Wavelength and Frequency

Under Construction By Allie Johnson

Wavelength and Frequency are used to describe a sinusoidal electromagnetic wave. Frequency is the number of peaks per second that pass a given location. Wavelength is the distance between two peaks.

The Main Idea

The frequency of a sinusoidal electromagnetic wave is also the inverse of the period. When that wave is plotted over time, a period will be the distance between two peaks. Frequency is measured in inverse seconds or hertz (Hz). Wavelength is the distance between two peaks of a sinusoidal electromagnetic wave when plotted over a direction.

Example of a Wavelength measurement on a simple sine wave

Wavelength is directly proportional to frequency. Wavelength is the speed of light divided by the frequency. Therefore, as frequency increases wavelength decreases. This is because over a specific amount of time, the wave will move at the speed of light. Electromagnetic radiation is categorized by its wavelength, spanning from gamma rays to radio waves.

Types of Electromagnetic Radiation Characterized by Wavelength

Relationship Between Frequency and Period

[math]\displaystyle{ f = \frac{1}{T} }[/math] where F is the frequency and T is the period.

Relationship Between Frequency and Wavelength

[math]\displaystyle{ f = \frac{c}{\lambda} }[/math] where F is the frequency, c is the speed of light constant (c = 2.998×108 m s−1), and lambda is the wavelength.

Relationship Between Frequency and Angular Frequency

[math]\displaystyle{ \omega = {{2 \pi} \over T} = {2 \pi f} , }[/math] where ω is the angular frequency or angular speed (radians per second),T is the frequency over period (measured in seconds), and f is the ordinary frequency (measured in hertz).

A Computational Model

Types of Electromagnetic Radiation Characterized by Wavelength

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Simple

Problem #1a: Calculate the frequency of radiation with a wavelength of 442 nm. Example #1b: The wavelength of an argon laser's output is 488.0 nm. Calculate the frequency of this wavelength of electromagnetic radiation.

Solution to 1a:

1) Convert nm to m:

442 nm x (1 m / 109 nm) = 4.42 x 10¯7 m 2) Substitute into λν = c:

(4.42 x 10¯7 m) (x) = 3.00 x 108 m s¯1 x = 6.79 x 1014 s¯1

Solution to 1b:

1) Convert nm to m:

488 nm x (1 m / 109 nm) = 4.88 x 10¯7 m Then, substitute into λν = c:

(4.88 x 10¯7 m) (x) = 3.00 x 108 m s¯1 x = 6.15 x 1014 s¯1

The use of nm for wavelength is quite common.

Connectedness

1. How is this topic connected to something that you are interested in?
2. How is it connected to your major?
3. Is there an interesting industrial application?

History

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

See also

For more interest in the movement of sinusoidal waves over time visit: http://www.physicsbook.gatech.edu/Electromagnetic_Propagation

For interest in how wave length and frequency relate to radiation visit: http://www.physicsbook.gatech.edu/Sinusoidal_Electromagnetic_Radiaton

Further reading

Chabay, Ruth W., and Bruce A. Sherwood. Matter & Interactions. 3rd ed. Hoboken, NJ: Wiley, 2011. Print

External links

https://www.youtube.com/watch?v=tJW_a6JeXD8

References

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