A spy satellite is in orbit at a distance of `1.0 xx 10^(6)` m above the ground. It carries a telescope that can resolve the two rails of a railroad track that are 1.4 m apart using light of wavelength 600 nm. Which one of the following statements best describes the diameter of the lens in the telescope?

Options

A. It is less than 0.14 m.
B. It is greater than 0.14 m and less than 0.23 m.
C. It is greater than 0.23 m and less than 0.35 m.
D. It is greater than 0.52 m.

Answer

To solve the problem, we need to determine the diameter of the lens in the telescope that can resolve two points (the rails of a railroad track) that are 1.4 m apart, using light of wavelength 600 nm. We will use the formula for the resolving power of a circular aperture, which is given by:

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

where:
- \(\theta\) is the angular resolution in radians,
- \(\lambda\) is the wavelength of light (600 nm = \(600 \times 10^{-9}\) m),
- \(D\) is the diameter of the lens.

### Step 1: Calculate the angular resolution \(\theta\)

The angular resolution \(\theta\) can also be expressed in terms of the distance to the object and the separation between the two points:

\[ \theta = \frac{y}{R} \]

where:
- \(y\) is the separation between the two points (1.4 m),
- \(R\) is the distance from the telescope to the object (1.0 x \(10^6\) m).

Substituting the values:

\[ \theta = \frac{1.4 \, \text{m}}{1.0 \times 10^6 \, \text{m}} = 1.4 \times 10^{-6} \, \text{radians} \]

### Step 2: Set the two expressions for \(\theta\) equal to each other

Now we can set the two expressions for \(\theta\) equal to each other:

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

Substituting the known values:

\[ \frac{1.22 \times 600 \times 10^{-9}}{D} = \frac{1.4}{1.0 \times 10^6} \]

### Step 3: Solve for \(D\)

Now, we can rearrange the equation to solve for \(D\):

\[ D = \frac{1.22 \times 600 \times 10^{-9} \times 1.0 \times 10^6}{1.4} \]

Calculating the numerator:

\[ 1.22 \times 600 \times 10^{-9} \times 1.0 \times 10^6 = 1.22 \times 600 \times 10^{-3} = 732 \]

Now, substituting back into the equation for \(D\):

\[ D = \frac{732}{1.4} \approx 522.86 \, \text{m} \]

### Step 4: Conclusion

The diameter of the lens in the telescope is approximately 0.523 m, which is greater than 0.52 m.