Question: **Two very narrow slits are spaced apart and are
placed 35.0 cm from a screen. What is the distance between the
first and second dark lines of the interference pattern when the slits
are illuminated with coherent light with λ= 550nm**

I'm not asking for an answer for this because I understand the method. I am confused by the theory behind it. When I consulted the solutions manual for this question, they say the first dark line occurs at m = 0 (where m is the order of the image), whereas I thought it occurred at m = 1. Why is this so?

Answer: 
The definition of s-polarised light is that the electric field is polarised so that it is perpendicular to the [plane of incidence](https://en.wikipedia.org/wiki/Plane_of_incidence). Where there is a specular reflection, the plane of incidence contains the k-vector of the incoming wave and the reflected wave.

Since the electric field of an EM wave must be perpendicular to the k-vector. This then leaves the possibility that you have a component of the reflected E-field that is p-polarised (as in the second diagram).

As the tangential component of the electric field must be continuous across the boundary, and this relationship must apply to the components of the electric field perpendicular to the plane of the diagram (the plane of incidence) *and* tangential to the boundary in the plane of incidence, then if we call the p-polarised component of the reflected wave $E\_{p,r}$ and the corresponding (and necessary) component in the transmitted wave $E\_{p,t}$, then we know that
$$ E\_{p,r} \cos \theta\_r = E\_{p,t} \cos \theta\_t \tag{1}$$
and because the s-polarised components have no contribution perpendicular to the interface, we know that
$$ \epsilon\_1 E\_{p,r} \sin \theta\_r = \epsilon\_2 E\_{p,t} \sin \theta\_t. \tag{2}$$

This forms a complete separate and independent pair of equations from those that determine the relationship between the angle of incidence and angle of reflection using the s-polarised components. Therefore we already know from these and the standard derivation, that $\theta\_i = \theta\_r$ and that $\epsilon\_1^{2}\sin\theta\_r = \epsilon\_2^{2}\sin\theta\_t$ (Snell's law for non-magnetic media).

Using Snell's law and equation (2) we can thus derive that
$$\epsilon\_2 E\_{p,r} = \epsilon\_1 E\_{p,t} \tag{3}$$
and if $\epsilon\_2 > \epsilon\_1$, then $E\_{p,r}<E\_{p,t}$.

Substituting Snell's law into equation (1) and using $\cos \theta = (1 - \sin^2 \theta)^{1/2}$ it can be shown that
$$ E\_{p,r}^{2} = E\_{p,t}^{2} \left[ \frac{1 - \sin^2 \theta\_t}{1 - (\epsilon\_2/\epsilon\_1)^4 \sin^2 \theta\_t}\right] \tag{4}$$
But in this case, if $\epsilon\_2 > \epsilon\_1$, we see that $E\_{p,r} > E\_{p,t}$ and so there is no simultaneous solution for equations (3) and (4), *except* $E\_{p,r}=E\_{p,t}=0$, for any real value of $\theta\_t$.

I feel sure there must be a more elegant way to show this.
Here's a simple explanation based on the dipolar nature of the medium:

For *most* of the materials, we can assume that the source of the reflected and refracted waves are the induced tiny dipoles in the dielectric medium. In an *isotropic* medium, polarization vector is proportional to the (total) electric field vector with a constant (as opposed to a tensor for anisotropic media). Therefore they are always in the same direction.
Also, the far-field radiation of a dipole is polarized in the same direction as the dipole moment vector and hence the incident (exciting) wave. 

Therefore, if the material is isotropic, the polarizations of reflected and refracted waves are the same as that of the incident wave. In general, this is not true for birefringent crystals, for which the direction of the induced polarization may not be the same as the incident electric field.