Question

36.3 diffraction by a circular aperture
the radio telescope at the arecibo observatory in puerto rico had a diameter of 1,000 ft (305 m). when this telescope was focusing on a signal of 9.26 ghz, what would have to be the diameter (in mm) of a visible light telescope that is focusing on light of wavelength 565 nm so that both telescopes have the same resolving power? assume the resolving power is diffraction limited.
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Explanation:

Step1: Recall Rayleigh Criterion

The Rayleigh criterion for diffraction - limited resolving power is given by $\theta = 1.22\frac{\lambda}{D}$, where $\theta$ is the minimum angular resolution, $\lambda$ is the wavelength of the radiation, and $D$ is the diameter of the aperture (telescope dish for radio, objective for optical). For the two telescopes to have the same resolving power, their $\theta$ must be equal. So, $1.22\frac{\lambda_1}{D_1}=1.22\frac{\lambda_2}{D_2}$, and we can cancel out 1.22, giving $\frac{\lambda_1}{D_1}=\frac{\lambda_2}{D_2}$.

Step2: Find $\lambda_1$ (wavelength of radio signal)

The frequency $f$ of the radio signal is $f = 9.26\ \text{GHz}=9.26\times 10^{9}\ \text{Hz}$. The speed of light $c = 3\times 10^{8}\ \text{m/s}$. Using the relation $c=\lambda f$, we can solve for $\lambda_1$: $\lambda_1=\frac{c}{f}$. Substituting the values, $\lambda_1=\frac{3\times 10^{8}\ \text{m/s}}{9.26\times 10^{9}\ \text{Hz}}\approx 0.0324\ \text{m}$.

Step3: List known values for optical telescope

$\lambda_2 = 565\ \text{nm}=565\times 10^{-9}\ \text{m}$, $D_1 = 305\ \text{m}$. We need to find $D_2$. From $\frac{\lambda_1}{D_1}=\frac{\lambda_2}{D_2}$, we can re - arrange to $D_2=\frac{\lambda_2 D_1}{\lambda_1}$.

Step4: Substitute values into the formula

Substitute $\lambda_1 = 0.0324\ \text{m}$, $\lambda_2 = 565\times 10^{-9}\ \text{m}$, and $D_1 = 305\ \text{m}$ into the formula for $D_2$:

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Convert this to millimeters: since $1\ \text{m}=1000\ \text{mm}$, $D_2 = 5.32\times 10^{-3}\times1000\ \text{mm}=5.32\ \text{mm}$.

Answer:

\boxed{5.32}