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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

Monochromatic light of wavelength \(\lambda=620 \mathrm{nm}\) from a distant source passes through a slit \(0.450 \mathrm{~mm}\) wide. The diffraction pattern is observed on a screen \(3.00 \mathrm{~m}\) from the slit. In terms of the intensity \(I_{0}\) at the peak of the central maximum, what is the intensity of the light at the screen the following distances from the center of the central maxi- mum: (a) \(1.00 \mathrm{~mm}\) (b) \(3.00 \mathrm{~mm}\) (c) \(5.00 \mathrm{~mm} ?\)

Light of wavelength \(633 \mathrm{nm}\) from a distant source is incident on a slit \(0.750 \mathrm{~mm}\) wide, and the resulting diffraction pattern is observed on a screen \(3.50 \mathrm{~m}\) away. What is the distance between the two dark fringes on either side of the central bright fringe?

An interference pattern is produced by light of wavelength \(580 \mathrm{nm}\) from a distant source incident on two identical parallel slits separated by a distance (between centers) of \(0.530 \mathrm{~mm} .\) (a) If the slits are very narrow, what would be the angular positions of the first-order and second- order, two-slit interference maxima? (b) Let the slits have width \(0.320 \mathrm{~mm} .\) In terms of the intensity \(I_{0}\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

Laser light of wavelength \(632.8 \mathrm{nm}\) 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.

Sound of frequency \(1250 \mathrm{~Hz}\) leaves a room through a 1.00-m-wide doorway (see Exercise 36.5 ). At which angles relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use \(344 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections.
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