Wavelength and Frequency: Difference between revisions

Revision as of 12:44, 3 December 2015

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

A Mathematical Model

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

Examples

Conversions to meters per second:

Prefix

p (pico) 10e-12

n (nano) 10e-9

µ (micro) 10e-6

m (milli) 10e-3

k (kilo) 10e3

M (mega) 10e6

G (Giga) 10e9

T (Tera) 10e12

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.

Middling

Problem #2a: Calculate the frequency of electromagnetic radiation that has a wavelength of 1.315 micrometers. Problem #2b: What is the frequency of infrared radiation of wavelength 67.5 μm?

Solution to 2a:

1) Convert μm to m:

1.315 μm x (1 m / 106 μm) = 1.315 x 10¯6 m 2) Substitute into λν = c:

(1.315 x 10¯6 m) (x) = 3.00 x 108 m s¯1 x = 2.28 x 1014 s¯1

Solution to 2b:

1) Convert μm to m:

67.5 μm = 67.5 x 10-6 m 2) Use λν = c to determine the frequency:

(67.5 x 10-6 m) (x) = 3.00 x 108 m/s x = 4.44 x 1012 s-1

Difficult

Problem #3a: Calculate the frequency of radiation with a wavelength of 4.92 cm. Problem #3b: Calculate the frequency of radiation with a wavelength of 4.55 x 10¯9 cm.

Comment: since the wavelengths are already in cm, we can use c = 3.00 x 1010 cm s¯1 and not have to do any conversions at all.

Solution to 3a:

(4.92 cm) (x) = 3.00 x 1010 cm s¯1 x = 6.10 x 109 s¯1

Solution to 3b:

(4.55 x 10¯9 cm) (x) = 3.00 x 1010 cm s¯1 x = 6.59 x 1018 s¯

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