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Submarine Communication Normal radiofrequency waves cannot penetrate more than a few meters below the surface of the ocean. One method of communicating with submerged submarines uses very low frequency (VLF) radio waves. What is the wavelength (in air) of a \(10.0-\mathrm{kHz}\) VLF radio wave?

Short Answer

Expert verified
The wavelength is 30,000 meters.

Step by step solution

01

Identify the Given Values

We are given the frequency of the VLF radio wave, which is \( f = 10.0 \) kHz. Note that \(1 \text{ kHz} = 1,000 \text{ Hz}\). Therefore, the frequency \( f = 10,000 \text{ Hz} \).
02

Recall the Speed of Light Formula

The speed of light \( c \) in air is approximately \( 3.00 \times 10^8 \text{ m/s} \). The relation between speed of light \( c \), frequency \( f \), and wavelength \( \lambda \) is given by the formula: \( c = \lambda \times f \).
03

Rearrange the Formula

To find the wavelength \( \lambda \), rearrange the formula \( c = \lambda \times f \) to \( \lambda = \frac{c}{f} \).
04

Substitute Given Values into the Formula

Substitute the values into \( \lambda = \frac{c}{f} \): \( \lambda = \frac{3.00 \times 10^8 \text{ m/s}}{10,000 \text{ Hz}} \).
05

Calculate the Wavelength

Perform the division to find \( \lambda \): \( \lambda = 3.00 \times 10^4 \text{ meters} \).
06

Finalize the Answer

The wavelength of a \( 10.0 \) kHz VLF radio wave in air is \( 30,000 \) meters.

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Key Concepts

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

Submarine Communication
Underwater communication can be quite complex. Regular radio waves have trouble penetrating water beyond just a few meters. This poses a challenge when it comes to communicating with submarines that are deep under the ocean. Standard radio frequencies don't work well because they can't travel far through salty seawater. This is where Very Low Frequency (VLF) radio waves come in.
  • VLF radio waves have frequencies of 3 kHz to 30 kHz.
  • These waves have long wavelengths, ranging from 10 to 100 kilometers.
VLF radio waves can penetrate seawater much more effectively than higher frequency waves. They allow signals to reach submarines that are submerged far below the surface. Though the information that can be transmitted is limited compared to higher frequency waves, VLF is essential for underwater communication.
Speed of Light
The concept of the speed of light is a fundamental aspect of many physics equations. The speed of light in a vacuum is approximately \(3.00 \times 10^8\) meters per second. This speed is considered a universal constant and is often denoted by the letter \(c\). In air, the speed remains close to this value, which is why it is used in many calculations that involve radio waves moving through the atmosphere.
When working with the speed of light, it is crucial to remember that it links frequency and wavelength. This relationship is given by the formula \(c = \lambda \times f\), where \(\lambda\) is the wavelength and \(f\) is the frequency of the wave. Understanding this formula helps in many fields, from physics to engineering, and is essential for calculating the wavelength of radio waves in different mediums.
Wavelength Calculation
Calculating the wavelength of a radio wave involves using its frequency and the speed of light. Based on the relationship \(c = \lambda \times f\), we can rearrange the equation to solve for wavelength \(\lambda\): \[ \lambda = \frac{c}{f} \]
This formula tells us how to find the wavelength if we know the frequency and the speed of the wave. Let's take the example of a VLF radio wave with a frequency of \(10.0\) kHz:
  • First, convert kHz to Hz: \(10.0\) kHz is \(10,000\) Hz.
  • Using the speed of light \(c = 3.00 \times 10^8\) m/s, substitute these into the formula:
  • \( \lambda = \frac{3.00 \times 10^8}{10,000} = 30,000\) meters.
Thus, the wavelength is found to be 30,000 meters. This shows how long a VLF radio wave is in the air, which bears relevance to both communication and scientific calculations.
Radio Frequency Penetration
Different radio frequencies penetrate materials to varying degrees, and this is dictated by their wavelength and frequency. Water, especially seawater, offers more resistance to radio waves than the air.
Higher frequencies (shorter wavelengths) struggle to penetrate water, reflecting or being absorbed rather quickly. In contrast, lower frequencies with longer wavelengths, such as VLF radio waves, can penetrate water more effectively. This is because:
  • VLF waves have enough length to move through conductive materials like seawater.
  • The energy of these longer waves dissipates more slowly in denser media.
This makes VLF radio waves an adequate choice for submarines and other underwater applications. Understanding the degree of penetration for different frequencies helps in choosing the right communication method for specific environments, especially in marine contexts.

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Most popular questions from this chapter

A beam of horizontally polarized light passes through a polarizer whose transmission axis is at an angle of \(35.0^{\circ}\) with the vertical. If the intensity of the transmitted light is \(0.55 \mathrm{~W} / \mathrm{m}^{2}\), what was the initial intensity of the beam?

In 1926 Albert Michelson measured the speed of light with a technique similar to that used by Fizeau. In place of a toothed wheel, Michelson used an eight- sided mirror rotating at \(528 \mathrm{rev} / \mathrm{s}\), as illustrated in Figure 15.38. The distance from the rotating mirror to the fixed reflector was \(35.5 \mathrm{~km}\). If the light completed the \(71.0-\mathrm{km}\) round-trip in the time it took the mirror to complete one-eighth of a revolution, what is the speed of light?

Four types of electromagnetic waves are listed below. Rank the waves in order of increasing wavelength. Indicate ties where appropriate. A. infrared waves C. green light B. X-rays D. radio waves

Consider a spiral galaxy that is moving directly away from Earth with a speed \(V=3.600 \times 10^{5} \mathrm{~m} / \mathrm{s}\) as measured from its center, as shown in Figure 15.39. The galaxy is also rotating about its center, so that points in its spiral arms are moving with a speed \(v=6.400 \times 10^{5} \mathrm{~m} / \mathrm{s}\) relative to the center. If light with a frequency of \(8.230 \times 10^{14} \mathrm{~Hz}\) is emitted in both arms of the galaxy, what frequency is detected by astronomers observing the arm that is moving (a) toward and (b) away from Earth? (Measurements of this type are used to map the speeds of various regions in distant, rotating galaxies.)

Triple Choice Is the wavelength of infrared waves greater than, less than, or equal to the wavelength of ultraviolet waves? Explain.

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