191k views
1 vote
What is the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth? Assume λ = 506 nm and a pupil diameter is 4.90 mm. (In this problem, you may use the Rayleigh criterion for the limiting angle of resolution of an eye.)

User Tsbertalan
by
3.9k points

2 Answers

5 votes

Final answer:

The approximate size of the smallest object that astronauts can resolve by eye when orbiting 250 km above the Earth is approximately 628 meters.

Step-by-step explanation:

The approximate size of the smallest object on Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth can be determined using the Rayleigh criterion. According to the Rayleigh criterion, the resolution of the eye is determined by the angle of resolution, which is given by:

angle of resolution = 1.22 * (wavelength / pupil diameter)

Using the given values, we can calculate the angle of resolution:

angle of resolution = 1.22 * (506 nm / 4.90 mm) = 1.2611957657 x 10^-3 radians

To determine the approximate size of the smallest object, we need to find the linear size corresponding to this angular resolution at a distance of 250 km:

linear size = 2 * distance * tan(angle of resolution)

linear size = 2 * 250000 m * tan(1.2611957657 x 10^-3 radians) = 628.1079000855 m

Therefore, the approximate size of the smallest object that astronauts can resolve by eye when they are orbiting 250 km above the Earth is approximately 628 meters.

User Venu Immadi
by
4.1k points
5 votes

Answer:

the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m

Step-by-step explanation:

Using Rayleigh criterion for the limiting angle of resolution of an eye


\theta = (1.22\lambda )/(D ) \\ \\ \theta = (1.22*506 *10^(-9) )/(4.90*10^(-3)m)


\theta = 1.2598*10^(-4) rad


\theta = 125.98*10^(-6) \ rad

Thus; the separation between the two sources is expressed as:


\theta = (y)/(L) \\ \\ y = L \theta \\ \\ y = (250*10^3 )(125.98*10^(-6) \ rad) \\ \\ y = 31.495 \ m

Thus; the approximate size of the smallest object on the Earth that astronauts can resolve by eye when they are orbiting 250 km above the Earth is y = 31.495 m

User Alexfernandez
by
3.2k points