An astronaut is looking down on earth's surface from a space shuttle at an altitude of `400 km. ` Assuming that the astronaut's pupil diameter is `5 mm` and the wavelength of visible light is `500 nm. ` The astronaut will be able to resolve linear object of the size of about .

Options

A. 0.5m
B. 5m
C. 50m
D. 500m

Explanation

To solve the problem of determining the smallest linear object that an astronaut can resolve while looking down at the Earth's surface from a space shuttle at an altitude of 400 km, we will use the formula for resolving power.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Altitude of the astronaut (D) = 400 km = \(400 \times 10^3\) m
- Diameter of the astronaut's pupil (d) = 5 mm = \(5 \times 10^{-3}\) m
- Wavelength of visible light (λ) = 500 nm = \(500 \times 10^{-9}\) m

2. **Use the Resolving Power Formula:**
The formula for resolving power (R_p) is given by:
\[ R_p = \frac{1.22 \cdot \lambda \cdot D}{d} \]

3. **Substitute the Values into the Formula:**
Plugging in the values we have:
\[ R_p = \frac{1.22 \cdot (500 \times 10^{-9}) \cdot (400 \times 10^3)}{5 \times 10^{-3}} \]

4. **Calculate the Numerator:**
First, calculate the numerator:
\[ 1.22 \cdot 500 \times 10^{-9} \cdot 400 \times 10^3 = 1.22 \cdot 500 \cdot 400 \cdot 10^{-6} \]
\[ = 1.22 \cdot 200000 = 244000 \text{ (approximately)} \]

5. **Calculate the Denominator:**
The denominator is:
\[ 5 \times 10^{-3} = 0.005 \]

6. **Final Calculation:**
Now, divide the numerator by the denominator:
\[ R_p = \frac{244000}{0.005} = 48800000 \text{ (approximately)} \]

7. **Convert to Meters:**
Since the resolving power is in meters, we can express it as:
\[ R_p \approx 48.8 \text{ m} \]

8. **Conclusion:**
The astronaut will be able to resolve linear objects of size approximately 50 meters.

### Final Answer:
The astronaut will be able to resolve linear objects of size about **50 meters**.