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Chapter 36: Problem 44

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 diameter of the visible-light telescope needs to be 4.4 meters.

Step by step solution

01

Understand the concept of resolution

The resolution of a telescope is determined by the formula: \( \theta = \frac{1.22 \cdot \lambda}{D} \), where \( \theta \) is the resolution, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the telescope's aperture. For two telescopes to have the same resolution, \( \theta \) must be equal for both, meaning that \( \frac{1.22 \cdot \lambda_{radio}}{D_{radio}} = \frac{1.22 \cdot \lambda_{light}}{D_{light}} \).
02

Set the equation for equal resolutions

Using the resolution formula, equate the two resolutions: \( \frac{1.22 \cdot 2.0 \times 10^{-2} \text{ m}}{8000 \times 10^{3} \text{ m}} = \frac{1.22 \cdot 550 \times 10^{-9} \text{ m}}{D_{light}} \). This equation allows us to find the diameter \( D_{light} \) for the visible-light telescope.
03

Solve for the diameter of the visible-light telescope

Rearrange the equation to solve for \( D_{light} \): \( D_{light} = \frac{1.22 \cdot 550 \times 10^{-9} \text{ m} \times 8000 \times 10^{3} \text{ m}}{1.22 \cdot 2.0 \times 10^{-2} \text{ m}} \). Simplify this expression to find \( D_{light} \).
04

Calculate the result

Calculating the above expression, we have: \( D_{light} = \frac{550 \times 10^{-9} \times 8000 \times 10^{3}}{2.0 \times 10^{-2}} = \frac{4.4 \text{ m}}\). Therefore, the diameter of the mirror of the visible-light telescope needs to be 4.4 meters.

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Key Concepts

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

Radio Telescopes
Radio telescopes are fascinating instruments used to study celestial objects by detecting radio waves emitted or reflected by them. Unlike traditional optical telescopes that rely on visible light, radio telescopes capture longer wavelengths ranging from a millimeter to several meters. This allows astronomers to explore things that would otherwise remain invisible.
One of the main advantages of radio telescopes is their ability to observe in all weather conditions and at any time of the day or night. This is possible because radio waves can penetrate clouds and are not hindered by sunlight.
  • Radio telescopes typically have large parabolic dishes that collect and focus the radio waves onto a receiver.
  • The Very Long Baseline Array (VLBA) mentioned is a notable example, combining multiple radio antennas across a massive distance鈥攎aking it equivalent to having an enormous single dish of about 8000 km!
  • By linking these telescopes together, they can achieve extremely high resolution, allowing astronomers to see fine details in distant objects.
Using radio telescopes, scientists have gained insights into phenomena like pulsars, galaxies, and the cosmic microwave background, expanding our understanding of the universe.
Visible-Light Telescopes
Visible-light telescopes are what most people picture when they think of telescopes. These devices capture and magnify light in the visible spectrum, which ranges from about 400 nm to 700 nm. They're primarily used to observe planets, stars, and galaxies by collecting the light that is naturally emitted or reflected by these objects.
There are different types of visible-light telescopes, but they generally fall into two categories:
  • Refracting Telescopes - Use lenses to bend and focus light; iconic for their long tubes and glass objectives.
  • Reflecting Telescopes - Utilize mirrors to gather light; these are often used in professional settings due to their ability to handle larger mirrors without the weight issues of lenses.
Visible-light telescopes are powerful tools in astronomical research, especially in studying the composition and motion of stars through spectroscopy. They require clear skies and are restricted by weather conditions. However, when appropriately utilized, they unlock views into our universe's brilliant and colorful aspects.
Wavelength Comparison
Wavelength comparison is crucial in understanding how different telescopes work and what they can observe. The wavelength of the light or waves used by a telescope greatly influences its design and the details it can resolve.
Radio waves have much longer wavelengths than visible light. For instance, in the provided exercise, the radio waves observed by the VLBA are 2 cm in wavelength, while visible light is just 550 nm. This is a significant difference:
1 cm = 10 mm = 10,000,000 nm 550 nm = 550 billionths of a meter
  • Because radio waves are longer, radio telescopes need to be much larger than visible-light telescopes to achieve similar resolution.
  • This size allows radio telescopes to distinguish between finer details despite the broader waves they work with.
  • Visible-light telescopes, working with much shorter wavelengths, can achieve high resolution with smaller apertures.
The key takeaway is that the wavelength directly affects a telescope's design requirements and capabilities. In creating telescopes that can achieve similar resolution with different wavelengths, engineers must carefully adjust their size and structure.

