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Chapter 36: Problem 41

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter of about \(8000 \mathrm{~km}\). When this radio telescope is focusing radio waves of wavelength \(2.0 \mathrm{~cm}\), what would have to be the diameter of the mirror of a visible-light telescope focusing light of wavelength \(550 \mathrm{nm}\) so that the visible-light telescope has the same resolution as the radio telescope?

Short Answer

Expert verified
The required diameter of the mirror of a visible-light telescope so that it has the same resolution as the radio telescope is approximately 12.03 meters.

Step by step solution

01

Understand the Problem and Gather Provided Information

In order to solve this problem, it is important to first understand the given information. The diameter of the VLBA radio telescope is given as 8000 km and the wavelength of the radio waves it is focusing is 2.0 cm. We are asked to find the diameter of the visible-light telescope that focuses light of wavelength 550 nm while having the same resolution.
02

Convert All Wavelengths to the Same Units

To solve the problem, it must be ensured that the wavelengths of the radio and visible-light waves are in same units. Radio wavelength is given in cm and visible-light wavelength in nm. Let's first convert 2.0 cm into nm. As 1 cm = 10^-7 nm, after conversion we get \(2.0 \times 10^{7}\) nm.
03

Apply the Telescope Resolution Formula and Solve for the Diameter of the Visible-Light Telescope Mirror

The telescope resolution formula is given by Resolution = 1.22 * (wavelength / diameter). As the resolution of both telescopes should be the same, we have 1.22 * ((2 cm) / (8000 km)) = 1.22 * ((550 nm) / D), where D is the diameter of the visible-light telescope mirror. After substituting the given values and making the appropriate unit conversions (into kilometers for D), we can solve for D. After resolving all these calculations, the diameter of the visible-light telescope mirror is found to be approximately 0.01203 km or 12.03 meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

VLBA (Very Long Baseline Array)
The VLBA is a sophisticated radio astronomy observatory. It combines data from multiple radio telescopes spread across long distances to simulate a giant telescope. This technique is called interferometry. It enables astronomers to see details in radio waves from distant cosmic objects with extraordinary precision.

  • Equivalent Diameter: The combined array of telescopes acts like a single telescope with an enormous diameter, in this case, about 8000 km.
  • High-Resolution Imagery: VLBA can create very detailed images by enhancing its resolving power.
Being able to achieve such resolution helps scientists study phenomena like black holes, quasars, and other distant galaxies in detail.
Radio Waves
Radio waves are a type of electromagnetic radiation with long wavelengths. They are used in astronomy to explore and understand the universe. Unlike light waves, radio waves can penetrate through dust and other opaque materials, making it easier to observe hidden cosmic events.

  • Wavelengths of Radio Waves: They vary from around 1 millimeter to over 100 kilometers.
  • Advantages in Astronomy: Radio waves offer insights into the dynamics and compositions of celestial bodies.
In the context of the VLBA, the specific wavelength used is 2.0 cm. This allows for excellent resolution when collecting data from vast astronomical distances.
Visible-Light Telescope
Visible-light telescopes observe the universe in wavelengths that are visible to the human eye. These telescopes help us see stars and galaxies as we perceive them naturally.

  • Wavelength Range: Typically, visible light ranges from about 380 nm to 740 nm.
  • Function: They gather and focus light with lenses or mirrors to form clear images.
For this exercise, the visible-light telescope focuses light with a wavelength of 550 nm. To match the resolution of the VLBA's radio telescope, it requires a specifically calculated mirror diameter, around 12.03 meters.
Wavelength Conversion
Understanding wavelengths and converting between units are crucial in resolving differences between telescopes. It ensures precision in calculations and comparisons.

  • Unit Conversion: Sometimes necessary to convert from centimeters to nanometers, or vice versa, to standardize units across different calculations.
  • Application in Telescopes: Helps match resolutions between different types of telescopes focusing on different wavelengths.
In this exercise, the radio wavelength of 2.0 cm is converted to 20,000,000 nm to facilitate direct comparison with the visible-light telescope's wavelength of 550 nm.

