Assume that Rayleigh’s criterion gives the limit of resolution of an astronaut’s eye looking down on Earth’s surface from a typical space shuttle altitude of $\mathbf{400}\mathbf{km}$. (a) under that idealized assumption, estimate the smallest linear width on Earth’s surface that the astronaut can resolve. Take the astronaut’s pupil diameter to be $\mathbf{5}\mathbf{}\mathbf{mm}$and the wavelength of visible light to be $\mathbf{550}\mathbf{nm}$. (b) Can the astronaut resolve the Great Wall of China (Fig. 36-40), which is more than$\mathbf{3000}\mathbf{km}$ long ,$\mathbf{5}\mathbf{to}\mathbf{10}\mathbf{m}$ thick at its base, $\mathbf{\text{4\hspace{0.17em}}}\mathbf{m}$thick at its top, and $\mathbf{\text{8\hspace{0.17em}}}\mathbf{m}$ in height? (c) Would the astronaut be able to resolve any unmistakable sign of intelligent life on Earth’s surface?

Shuttle altitude, $\mathrm{L}$ = $400\text{\hspace{0.17em}km}$

Thepupil diameter, $\mathrm{d}$ = $5\text{\hspace{0.17em}mm}$

Wavelength of light, $\mathrm{\lambda }$ = $550\text{\hspace{0.17em}nm}$

The Great Wall of China,

Length = $3000\text{\hspace{0.17em}km}$

Base thickness = $5\text{\hspace{0.17em}to\hspace{0.17em}}10\text{\hspace{0.17em}m}$

Top thickness = $4\text{\hspace{0.17em}m}$

Height = $8\text{\hspace{0.17em}m}$

The Rayleigh criterion specifies the minimum separation between two light sources that may be resolved into distinct objects. When a point source, such as a star, is observed through a telescope with a circular aperture, the image is not a point source – it is a disk surrounded by a number of very faint rings.

Since the diffraction is due to a circular aperture, then we will use the Rayleigh criterion that is:

$\mathrm{\theta }=\frac{1.22\mathrm{\lambda }}{\mathrm{d}}$

Here, $\lambda \text{\hspace{0.17em}}=$wavelength of light,

$\mathrm{d}=$diameter of the pupil

The angular separation $\mathrm{\theta }$ is given in equation$36-16$that is:

$\mathrm{\theta }=\frac{\mathrm{D}}{\mathrm{L}}$

Here, $\mathrm{D}=$diameter of the structure

$\mathrm{L}=$ shuttle altitude

Solving for $\mathrm{D}$using both equations:

$\mathrm{D}=\frac{1.22\mathrm{\lambda L}}{\mathrm{d}}$

Substituting the given values:

$\begin{array}{l}\mathrm{D}=\frac{1.22(550\times {10}^{-9})(400\times {10}^3)}{5\times {10}^{-3}}\\ \mathrm{D}=53.68\text{\hspace{0.17em}m}\\ \boxed{\mathrm{D}=54\text{\hspace{0.17em}m}}\end{array}$

Hence, the linear diameter of the structure producing the diffraction is $54\text{\hspace{0.17em}}\mathrm{m}$.

No, the astronaut cannot resolve at its base and at its top. But he can resolve the great wall along its length which is 300 km. Because the smallest linear width the astronaut can resolve is $54\text{\hspace{0.17em}}\mathrm{m}$.

Therefore, the astronaut can resolve object that has a minimum of $54\text{\hspace{0.17em}}\mathrm{m}$and higher.

Hence, the astronaut cannot resolve the Great Wall of China.

The astronaut can resolve anything that has a width that is more thanon the opposite side of the sunlight since intelligent life can be noticed with the artificial light.