Electromagnetic waves
Level 0 (green)- this is basic material that you have probably encountered already, although the approach may be slightly different. No prior knowledge is assumed.
Consider a wave travelling in the +z direction.
Consider a snapshot of the wave (i.e. fix the time t). The wave has a wavelength λ because if z is increased by λ, the argument of ths cos function changes by 2π, corresponding to a complete cycle of the cos function. The wavelength λ is the peak to peak distance in this snapshot. SI units m. |
 |
Now fix z and watch the wave pass as a function of time. If t increases by 1/ν, the argument of the cos function changes by 2π. Hence the time for one cycle to pass is 1/ν, or equivalently ν cycles pass per unit time. The frequency ν is the number of cycles that pass in unit time (e.g. in a second). SI units Hz = s–1. |
 |
To understand why this wave is travelling in the +z direction, consider the motion of a peak. At time zero the peaks of the cos functions occur whenever
i.e. whenever
.
As time proceeds these peaks occur whenever
i.e. whenever
We therefore conclude that the peaks move towards larger z with speed λν. Physicists call this the phase velocity of the wave.
Note that it is extremely important to be able to use this formula to calculate wavelength from frequency or vice versa. A frequent source of error is the misuse of this conversion. Please note that the units are particularly simple, and if you keep track of them properly you will never get this wrong.
Having established that our travelling wave is indeed a travelling wave we proceed to show that it obeys the wave equation. A wave equation is a second order partial differential equation that relates the dependence of the wave on position and on time.
Taking the travelling wave and differentiating twice with respect to the position variable z,
and similarly, differentiating twice with respect to t,
These two quantities only differ by a factor of c2, and hence
This is the 1-dimensional wave equation. We have shown that this equation is obeyed by the simple travelling wave of the tutorial, in order to make it plausible but it may also be shown more generally. See for example the more advanced description of Maxwell's equations and electromagnetic radiation.
Later in the maths course you will learn methods for solving this equation. It is an elementary exercise in partial differentiation to show that any function of the form
solves this equation. This is known as d'Alembert's solution, and you may have met it as a problem in the maths course.
This section introduces the main properties of electromagnetic radiation. Chemistry students do not need in general to understand where these properties come from, however an addition downloadable handout is available which shows how the wave equation and many of these properties arise as a consequence of the theory of electricity and magnetism, and which will give a deeper insight into the prequantum picture of light and electromagnetic radiation.
- Light is an electromagnetic wave. It obeys the wave equation and the equations of electromagnetism. It consists of an oscillation of the electric field and the magnetic field. These oscillations are linked because a changing electric field induces a magnetic field and vice versa.
- Electromagnetic waves travel through vacuum with a speed which is a fundamental constant,
-
Electromagnetic radiation is a transverse wave. This means that its oscillations (both electric and magnetic vectors) are perpendicular to the direction of propagation, similar to water waves and unlike sound waves in a gas.
The electric and magnetic vectors are perpendicular to each other. They are also in phase with one another.
If the electric vector always oscillates in the same direction, the wave is plane polarised.
- The intensity of light is defined to be the energy carried by the light per unit time per unit area, i.e. it is the energy flux of the light. The energy is carried in the direction of propagation of the wave, perpendicular to both E and B, i.e. in the direction of E×B (this is called the Poynting vector). The intensity is proportional to the square of the amplitude of the wave. If E0 is the amplitude of the electric vector the average intensity (averaged over a complete oscillation) is
(It may also be expressed in terms of the amplitude of the magnetic vector B0.) - Electromagnetic waves cover a very wide spectrum of wavelengths and frequencies
When light passes through a medium other than vacuum it generally slows down. The ratio of the speed of light in vacuum to that in the medium is known as the refractive index.
When the light arrives at the interface the frequency of arrival of the wavefronts does not change, but since the speed of propagation does change, the wavelength of the light is shortened.
The consequence of this is that the direction of the light changes. The light is said to be refracted. θ1 is the angle of incidence |
 |
To analyse this phenomenon consider the distance xbetween wavefronts measured along the interface. Since the direction of propagation is normal to the wavefronts we have similar triangles and |
 |
Hence
or equivalently
this equation is known as Snell's law.
If n2 > n1 the ray is refracted towards the normal and
θ2 < θ1.
If n2 < n1 the ray is refracted away from the normal and
θ2 > θ1.
In the latter case there will be some value of θ1, the critical angle, at which sin θ2 = 1, for angles of incidence larger than this Snell's law would give an impossible angle of refraction (sin θ2 > 1). The light is then completely reflected from the surface and does not penetrate it. This phemomenon is known as total internal reflection.
Example. The critical angle for total internal reflection from the interface between hexane and air is found to be 46.7°. Determine the refractive index of hexane.
Solution. The critical angle for total internal reflection is
and hence
assuming that the refractive index of air is 1 (it is about 1.003).
The phenomenon that distinguishes waves from other types of motion is interference. Interference occurs when two waves overlap and propagate through the same region.
The rule for combining the waves is to add their amplitudes at each point. This is called the principle of superposition.
Superposition only leads to interesting results if there is a well defined phase relationship between the waves being added.
There are two extreme cases of interference
Constructive interference. The peaks of the two waves coincide. The waves are said to be in phase. If the amplitude is doubled, and the intensity is multiplied by 4. |
 |
Destructive interference. The peaks of the one wave coincide with the troughs of the other. The waves are said to be out of phase. The amplitudes cancel one another out. |
 |
The whole range of relative phases is illustrated in the following diagram, in which the lower curves are the waves, and the upper curve (offset) is the intensity.
The classic experiment that showed that light was a wave was performed in Cambridge by Young in 1801. This experiment resolved a century-old dispute between followers of Newton, who believed that light was corpuscular and those of Huygens, who believed that it was a wave.
Young arranged for a unidirectional source of light to fall on a screen in which two narrow parallel slits were cut. The resulting image was then viewed as a projection onto another screen.
 |
In the corpuscular theory an image of the two slits should be seen, or if the images overlap, a single combined image. In the wave theory if the slits are narrow enough each one acts as a line source from which waves will spread in circles, and the image will be an interference pattern between these circles. The observation of interference, with clear directions of constructive and destructive interference clinched it for the follower of Huygens, at least for the next 100 years and the discovery of the dual nature of photons. |
 |
The slits are distance d apart and the screen is a distance z away from the slits. The diagram shows the distance travelled by light from each slit to a point in the image. For this point to be a point of constructive interference, a peak arriving from one slit must concide with a peak from the other slit. This can only happen if the difference in path lengths is an integer number of wavelengths, nλ. |
 |
It is not hard to solve this in general, for example using the cosine rule, but there is a great simplification if z >> d, as the two paths are then approximately parallel. If the angle of the point in the image relative to the normal is θ, the path difference is d sin θ, i.e. the criterion for constructive interference is |
Example: Light of wavelength 532 nm falls on a pair of slits. Two adjacent bright fringes are observed at 19.4° and 26.3°. Deduce
(i) the distance between the slits,
(ii) the values of n for these fringes.
Solution: (i) For the fringes to be adjacent their corresponding values of n must differ by 1, hence
from which we can deduce that d = 4.80 µm.
(ii) Once we have d it is simple to deduce the appropriate values of n, which are 3 and 4.