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Chapter 36: Problem 39

Two satellites at an altitude of \(1200 \mathrm{~km}\) are separated by \(28 \mathrm{~km} .\) If they broadcast \(3.6 \mathrm{~cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

Short Answer

Expert verified
The minimum diameter of the receiving dish that would be needed to resolve the transmissions from the two satellites according to Rayleigh's criterion can thus be calculated by substituting the given values into the formula. This involves prior conversion of the given parameters into consistent units (meters in this case).

Step by step solution

01

Identify the Given Parameters

In this task, several parameters are given. These include: \n- The wavelength of the microwaves (\(位 = 3.6 \) cm), \n- The separation of the satellites (d, which is equal to \(28 \) km), and \n- The altitude of the satellites (D, which is \(1200 \) km). These measurements may need to be converted to consistent units.
02

Convert All Parameters to Consistent Units

To achieve consistent calculations, convert all parameters to the same unit. It's practical to convert all parameters to meters. Therefore, the wavelength \(位 = 3.6 \) cm will be \(3.6 \times 10^{-2} \) meters, and the distance D = \(28 \) km will be \(28 \times 10^{3} \) meters.
03

Apply Rayleigh's Criterion for Resolution

According to Rayleigh's criterion for resolution, the minimum resolvable angle \(胃\) is given by the equation \(胃 = 1.22位/D \). In this case, \(胃\) can be found by dividing the separation distance d by the altitude D of the satellites, \(胃 = d/D \). By setting these two equal, we can generally write the equation as \( d/D = 1.22位/D \).
04

Solve for D (The Minimum Diameter of the Receiving Dish)

In step 4, rearrange the equation from step 3 and solve for D, the minimum diameter of the receiving dish needed to resolve the transmissions from the two satellites. This gives: \( D = 1.22位d/D \). Upon rearranging, the formula becomes \( D = sqrt{1.22位d} \).
05

Substitute Given Values into the Formula and Calculate D

Finally, we substitute the given values into the formula from the last step (not forgetting to convert them to meters) and compute D, the diameter of the receiving dish, to get the numerical answer.

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Key Concepts

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

Wavelength Conversion
Wavelength conversion is a crucial step in solving any problem related to wave propagation and resolution. Wavelength is the distance between two successive peaks of a wave and is an essential factor in calculating angular resolution.

In the exercise, we are given the wavelength of microwaves as 3.6 cm. For scientific calculations, it is vital to convert all measurements to a consistent unit system, typically meters.
Therefore, the wavelength of 3.6 cm is converted to meters by multiplying by 0.01, resulting in 0.036 meters.

Consistent units ensure accuracy in further calculations, such as when applying formulas like Rayleigh's Criterion. Remembering this simple conversion can help avoid errors in physics and engineering problems.
Satellite Communications
Satellite communications involve the transmission and reception of signals via satellites orbiting the Earth. These systems rely on electromagnetic waves to send data over vast distances.

An essential factor in satellite communications is the ability to differentiate between signals from different sources, particularly when satellites are closely positioned. In the given exercise, the two satellites are positioned 28 km apart at a height of 1200 km.
  • This separation distance requires precise angular resolution to receive distinct signals from each satellite.
  • The ability to resolve signals is dictated by factors like wavelength and receiving dish diameter.
Understanding how satellite communication systems operate at such high altitudes and separations is beneficial for improving signal clarity and reliability.
Angular Resolution
Angular resolution refers to the smallest angle over which separate objects can be distinguished. It's crucial in fields like astronomy and telecommunications.
In the context of satellite communications, angular resolution determines how distinctly a satellite can transmit signals that can be identified separately.

Rayleigh's Criterion helps calculate the minimum angular resolution needed to distinguish between two point sources of light or electromagnetic waves, such as satellites.
The formula used is \[胃 = \frac{1.22位}{D}\]
where:
  • \( 胃 \) is the angular resolution.
  • \( 位 \) is the wavelength of the transmitted signal.
  • \( D \) is the receiving aperture diameter.
In the exercise, bearing the separation and altitude into account helps in understanding how angular resolution operates practically within Rayleigh's Criterion.
Microwave Transmission
Microwave transmission is a common method used for transmitting information over long distances via electromagnetic waves. It leverages the microwave spectrum, which generally includes frequencies from 1 GHz to 30 GHz.

In this exercise, we deal with 3.6 cm microwaves, which are within this frequency range and ideal for satellite communication due to their stability and ability to carry substantial data.
  • They travel through space, experiencing minimal interference, unlike lower frequency radio waves.
  • These properties make microwaves a reliable choice for time-sensitive and high-data rate transmissions.
Understanding the characteristics and benefits of microwave transmission can significantly aid in grasping how satellite communication works, particularly in maintaining clear and effective data transfer.

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Most popular questions from this chapter

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about \(5.0 \mu \mathrm{m}\) in diameter) limits the size of an object at the near point \((25 \mathrm{~cm})\) of the eye to a height of about \(50 \mu \mathrm{m}\). (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye. eter of the human pupil is about \(2.0 \mathrm{~mm}\), does the Rayleigh criterion allow us to resolve a \(50-\mu \mathrm{m}\) -tall object at \(25 \mathrm{~cm}\) from the eye with light of wavelength \(550 \mathrm{nm} ?\) (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the \(25 \mathrm{~cm}\) near point with light of wavelength \(550 \mathrm{nm} ?\) (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes \(\left(60 \mathrm{~min}=1^{\circ}\right),\) and compare it with the experimental value of about \(1 \mathrm{min.}\) (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

When laser light of wavelength \(632.8 \mathrm{nm}\) passes through a diffraction grating, the first bright spots occur at \(\pm 17.8^{\circ}\) from the central maximum. (a) What is the line density (in lines/cm) of this grating? (b) How many additional bright spots are there beyond the first bright spots, and at what angles do they occur?

The wavelength range of the visible spectrum is approximately \(380-750 \mathrm{nm} .\) White light falls at normal incidence on a diffraction grating that has 350 slits \(/ \mathrm{mm} .\) Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (Note: An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example \(36.4 .\) )

Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, \(a\) is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength \(632.8 \mathrm{nm}\) was shone on a single strand of hair, and the diffracted light was viewed on a screen \(1.25 \mathrm{~m}\) away, the first dark fringes on either side of the central bright spot were \(5.22 \mathrm{~cm}\) apart. How thick was this strand of hair?

A slit \(0.360 \mathrm{~mm}\) wide is illuminated by parallel rays of light that have a wavelength of \(540 \mathrm{nm}\). The diffraction pattern is observed on a screen that is \(1.20 \mathrm{~m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)
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