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Chapter 36: Problem 43

Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 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
Minimum dish diameter needed: 1.88 m.

Step by step solution

01

Understanding Rayleigh's Criterion

Rayleigh's criterion states that two sources can be just resolved when the central maximum of the diffraction pattern of one source coincides with the first minimum of the diffraction pattern of the other source. Mathematically, this is given by: \( \theta = 1.22 \frac{\lambda}{D} \), where \( \theta \) is the angular separation, \( \lambda \) is the wavelength, and \( D \) is the diameter of the receiving dish.
02

Convert Units

First, convert the wavelength from centimeters to meters. Since the wavelength is \(3.6\,\mathrm{cm}\), it is: \( \lambda = 3.6 \times 10^{-2}\,\mathrm{m} \).
03

Calculate the Angular Separation

To find the angular separation, \( \theta \), use the formula: \( \theta = \frac{s}{L} \) where \( s \) is the separation distance and \( L \) is the distance to the satellites. Here, \( s = 28\,\mathrm{km} = 28000\,\mathrm{m} \) and \( L = 1200\,\mathrm{km} = 1200000\,\mathrm{m} \). So, \( \theta = \frac{28000}{1200000} \approx 0.0233 \) radians.
04

Solve for Diameter of Receiving Dish

Rearrange the Rayleigh's criterion formula to solve for \( D \): \( D = 1.22 \frac{\lambda}{\theta} \). Plug in the known values: \( \lambda = 3.6 \times 10^{-2}\,\mathrm{m} \) and \( \theta \approx 0.0233\,\mathrm{radians} \). Thus, \( D = 1.22 \frac{3.6 \times 10^{-2}}{0.0233} \approx 1.88\,\mathrm{m} \).
05

Conclusion - Minimum Dish Diameter

The minimum diameter of the receiving dish, as calculated using the previous steps, is approximately \(1.88\,\mathrm{m}\). This is the smallest size necessary to resolve the two transmissions according to Rayleigh's criterion.

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Key Concepts

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

Angular Separation
Angular separation is a critical concept when considering resolving two point sources, such as satellites in space. It refers to the angle formed between the lines connecting the observer and each of the two sources. This angle helps determine how close two objects can be such that they still appear as distinct from one another rather than merging into one.

In the context of Rayleigh's criterion, angular separation is key because it influences the ability to resolve separate entities. In our example, we calculated the angular separation using the formula:
  • \( \theta = \frac{s}{L} \)
Here, \( s \) represents the separation distance between the satellites, and \( L \) is the altitude of the satellites from the observer. The calculated angle, \( \theta \approx 0.0233 \) radians, guides us to understand how finely we need to calibrate our observation instrument, such as a receiving dish, to distinguish between the two signals.
Diffraction Pattern
A diffraction pattern emerges when a wave encounters an obstacle or opening, causing the wave to spread out. This spreading forms a series of light and dark bands or fringes, known as a diffraction pattern. When applied to astronomy or satellite communications, observing diffraction patterns is essential in determining clarity and resolution for signals.

According to Rayleigh's criterion, a wave's central maximum鈥攖he brightest part of the diffraction pattern鈥攕hould align with the first minimum of another wave's pattern for the waves to be just resolved. This criterion ensures that the diffraction effects do not mask the difference between the two sources.
  • The pattern is influenced by the wavelength of the wave and the diameter of the instrument's aperture.
  • The larger the aperture, the smaller the diffraction pattern, allowing for higher resolution observations.
In the context of the satellites broadcasting microwaves, the diffraction pattern helps determine the necessary receiving-dish diameter to sharply resolve the two incoming transmissions.
Wavelength Conversion
Converting the unit of wavelength is often an essential step in solving physics problems, ensuring that all measurements are cohesive and compatible with each other. Wavelength is frequently provided in different units, such as centimeters or nanometers, and must be converted to meters for use in formulas like Rayleigh's criterion.

For the satellite exercise, the microwave's wavelength was given as 3.6 cm. To convert this to meters鈥攁ppropriate for calculations involving international units鈥攖he conversion is:
  • \( 3.6 \times 10^{-2} \) meters
Understanding how to seamlessly convert between units allows for accurate computations and prevents errors related to unit inconsistency. This precise conversion is vital in calculating parameters such as angular separation and receiving dish diameter accurately.
Receiving Dish Diameter
The diameter of a receiving dish is a crucial factor in determining how well it can resolve signals from closely spaced sources. According to Rayleigh's criterion, the minimum diameter necessary to resolve two signals depends on the microwave wavelength and the angular separation between the sources.

In our exercise, we derived the formula for determining the minimum diameter \( D \) of the dish using:
  • \( D = 1.22 \frac{\lambda}{\theta} \)
By using the calculated angular separation \( \theta \approx 0.0233 \) radians and the converted wavelength \( \lambda = 3.6 \times 10^{-2} \) meters, it was determined that a minimum dish diameter of approximately 1.88 meters is needed.

This calculation ensures that the receiving dish is adequately sized to separate and identify signals from the two satellites, fulfilling the conditions set by Rayleigh's criterion.

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Most popular questions from this chapter

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 \(\mathrm{cm}\) from the slits. The centers of the slits are 0.640 \(\mathrm{mm}\) apart and the width of each slit is 0.434 \(\mathrm{mm}\) . If the intensity at the center of the central maximum is \(5.00 \times\) \(10^{-4} \mathrm{W} / \mathrm{m}^{2},\) what is the intensity at a point on the screen that is 0.900 \(\mathrm{mm}\) from the center of the central maximum?

Identifying Isotones hy Snectra Different isotones of the same element emit light at slightly different wavelengths. A wavelength in the emission spectrum of a hydrogen atom is 656.45 nm; for deuterium, the corresponding wavelength is 656.27 \(\mathrm{nm} .\) (a) What minimum number of slits is required to resolve these two wavelengths in second order? (b) If the grating has 500.00 slits/mm, find the angles and angular separation of these two wavelengths in the second order.

Monochromatic light from a distant source is incident on a slit 0.750 \(\mathrm{mm}\) wide. On a screen 2.00 \(\mathrm{m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 \(\mathrm{mm}\) . Calculate the wavelength of the light.

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply.(a) If the first diffraction minima are at \(\pm 90.0^{\circ},\) so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta=45.0^{\circ}\) to the intensity at \(\theta=0 ?\)

Intensity Pattern of \(N\) Slits. (a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda .\) Show that at a time \(t,\) the electric field at a distant point \(P\) is $$\begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned}$$ where \(E_{0}\) is the amplitude at \(P\) of the electric field due to an indi- vidual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrange- ment), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\) . (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship $$e^{i z}=\cos z+i \sin z$$ where \(i=\sqrt{-1}\) . In this expression, cos \(z\) is the real part of the complex number \(e^{i z},\) and \(\sin z\) is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity $$\sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega t+n \phi)}$$ (c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A},\) show that the sum in part (b) can be written as $$E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)}\( \)\quad=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i[k R-\omega t+(N-1) \phi}$$ Then, using the relationship \(e^{i z}=\cos z+i \sin z,\) show that the (real) electric field at point \(P\) is $$I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2}$$ where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N=2 .\) It will help to recall that \(\sin 2 A=2 \sin A \cos A .\) Explain why your result differs from Eq. \((35.10),\) the expression for the intensity in two-source interfer- ence, by a factor of \(4 .\) Hint: Is \(I_{0}\) defined in the same way in both expressions?
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