Question

Estimate the linear separation of two objects on Mars that can just be resolved under ideal conditions by an observer on Earth (a) using the naked eye and (b) using the $\mathbf{200}\mathbf{}\mathbf{\text{in}}\mathbf{.}\left(=5.1\text{m}\right)$ Mount Palomar telescope. Use the following data: distance to Mars = $\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{\text{km}}$, diameter of pupil = $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{mm}$, wavelength of light $\mathbf{550}\mathbf{nm}$.

Short answer

1. The linear separation of two objects on Mars is$1.1\times {10}^4\text{km}$ .
2. The linear separation of two objects on Mars is $11\text{km}$.

Step-by-step solution

Introduction

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Calculation

Distance from the observer to the object

$$\begin{gathered} L=8\times {10}^7\text{km} \\ =8\times {10}^{10}\text{m} \end{gathered}$$

Wave length

$$\begin{gathered} \lambda =550\text{nm} \\ =\left(550\text{nm}\right)\left({10}^{-9}\text{m/nm}\right) \\ =550\times {10}^{-9}\text{m} \end{gathered}$$

Distance from the pupil eye

$$\begin{gathered} d=5\text{mm} \\ =\left(5\text{mm}\right)\left({10}^{-3}\text{m/mm}\right) \\ =5\times {10}^{-3}\text{m} \end{gathered}$$

D is separation between objects.

(a) Find the linear separation of two objects on Mars

We have the condition$\tan {\theta }_R=\frac{D}{L}$

And from criteria $\sin {\theta }_R=\frac{1.22\lambda }{d}$

As ${\theta }_R$is small $\tan {\theta }_R\approx \sin {\theta }_R\approx {\theta }_R$

Therefore,

$$\begin{gathered} \frac{D}{L}=\frac{1.22\lambda }{d} \\ D=\frac{1.22L\lambda }{d} \\ D=\frac{\left(1.22\right)\left(8\times {10}^{10}\text{m}\right)\left(550\times {10}^{-9}\text{m}\right)}{5\times {10}^{-3}\text{m}} \\ =1073.6\times {10}^4\text{m} \\ =\left(1.1\times {10}^7\text{m}\right)\left({10}^{-3}\text{km/m}\right) \\ =1.1\times {10}^4\text{km} \end{gathered}$$

Hence, the linear separation of two objects on Mars is $1.1\times {10}^4\text{km}$.

(b) Find the linear separation of two objects on Mars

In this case $d=5.1\text{m}$;

$$\begin{gathered} D=\frac{\left(1.22\right)\left(8\times {10}^{10}\text{m}\right)\left(550\times {10}^{-9}\text{m}\right)}{5.1\text{m}} \\ =1052.549\times {10}^1\text{m} \\ =\left(10525.49\text{m}\right)\left({10}^{-3}\text{km/m}\right) \\ =11\text{km} \end{gathered}$$

Hence, the linear separation of two objects on Mars is $11km$.