# Materials in Electronics/Wave-Particle Duality/The Two-Slit Experiment/Ideal/Irradiance

This page will dear with the effects produced as a result of two vector waves, which is more complex than simple addition of scalar waves. According to the principle of superposition, the electric field intensity, E, (a vector quantity) at a point P is given by the sum of all n separate fields at that point:

$\mathbf {E} =\sum _{i=1}^{n}{\mathbf {E} _{i}}$ Generally, due to the rapid oscillations of EM waves (hundreds of terahertz), it is not possible to measure the actual field. Therefore, we will approach the problem by looking at the irradiance produced, rather than the field.

We will consider for now two point sources, S1 and S1, and will deal with only linearly polarised waves. Now, the two waves (one from each source) incident at P are each given by the following equations:

$\mathbf {E} _{1}\left(\mathbf {r} ,t\right)=\mathbf {E} _{0,1}\cos \left(\mathbf {k} _{1}\cdot \mathbf {r} -\omega t+\phi _{1}\right)$ $\mathbf {E} _{2}\left(\mathbf {r} ,t\right)=\mathbf {E} _{0,2}\cos \left(\mathbf {k} _{2}\cdot \mathbf {r} -\omega t+\phi _{2}\right)$ where:

• En is the light field due to that wave
• E0,n is the amplitude vector of that wave
• kn is the wavevector representing the direction of propagation and wavenumber, k
• r is the position vector of the point P
• ω is the angular frequency of the wave
• t is time
• φn is the initial phase of that wave

All these elements are shown in the diagram below:

$I=\epsilon v\langle \mathbf {E} ^{2}\rangle$ where $\langle \mathbf {E} ^{2}\rangle$ means the time-average of the magnitude of the electric field squared. This is an elementary result of fields theory and this is not the place to prove it. For the moment, as we are dealing with relative intensities, we will leave out the constants and set

$I=\langle \mathbf {E} ^{2}\rangle$ From the principle of superposition,

$\mathbf {E} =\mathbf {E} _{1}+\mathbf {E} _{2}$ where E1 and E2 are the components of the field resulting from each slit. We can now write:

$\mathbf {E} ^{2}=\left(\mathbf {E} _{1}+\mathbf {E} _{2}\right)^{2}$ $\mathbf {E} ^{2}=\mathbf {E} _{1}^{2}+\mathbf {E} _{2}^{2}+2\mathbf {E} _{1}\mathbf {E} _{2}$ Time-avaraging both sides of this equation gives us:

$\langle \mathbf {E} ^{2}\rangle =\langle \mathbf {E} _{1}^{2}\rangle +\langle \mathbf {E} _{2}^{2}\rangle +2\langle \mathbf {E} _{1}\mathbf {E} _{2}\rangle$ We will write this as a sum of irradiances:

$I=I_{1}+I_{2}+I_{1,2}\,$ The last term in this expression is called the interference term. Recalling our form for the two waves (just note that we are describing the wave by the electric field now), this can be written as:

$\mathbf {E} _{1}\mathbf {E} _{2}=A_{1}\sin \left(k_{1}x-\omega t+\phi _{1}\right)\times A_{2}\sin \left(k_{2}x-\omega t+\phi _{2}\right)$ Applying the double angle formula, we get:

$\mathbf {E} _{1}\mathbf {E} _{2}=A_{1}A_{2}\left[\sin \left(k_{1}x+\phi _{1}\right)\sin \left(\omega t\right)+\cos \left(k_{1}x+\phi _{1}\right)\cos \left(\omega t\right)\right]\times \left[\sin \left(k_{2}x+\phi _{2}\right)\sin \left(\omega t\right)+\cos \left(k_{2}x+\phi _{2}\right)\cos \left(\omega t\right)\right]$ 