text{ m} \]

### Step 2: Convert the wavelength from angstroms to meters.
Given:
- Wavelength (λ) = 5500 Å

To convert angstroms to meters, we use the conversion factor:
1 Å = \( 10^{-10} \) meters.

So,
\[ \lambda = 5500 \text{ Å} \times 10^{-10} \text{ m/Å} = 5.5 \times 10^{-7} \text{ m} \]

### Step 3: Calculate the limit of resolution (θ) using the formula.
The formula for the limit of resolution (θ) is given by:
\[ \theta = \frac{1.22 \lambda}{D} \]

Substituting the values we have:
\[ \theta = \frac{1.22 \times 5.5 \times 10^{-7}}{0.127} \]

### Step 4: Calculate θ.
Calculating the numerator:
\[ 1.22 \times 5.5 = 6.71 \times 10^{-7} \]

Now substituting into the formula:
\[ \theta = \frac{6.71 \times 10^{-7}}{0.127} \]
\[ \theta \approx 5.28 \times 10^{-6} \text{ radians} \]

### Step 5: Calculate the resolving power (R) of the telescope.
The resolving power (R) is given by:
\[ R = \frac{1}{\theta} \]

Substituting the value of θ:
\[ R = \frac{1}{5.28 \times 10^{-6}} \]
\[ R \approx 1.89 \times 10^{5} \text{ m}^{-1} \]

### Final Result:
The resolving power of the telescope is approximately:
\[ R \approx 1.89 \times 10^{5} \text{ m}^{-1} \]

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