Question

If Superman really had x-ray vision at $\mathbf{0}\mathbf{.}\mathbf{10}\mathbf{nm}$ wavelength and a $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{mm}$ pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{cm}$ to do this?

Short answer

The altitude using the angular separation formula is $\text{1.6}\times {\text{10}}^3\text{km}$.

Step-by-step solution

Given data

Wavelength of x-Ray,$\lambda =0.10\text{nm}$

The pupil diameter $d=4.00\text{mm}$

The resolving separation $D=5.0\text{cm}$

Definition of X-rays

X-rays is defined as the form of electromagnetic radiation, similar to light but of shorter wavelength and capable of penetrating solids and of ionizing gases. Such radiation having wavelengths in the range of approximately $\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{-}\mathbf{10}$Nanometers.

Determining the altitude using the angular separation formula

Let the altitude be 'L'

Since angular separation ${\varphi }_R\text{= 1.22}\frac{\lambda }{d}$

But ${\varphi }_R=\text{D/L}$

So $D=\frac{\text{1.22}\lambda L}{d}$

Or,

$$\begin{gathered} L=\frac{Dd}{1.22\lambda } \\ =\frac{5.0\times {10}^{-2}\text{m}\times 4.0\times {10}^{-3}\text{m}}{1.22\times 0.10\times {10}^{-9}\text{m}} \\ =1.639\times {10}^6\text{m} \\ L=\text{1.6}\times {\text{10}}^3\text{km} \end{gathered}$$

Hence, the altitude using the angular separation formula is $\text{1.6}\times {\text{10}}^3\text{km}$.