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Most popular questions from this chapter

A slit 0.240 \(\mathrm{mm}\) wide is illuminated by parallel light rays of wavelength 540 \(\mathrm{nm}\) . The diffraction pattern is observed on a screen that is 3.00 \(\mathrm{m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

A diffraction grating has 650 slits/mm. What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm.}\) )

A glass sheet is covered by a very thin opaque coating. In the middle of this sheet there is a thin scratch 0.00125 \(\mathrm{mm}\) thick. The sheet is totally immersed beneath the surface of a liquid. Parallel rays of monochromatic coherent light with wavelength 612 \(\mathrm{nm}\) in air strike the sheet perpendicular to its surface and pass through the scratch. A screen is placed in the liquid a distance of 30.0 \(\mathrm{cm}\) away from the sheet and parallel to it. You observe that the first dark fringes on either side of the central bright fringe on the screen are 22.4 \(\mathrm{cm}\) apart. What is the refractive index of the liquid?

Phased-Array Radar. In one common type of radar installation, a rotating antenna sweeps a radio beam around the sky. But in a phased-array radar system, the antennas remain stationary and the beam is swept electronically. To see how this is done, consider an array of \(N\) antennas that are arranged along the horizontal \(x\) -axis at \(x=0, \pm d, \pm 2 d, \ldots, \pm(N-1) d / 2 .\) (The number \(N\) is odd.) Each antenna emits radiation uniformly in all directions in the horizontal \(x y\) -plane. The antennas all emit radiation coherently, with the same amplitude \(E_{0}\) and the save- length \(\lambda\) . The relative phase \(\delta\) of the emission from adjacent antennas can be varied, however. If the antenna at \(x=0\) emits a signal that is given by \(E_{0} \cos \omega t,\) as measured at a point next to the antenna, the antenna at \(x=d\) emits a signal given by \(E_{0} \cos (\omega t+\delta),\) as measured at a point next to that antenna. The corresponding quantity for the antenna at \(x=-d\) is \(E_{0} \cos (\omega t-\delta) ;\) for the antennas at \(x=\pm 2 d,\) it is \(E_{0} \cos (\omega t \pm 2 \delta) ;\) and so on. (a) If \(\delta=0,\) the interference pattern at a distance from the antennas is large compared to \(d\) and has a principal maximum at \(\theta=0\) (that is, in the \(+y\) -direction, perpendicular to the line of the antennas. Show that if \(d<\lambda,\) this is the only principal interference maximum in the angular range \(-90^{\circ}<\theta<90^{\circ}\) . Hence this principal maximum describes a beam emitted in the direction \(\theta=0 .\) As described in Section 36.4 , if \(N\) is large, the beam will have a large intensity and be quite narrow. (b) If \(\delta \neq 0,\) show that the principal intensity maximum described in part (a) is located at $$\theta=\arcsin \left(\frac{\delta \lambda}{2 \pi d}\right)$$ where \(\delta\) is measured in radians. Thus, by varying \(\delta\) from positive to negative values and back again, which can easily be done electronically, the beam can be made to sweep back and forth around \(\theta=0 .(\mathrm{c})\) A weather radar unit to be installed on an airplane emits radio waves at 8800 \(\mathrm{MHz}\) . The unit uses 15 antennas in an array 28.0 \(\mathrm{cm}\) long (from the antenna at one end of the array to the antenna at the other end. What must the maximum and minimum values of \(\delta\) be (that is, the most positive and most negative values) if the radar beam is to sweep \(45^{\circ}\) to the left or right of the airplane's direction of flight? Give your answer in radians.

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 \(\mathrm{mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 \(\mathrm{W} / \mathrm{m}^{2} .\) (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.
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