Question

The Very Large Array (VLA) is a set of 27 radio telescope dishes in Catron and Socorro counties, New Mexico (see figure below). 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 1.17 GHz, what is the angular resolution of the VLA?
8.689 µrad

(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, 24 000 light-years away, if they are to be resolved?
ly

(c) As the telescope looks up, a circling hawk looks down. Assume the hawk is most sensitive to green light having a wavelength 500 nm and has a pupil of diameter 10.8 mm. Find the angular resolution of the hawk's eye. (Assume the hawk's visual acuity is limited only by Rayleigh's criterion. )
µrad

(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?
mm

Answer 1

Question A:

The angular resolution is given by the formula:

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

Where:

• $\theta$ is the angular resolution (radians)
• $\lambda$ is the wavelength of light (m)
• D is the diameter of the lens aperture. (m)

Since we dont have the wavelength we calculate it using:

$\lambda =\frac{V}{f}$

Where:

• V is the speed of light (3x10⁸ m/s)
• f is the frequency (1/s)

So, we can do our calculations:

$\theta=1.22\frac{V}{f.D}\rightarrow \theta=1.22\frac{3\times10^8\frac{m}{s}}{(1.17\times10^9\frac{1}{s}).(3.6\times10^4m)}\rightarrow \mathbf{\theta=8.689\mu rad}$

Question B:

If we are being asked for the separation distance of two clouds at the center of the galaxy that radiate at the frequency used in part (a) and the distance to the galaxy is given we should use this formula:

$\theta=\frac{d}{L}$

Where:

• $\theta$ is the angular resolution (radians)
• L distance to the galaxy
• d distance between the two clouds.

The we solve for d:

$\theta=\frac{d}{L}\rightarrow d=\theta.L\rightarrow d= (8.689\times10^{-6} rad).(24,000 ly)\rightarrow \mathbf{d=0.2085ly}$

Question C:

We are again being asked for an angular resolution. This time is a hawk's eye and all the informaton is already given so all we have to do is fill the formula with the given data:

$\theta=1.22\frac{\lambda }{D}\rightarrow \theta=1.22 \left (\frac{500\times10^{-9}m}{10.8\times10^{-3}m} \right ) \rightarrow \theta=5.648\times10^{-5}rad$

$\mathbf{\rightarrow\theta=56.48 \mu rad }$

Question D:

Just like question B we are asked a distance using the angular resolution of the hawk's eye when the mouse is 30m below. So, we will use same procedure as in question B.

$d=\theta.L\rightarrow d= (5.648\times10^{-5}rad).(30m)\rightarrow d=1.6944\times10^{-3}m$

$\rightarrow \mathbf{d=1.6944mm}$

And those are the answers. I had fun solving this.

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