Question

The telescopes on some commercial surveillance satellites can resolve objects on the ground as small as across (see Google Earth), and the telescopes on military surveillance satellites reportedly can resolve objects as small as $\mathbf{10}\mathbf{}\mathit{c}\mathit{m}$ across. Assume first that object resolution is determined entirely by Rayleigh’s criterion and is not degraded by turbulence in the atmosphere. Also assume that the satellites are at a typical altitude of $\mathbf{400}\mathbf{nm}$and that the wavelength of visible light is role="math" localid="1663028559183" $\mathbf{550}\mathbf{nm}$. What would be the required diameter of the telescope aperture for (a) role="math" localid="1663028596951" $\mathbf{85}\mathbf{cm}$resolution and (b) role="math" localid="1663028635287" $\mathbf{10}\mathbf{cm}$resolution? (c) Now, considering that turbulence is certain to degrade resolution and that the aperture diameter of the Hubble Space Telescope is role="math" localid="1663028673584" $\mathbf{2}\mathbf{.}\mathbf{4}\mathbf{}\mathit{m}$, what can you say about the answer to (b) and about how the military surveillance resolutions are accomplished?

Short answer

1. The diameter of telescope is $0.32m$.
2. The diameter of telescope is $2.7m$.
3. Military satellites do not use Hubble space Telescope-sized apparatus.

Step-by-step solution

Given data

Wavelength of light$\lambda =550\text{mm}$

Attitude of satellite $L=400\text{km}$

Definition of Rayleigh criterion

The Rayleigh criteria specify the minimal distance between two light sources that must exist in order to resolve them into separate objects.

Concept used

Wavelength of light $\lambda =550\text{mm}$

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

Attitude of satellite $L=400\text{km}$

$$\begin{gathered} d=\left(400\text{km}\right)\left({10}^3\text{m/km}\right) \\ =400\times {10}^3\text{m} \end{gathered}$$

D is resolution of telescope,

d is diameter of telescope.

For circular aperture, use the condition

$\sin {\theta }_R=1.22\frac{\lambda }{d}$ And $\tan {\theta }_R=\frac{D}{L}$

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

$$\begin{gathered} 1.22\frac{\lambda }{d}=\frac{D}{L} \\ d=1.22\frac{\lambda L}{D} \end{gathered}$$

(a) Determining the required diameter of the telescope for 85 cm resolution

$$\begin{gathered} D=85\text{cm} \\ =\left(85\text{cm}\right)\left({10}^{-2}\text{m/cm}\right) \\ =85\times {10}^{-2}\text{m} \end{gathered}$$

$$\begin{gathered} d=\frac{\left(1.22\right)\left(550\times {10}^{-9}\text{m}\right)\left(400\times {10}^3\text{m}\right)}{\left(85\times {10}^{-2}\text{m}\right)} \\ =3157.6471\times {10}^{-4}m \\ =0.32\text{m} \end{gathered}$$

Hence, the diameter is 0.32 m

(b) Determining the required diameter of the telescope for 10 cm resolution

Here,

$$\begin{gathered} d=\frac{\left(1.22\right)\left(550\times {10}^{-9}\text{m}\right)\left(400\times {10}^3\text{m}\right)}{\left(0.1\text{m}\right)} \\ =2684000\times {10}^{-4}\text{m} \\ =2.684\text{m} \\ =2.7\text{m} \end{gathered}$$

Hence, the diameter is 2.7 m.

(c) Determining how the military surveillance resolutions are accomplished

Since the Hubble Telescope's equipment is smaller than the military telescope's apparatus, it cannot be utilised to resolve objects smaller than 10 cm .