There is no complete answer to what light really is. However in certain circumstances, light acts as a wave. One of those circumstances is in the classic, Two-Slit Interference experiment.
In Two-Slit Interference, a single source of light is split into two, to generate two coherent sources. The two slits are very closely spaced and narrow. Each slit acts as a new source of light. Since the slits are illuminated by the same wave front, these sources are in phase. Where the wave fronts from the two sources overlap, an interference pattern is formed. And to project it onto a screen, the interference pattern is observed. And it looks like this. http://img181.imageshack.us/img181/1952/rteb2.png (click the link). The bright bands are called anti-nodes, and the bands are called nodes. The brightest band on the middle is called the “central maximum”. The anti-nodes are represented by n1, n2, n3, etc. and are labeled starting from the nearest to the central maximum. The dark bands are where the waves from the two sources are out of phase and interfere destructively. The bright bands are where the waves from the two sources are in phase and interfere constructively.
The essential geometry of Two-Slit Interference is shown in this diagram. http://img176.imageshack.us/img176/3829/asdcx7.png (click the link). At the central maximum, light waves from slits A and B have traveled the same distance from the slits to the eye, so the waves are in phase with each other and interfere constructively in the retina of the eye. At n1, below the central maximum, on the diagram, light from slit B has traveled one wavelength farther than light from slit A. so the waves are again in phase, and also constructively interfere.
At every nth anti-node bellow the central maximum, the light from slit B has traveled n wavelengths farther than the light from slit A, and constructive interference occurs. The same goes for the anti-nodes above the central maximum, except for the nth anti-node above the central maximum; the light from slit A has traveled n wavelengths farther than the light from slit B.
In the diagram, the line AC is constructed perpendicular to the line AP. Since the slits are very close together (in the experiment, not the diagram), lines AP and BP are nearly parallel. Therefore, to a very close approximation, AP = CP. This means that for constructive interference to occur at P, it must be true that BC = n, where is the wavelength.
From the right angled triangle, ABC, it can be seen that:
Sin ‘ = n/d
Where d = AB (slit spacing). The dotted lines on the diagram show a projection of the interference pattern onto the Diffraction Scale as it appears when looking through the slits. Since it has already been assumed that BP is parallel to AP, then ‘ = . To find , it can be seen that:
Tan = x/L
Where L is the distance from the Diffraction Scale to the Diffraction Plate, and x is the spacing of the anti-nodes. Since is a very small angle, sin is extremely close to being equal to tan . Therefore, if sin = tan , then:
n/d = x/L
This can be rearranged to make the subject.
= dx/nL
where:
= wavelength
d = slit spacing
x = anti-node spacing
n = the anti-node
L = distance from diffraction plate to diffraction scale
This equation can be used to find the wavelengths of the different colors of light using the Two-Slit Interference experiment.
