Chapter 38, Problem 38.59AP

The Very Large Array (VLA) is a set of 27 radio telescope dishes in Catron and Socorro counties, New Mexico (Fig. P37.37). The antennas can be moved apart on railroad tracks, and their combined signals give the resolving power of a synthetic aperture 36.0 km in diameter. (a) If the detectors are tuned to a frequency of 1.40 GHz, what is the angular resolution of the VLA? (b) Clouds of interstellar hydrogen radiate at the frequency used in part (a). What must be the separation distance of two clouds at the center of the galaxy, 26 000 light-years away, if they are to be resolved? (c) What If? As the telescope looks up, a circling hawk looks down. Assume the hawk is most sensitive to green light having a wavelength of 500 nm and has a pupil of diameter 12.0 mm. Find the angular resolution of the hawk’s eye. (d) A mouse is on the ground 30.0 m below. By what distance must the mouse’s whiskers be separated if the hawk can resolve them?

Figure P37.37

To determine

The angular resolution of the Very Large Array (VLA).

Answer to Problem 38.59AP

The angular resolution of the VLA is $1.50\text{arc}\text{seconds}$.

Explanation of Solution

Given info: The diameter of the aperture is $36.0\text{km}$ and the antennas are tuned to frequency at $1.40\text{GHz}$.

The formula to calculate the angular resolution is,

${\theta }_{\min }=1.22\frac{\lambda }{D}$ (1)

Here,

${\theta }_{\min }$ is the minimum angular resolution.

$\lambda$ is wavelength of the signal.

$D$ is the aperture diameter.

First calculate $\lambda$, the formula to calculate the wavelength is ,

$\lambda =\frac{\text{v}}{f}$ (2)

Here,

$\text{v}$ is the velocity of the signal.

$f$ is the frequency of the signal.

Substitute $3.00\times {10}^8\text{m/s}$ for $\text{v}$ and $1.40\text{GHz}$ for $f$ in equation (2).

$\begin{array}{c}\lambda =\frac{\text{v}}{f}\\ =\frac{3.00\times {10}^8\text{m/s}}{1.40\text{GHz}\times \left(\frac{\text{1}{\text{s}}^{-1}}{{10}^{-9}\text{GHz}}\right)}\\ =0.214\text{m}\end{array}$ (3)

Thus, the wavelength is $0.214\text{m}$.

From equation (3), substitute $0.214\text{m}$ for $\lambda$ in equation (1).

$\begin{array}{c}{\theta }_{\min }=\left(1.22\right)\left(\frac{0.214}{36\text{km}}\right)\\ =\left(1.22\right)\left(\frac{0.214\text{m}}{36\text{km}\times \frac{1\text{m}}{{10}^{-3}\text{km}}}\right)\\ =7.26\times {10}^{-6}\text{rad}\\ \text{=}7.26\text{μ rad}\end{array}$

Convert ${\theta }_{\text{min}}$ in terms of arc seconds.

$\begin{array}{c}{\theta }_{\min }=\left(7.26\text{μrad}\right)\left(\frac{180\times 60\times 60}{\pi }\text{s}\right)\\ =1.50\text{arc seconds}\end{array}$

Conclusion:

Therefore, the angular resolution is $7.26\text{μ rad}$ $\approx$ $1.50\text{arc seconds}$.

To determine

The separation between the interstellar hydrogen clouds at the centre of the galaxy if they are to be resolved.

Answer to Problem 38.59AP

The separation between the clouds if they are $26000\text{light years}$ away from each other is $0.189\text{light years}$.

Explanation of Solution

Given info: The distance between the clouds is $26000\text{light years}$.

The formula to calculate the minimum separation between the clouds so that they can be well resolved is,

$d={\theta }_{\min }L$ (4)

Here,

${\theta }_{\min }$ is the minimum angular resolution of the telescope.

$d$ is the minimum separation between the clouds to have resolved image.

$L$ is the distance between the clouds

Substitute $7.26\times {10}^{-6}\text{rad}$ for ${\theta }_{\min }$ part (a), and $26000\text{light years}$ for $L$ in equation (4)

$\begin{array}{c}d={\theta }_{\min }L\\ =7.26\times {10}^{-6}\text{rad}\times 26000\text{light years}\\ =0.189\text{light years}\end{array}$

Conclusion:

Therefore, the minimum separation between the two clouds required to have resolved image is $0.189\text{light years}$.

To determine

The angular resolution of the hawk’s eye.

Answer to Problem 38.59AP

The angular resolution of the hawk’s eye is $50.8\text{μrad}$. or $10.5\text{arc second}$.

Explanation of Solution

Given info: The wavelength of the light to which hawk’s eye is sensitive is green light. The wavelength of the green light is $500\text{nm}$ and the pupil diameter of the hawk’s eye is $12\text{mm}$.

From equation (1) the formula to calculate the angular resolution is,

${\theta }_{\min }=1.22\frac{\lambda }{D}$

Here,

$\lambda$ is the wavelength of the light hawk eye is sensitive.

$D$ is the pupil diameter.

${\theta }_{\min }$ is the minimum angular resolution of the hawks eye.

Substitute $500\text{nm}$ for $\lambda$ and $12\text{mm}$ for $D$ in the above equation,

$\begin{array}{c}{\theta }_{\min }=1.22\frac{\lambda }{D}\\ =\left(1.22\right)\left(\frac{500\text{nm}\times \frac{\text{1}\text{m}}{{10}^9\text{nm}}}{12.0\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}}\right)\\ =50.8\times {10}^{-6}\text{rad}\\ \text{=}50.8\text{μrad}\end{array}$

Convert ${\theta }_{\min }$ into seconds.

$\begin{array}{c}{\theta }_{\min }=\left(50.8\text{μrad}\right)\left(\frac{180\times 60\times 60}{\pi }\text{s}\right)\\ =10.5\text{arc seconds}\end{array}$

Conclusion:

Therefore, angular resolution of the hawk’s eyes is $10.5\text{arc seconds}$.

To determine

The separation between the mouse whiskers so that it is well resolved for the hawk’s eyes.

Answer to Problem 38.59AP

The separation between mouse whiskers so that it is well resolved for the hawk’s eyes is $1.52\text{mm}$

Explanation of Solution

Given info: The distance between the hawk and the mouse is $30.0\text{m}$

The formula to calculate the minimum separation between the mouse whiskers so that they can be well resolved is,

$d={\theta }_{\min }L$ (4)

Here,

$d$ is the minimum separation between the whiskers.

$L$ is the distance between the hawk and the mouse.

${\theta }_{\min }$ is the minimum angular resolution of the hawks eye.

Substitute $50.8\times {10}^{-6}\text{rad}$ for ${\theta }_{\min }$ from part (c), and $30.0\text{m}$ for $L$ in equation (4)

$\begin{array}{c}d={\theta }_{\min }L\\ =50.8\times {10}^{-6}\text{rad}\times 30.0\text{m}\\ \text{=1}\text{.52}\times {\text{10}}^{-3}\text{m}\\ \text{=1}\text{.52}\times {\text{10}}^{-3}\text{m}\times \frac{\text{1 mm}}{{10}^{-3}\text{m}}\end{array}$

$d=1.52\text{mm}$

Conclusion:

Therefore, the minimum separation between the two mouse whiskers so that hawk cans e them is $1.52\text{mm}$.