Electric and magnetic fields fluctuating together can form a propagating electromagnetic wave. An electromagnetic wave is a transverse wave, the electric and magnetic fields are perpendicular to the direction the wave travels.

Electromagnetic waves can travel through a vacuum or a material substance. The speed of light in a vacuum is c = 3.00 x 10 8 m/s. This speed is generally slower in other materials.

AM radio waves have frequencies between 545 and 1605 kHz. FM is between 88 and 108 MHz. TV channels 2-6 are from 54 to 88 MHz, channels 7-13 use 174 to 216 MHz.

An electromagnetic wave, like any wave, follows this equation: v = f. In the case of EM waves: c = f. EM waves range from less than 10 4 Hz to greater than Hz.

Since c is constant, c = f may be used to find the range of wavelengths: from over 10 4 m to m.

The parts of the spectrum are: radio waves, infrared (heat waves), visible light (roygbiv), ultraviolet (electric arc), x-rays, and gamma rays (nuclear decay).

Higher frequency (shorter wavelangth) radio waves are called microwaves. Red martians invaded venus using x-ray guns. Radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays.

The spectrum ranges from long wavelengths with low frequency and low energy to short wavelengths with high frequency and high energy.

Visible light has a frequency range of about 4.0 x Hz to 7.9 x Hz. This corresponds to extremely small wavelengths that are usually expressed in nanometers (nm) 1 nm = m. An angstrom ( m) is an occasionally used but obsolete unit.

Ex. 1 - Find the range in wavelengths (in vacuum) for visible light in the frequency range between 4.0 x Hz (red light) and 7.9 x Hz (violet light). Express the answers in nanometers.

Ex. 2 - Diffraction is the ability of a wave to bend around an obstacle or the edges of an opening. Would you expect AM or FM radio waves to bend more readily around an obstacle such as a building?

The most accurate measurements of the speed of light before modern day measurements were made by Albert Michelson using an rotating mirror in Today the speed of light is set at m/s.

When we look at stars, we see them as they were thousands or more years ago.

Maxwell determined that EM waves travel through a vacuum at a speed given by: c = 1/ (√  0 µ 0 ).  0 is the electric permittivity of free space,  0 = 8.85 x C 2 /(Nm 2 ), µ 0 is the magnetic permittivity of free space, µ 0 = 4π x Tm/A.

Maxwell determined this formula theoretically, but when the values for  0 and µ 0 substituted, the result obtained is 3 x 10 8 m/s; which fits experimental evidence.

The energy of an EM wave u is carried by the electromagnetic and magnetic fields that comprise the wave. The total amount of energy per volume is found by adding the electric energy density and the magnetic energy density.

u =  0 E 2 /2 + B 2 / 2µ 0 Since the electric field and magnetic field carry equal amounts of energy, this equation reduces to: u =  0 E 2 / = B 2 /µ 0

The electric and magnetic fields are related by E = cB. The values of E and B fluctuate. If an average value for u is needed, average values for E 2 and B 2 must be found.

We use rms values for E and B in this case. E rms = E 0 / √2 B rms = B 0 / √2

Ex. 4 - Sunlight enters the top of the earth’s atmosphere with an electric field whose rms value is E rms = 720 N/C. Find (a) the average total energy density of this electromagnetic wave and (b) the rms value of the sunlight’s magnetic field.

The intensity is found by multiplying the speed of light by the total energy density of the wave: S = cu. From previous equations: S = cu = c  0 E 2 = c B 2 /µ 0 Using rms values gives average intensity.

Ex. 5 - A neodymium-glass laser has an electric field with an rms value of E rms = 2.0 x 10 9 N/C. Find the average power of each pulse that passes through a 1.6 x m 2 surface that is perpendicular to the laser beam. (P = SA)

The single equation for the doppler effect is: f’ = f(1 ±u/c) if u<<c. f’ is observed frequency f is emitted frequency u is relative speed c is speed of light

Electromagnetic waves are transverse, and can therefore be polarized. Polaroid material only allows transmission along the transmission axis. A filter such as this absorbs 1/2 of the intensity of the unpolarized light.

A polarized filter absorbs as much of the electric (and magnetic) field as it emits.

When two polarizing filters are used, the first is called the polarizer and the second the analyzer.

If the transmission axis of the analyzer is oriented at an angle θ to the transmission axis of the polarizer, the electric field strength is E cos θ.

Intensity S is proportional to cos 2 θ. Therefore: S avg = S 0 avg cos 2 θ S 0 avg is average light intensity entering the analyzer. This is Malus’ Law.

Ex. 7 - What value of θ should be used so the average intensity of light leaving the analyzer is one-tenth the average intensity of the unpolarized light?

When θ is 90°, the polarizer and analyzer are said to be crossed, and no light is emitted by the combination.

Ex. 8 - No light is emitted from a crossed polarizer-analyzer combination. Suppose a third piece of polarizing material is inserted between the combination at an axis angle of 45°. Is light now emitted from the combination?