Question

What is the approximate size of the smallest object on the earth that astronauts can resolve by eye when they are orbiting $250km$ above the earth? Assume$\lambda = 500nm$ and a pupil diameter of$5.00mm$?

Answer 1

In order to solve this question, we are going to firstly calculate the resolving power of the object from its size and distance. Then, the same is calculated by using the formula for resolving power containing the wavelength and the diameter of the viewer's lens. They are compared to calculate the length.

Formula used:
The resolving power of an object is given by the formula
$R.P. = \dfrac{{1.22\lambda }}{d}$
Where,$\lambda$ is the wavelength and$d$ is the diameter of the pupil.
If the length of the object is$l$, then the resolving power of that object is given as:
$R.P. = \dfrac{l}{s}$

Complete step-by-step solution:
In this question, we are given the case of Earth. It is given that an object is orbiting above a height of$250km$from the earth’s surface.
Thus, if the length of the smallest object is considered to be$l$, then the resolving power of that object is given as:
$R.P. = \dfrac{l}{s} = \dfrac{l}{{250}}$
Now, another way to find the resolving power of an object is given by the formula
$R.P. = \dfrac{{1.22\lambda }}{d}$
Where,$\lambda$ is the wavelength and$d$ is the diameter of the pupil.
Now, putting the values of the wavelength and the diameter in the above equation
$R.P. = \dfrac{{1.22 \times 500 \times {{10}^{ - 9}}}}{{5 \times {{10}^{ - 3}}}} = 1.22 \times {10^{ - 4}}$
Equating the two values of the resolving powers, we get
$\dfrac{l}{{250}} = 1.22 \times {10^{ - 4}}$
Thus, calculating the length, we get
$l = 1.22 \times {10^{ - 4}} \times 250 = 305 \times {10^{ - 4}} = 3.05cm$
Hence, the approximate size of the smallest object on the earth that astronauts can resolve by eye when they are orbiting $250km$ above the earth is equal to$3.05cm$.

Note: The resolving power of an objective lens is measured by its ability to differentiate two lines or points in an object. The human eye has an angular resolution of about $0.02$ degrees or $0.0003$radians which enables us to distinguish things that are $30$ centimeters apart at a distance of$1$ kilometer.