Question

An astronaut is looking down on earth’s surface from a space shuttle at an altitude of $400km$. Assuming that the astronaut’s pupil diameter is $5mm$ and the wavelength of visible light is $500nm$, the astronaut will be able to resolve linear objects of the size of about:
$A)0.5m$
$B)5m$
$C)50m$
$D)500m$

Answer 1

Hint: Resolving power of a human eye measures the ability of the human eye to distinctly differentiate between two lines. Resolving power of an astronaut’s eye, looking down to the earth’s surface measures the average size of linear objects, the astronaut can see distinctly from the space. Resolving power is proportional to the wavelength of light as well as the distance of the eye from the linear object.
Formula used:
$R=1.22\dfrac{\lambda }{b}\times D$

Complete answer:
Resolving power of a human eye is a measure of the ability of the human eye to clearly differentiate between two given lines. Resolving power of an astronaut’s eye calculates the size of linear objects, the astronaut can see clearly from space. Mathematically, resolving power of an eye lens is given by
$R=1.22\dfrac{\lambda }{b}\times D$
where
$R$ is the resolving power of an eye lens
$\lambda $ is the wavelength of light
$b$ is the diameter of eye lens
$D$ is the distance of a linear object from the eye lens
Let this be equation 1.
Coming to our question, an astronaut is away from the earth’s surface at a distance of $400km$. The diameter of the astronaut’s eye lens is given as $5mm$.
If we take the wavelength of light as $500nm$, as provided in the question, the resolving power of the astronaut’s eye lens is given by
$R=1.22\dfrac{\lambda }{b}\times D\Rightarrow R=1.22\left( \dfrac{500nm}{5mm} \right)\times 500km$
where
$R$ is the resolving power of the astronaut’s eye lens
$\lambda =500nm$ is the wavelength of light
$b=5mm$ is the diameter of the astronaut’s pupil
$D=400km$ is the distance of a linear object on the earth’s surface, from the astronaut
Let this be equation 2.
Simplifying equation 2, we have
$R=1.22\left( \dfrac{500nm}{5mm} \right)\times 400km\Rightarrow 1.22\times \left( \dfrac{500\times {{10}^{-9}}m}{5\times {{10}^{-3}}m} \right)\times 400\times {{10}^{3}}m=48.8m\approx 50m$
Therefore, the astronaut will be able to resolve objects of size equal to $50m$.

Hence, the correct answer is option $C$.

Note:
These types of questions check a student’s ability to deal with numbers and conversions. Students need to be thorough with conversion formulas. Conversion formulas used in the above solution are as follows.
$1mm={{10}^{-3}}m$
$1nm={{10}^{-9}}m$
$1km={{10}^{3}}m$
Also, it is always advisable to convert units into the SI unit system because it is the easiest and the safest way of doing calculations.