# 24.3 The electromagnetic spectrum  (Page 4/33)

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Television is also broadcast on electromagnetic waves. Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.) These TV channels are called VHF (for very high frequency ). Other channels called UHF (for ultra high frequency ) utilize an even higher frequency range of 470 to 1000 MHz.

The TV video signal is AM, while the TV audio is FM. Note that these frequencies are those of free transmission with the user utilizing an old-fashioned roof antenna. Satellite dishes and cable transmission of TV occurs at significantly higher frequencies and is rapidly evolving with the use of the high-definition or HD format.

## Calculating wavelengths of radio waves

Calculate the wavelengths of a 1530-kHz AM radio signal, a 105.1-MHz FM radio signal, and a 1.90-GHz cell phone signal.

Strategy

The relationship between wavelength and frequency is $c=\mathrm{f\lambda }$ , where $c=3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}/\text{s}$ is the speed of light (the speed of light is only very slightly smaller in air than it is in a vacuum). We can rearrange this equation to find the wavelength for all three frequencies.

Solution

Rearranging gives

$\lambda =\frac{c}{f}.$

(a) For the $f=\text{1530}\phantom{\rule{0.25em}{0ex}}\text{kHz}$ AM radio signal, then,

$\begin{array}{lll}\lambda & =& \frac{\text{3.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{1530}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{cycles/s}}\\ & =& \text{196 m.}\end{array}$

(b) For the $f=\text{105.1 MHz}$ FM radio signal,

$\begin{array}{lll}\lambda & =& \frac{\text{3.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{105.1}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{cycles/s}}\\ & =& \text{2.85 m.}\end{array}$

(c) And for the $f=\text{1.90 GHz}$ cell phone,

$\begin{array}{lll}\lambda & =& \frac{3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}{\text{1.90}×{\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{cycles/s}}\\ & =& \text{0.158 m.}\end{array}$

Discussion

These wavelengths are consistent with the spectrum in [link] . The wavelengths are also related to other properties of these electromagnetic waves, as we shall see.

The wavelengths found in the preceding example are representative of AM, FM, and cell phones, and account for some of the differences in how they are broadcast and how well they travel. The most efficient length for a linear antenna, such as discussed in Production of Electromagnetic Waves , is $\lambda /2$ , half the wavelength of the electromagnetic wave. Thus a very large antenna is needed to efficiently broadcast typical AM radio with its carrier wavelengths on the order of hundreds of meters.

One benefit to these long AM wavelengths is that they can go over and around rather large obstacles (like buildings and hills), just as ocean waves can go around large rocks. FM and TV are best received when there is a line of sight between the broadcast antenna and receiver, and they are often sent from very tall structures. FM, TV, and mobile phone antennas themselves are much smaller than those used for AM, but they are elevated to achieve an unobstructed line of sight. (See [link] .)

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Alright Thank you
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