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Most popular questions from this chapter

Coherent electromagnetic waves with wavelength \(\lambda=500 \mathrm{nm}\) pass through two identical slits. The width of each slit is \(a,\) and the distance between the centers of the slits is \(d=9.00 \mathrm{~mm}\). (a) What is the smallest possible width \(a\) of the slits if the \(m=3 \max -\) imum in the interference pattern is not present? (b) What is the next larger value of the slit width for which the \(m=3\) maximum is absent?

Your physics study partner tells you that the width of the central bright band in a single-slit diffraction pattern is inversely proportional to the width of the slit. This means that the width of the central maximum increases when the width of the slit decreases. The claim seems counterintuitive to you, so you make measurements to test it. You shine monochromatic laser light with wavelength \(\lambda\) onto a very narrow slit of width \(a\) and measure the width \(w\) of the central maximum in the diffraction pattern that is produced on a screen \(1.50 \mathrm{~m}\) from the slit. (By "width," you mean the distance on the screen between the two minima on either side of the central maximum.) Your measurements are given in the table. $$ \begin{array}{l|llllllllr} a(\mu \mathrm{m}) & 0.78 & 0.91 & 1.04 & 1.82 & 3.12 & 5.20 & 7.80 & 10.40 & 15.60 \\ \hline w(\mathrm{~m}) & 2.68 & 2.09 & 1.73 & 0.89 & 0.51 & 0.30 & 0.20 & 0.15 & 0.10 \end{array} $$ (a) If \(w\) is inversely proportional to \(a\), then the product \(a w\) is constant, independent of \(a\). For the data in the table, graph \(a w\) versus \(a\). Explain why \(a w\) is not constant for smaller values of \(a\). (b) Use your graph in part (a) to calculate the wavelength \(\lambda\) of the laser light. (c) What is the angular position of the first minimum in the diffraction pattern for (i) \(a=0.78 \mu \mathrm{m}\) and (ii) \(a=15.60 \mu \mathrm{m} ?\)

Quasars, an abbreviation for quasi-stellar radio sources, are distant objects that look like stars through a telescope but that emit far more electromagnetic radiation than an entire normal galaxy of stars. An example is the bright object below and to the left of center in Fig. \(\mathrm{P} 36.58 ;\) the other elongated objects in this image are normal galaxies. The leading model for the structure of a quasar is a galaxy with a supermassive black hole at its center. In this model, the radiation is emitted by interstellar gas and dust within the galaxy as this material falls toward the black hole. The radiation is thought to emanate from a region just a few light-years in diameter. (The diffuse glow surrounding the bright quasar shown in Fig. \(\mathrm{P} 36.58\) is thought to be this quasar's host galaxy.) To investigate this model of quasars and to study other exotic astronomical objects, the Russian Space Agency has placed a radio telescope in a large orbit around the earth. When this telescope is \(77,000 \mathrm{~km}\) from earth and the signals it receives are combined with signals from the ground-based telescopes of the VLBA, the resolution is that of a single radio telescope \(77,000 \mathrm{~km}\) in diameter. What is the size of the smallest detail that this arrangement can resolve in quasar \(3 \mathrm{C} 405,\) which is \(7.2 \times 10^{8}\) light-years from earth, using radio waves at a frequency of \(1665 \mathrm{MHz}\) ? (Hint: Use Rayleigh's criterion.) Give your answer in lightyears and in kilometers.

If you can read the bottom row of your doctor鈥檚 eye chart, your eye has a resolving power of 1 arcminute, equal to \(\frac{1}{60}\) degree. If this resolving power is diffraction limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume \(\lambda=550 \mathrm{nm}\).

The intensity of light in the Fraunhofer diffraction pattern of a single slit is given by Eq. (36.5). Let \(\gamma=\beta / 2\). (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma\). (b) Determine the two smallest positive values of \(\gamma\) that are solutions of this equation. (Hint: You can use a trial-and-error procedure. Guess a value of \(\gamma\) and adjust your guess to bring \(\tan \gamma\) closer to \(\gamma\). A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.) (c) What are the positive values of \(\gamma\) for the first, second, and third minima on one side of the central maximum? Are the \(\gamma\) values in part (b) precisely halfway between the \(\gamma\) values for adjacent minima? (d) If \(a=12 \lambda,\) what are the angles \(\theta\) (in degrees) that locate the first minimum, the first maximum beyond the central maximum, and the second minimum?
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