Do surface charges exist when EM wave gets reflected by dielectric boundary? If both the materials on a boundary are dielectric then do surface charge exists. While soving for reflection coeffecient it wass sufficient to satisfy tangential componenta, I was solving using pillbox it showed that there should be free surface charges on boundary to satisfy change in normal component of Electric flux density vector. But I am confused how so can happen.
 A: At a dielectric boundary, there will be an incident, reflected and transmitted wave. Surface charges are where lines of electric field begin or end at the boundary. If the electric field is purely tangential to the interface, for example at normal incidence, then the polarisation charges in the medium will oscillate parallel to the interface and there will be zero net polarisation charge at the interface.
However, if incidence is at an oblique angle, then there will be components of the electric field normal to the boundary. These are not continuous, and where there is a step change in the electric field, there will be polarisation charges, since electric field lines begin and end on charges.
If by electric flux density you mean the displacement field, then the normal component of this is continuous at the interface because there are no free charges.
I'm guessing from your question that you tried to find the D-field either side of a dielectric boundary and found that it was discontinuous. Here is my proof that it is continuous - see if you can find where you made your error.
Take the example of p-polarised light. The reflection and transmission (amplitude) coefficients can be written
$$r = \frac{E_r}{E_i} = \frac{n_2 \cos \theta_i - n_1 \cos \theta_t}{n_1 \cos \theta_t + n_2 \cos \theta_i}\ , \tag*{(1)}$$
$$ t = \frac{E_t}{E_i} = \frac{2 n_1 \cos \theta_i}{n_1 \cos \theta_t + n_2 \cos \theta_i}\ , \tag*{(2)}$$
where $n_1$ and $n_2$ are the refractive indices of the two dielectric media, $\theta_i$ and $\theta_t$ are the angle of incidence and the angle of transmission (refraction) and a convention has been adopted that positive $r$ means that $E_r$ is phase shifted with respect to $E_i$. We are assuming linear dielectrics with $\mu_r=1$, so that $n_1^2 = \epsilon_{r,1}$ and $n_2^2 = \epsilon_{r,2}$, where $\epsilon_r$ are the relative permittivities.
The normal component of D-field in medium 1 is the sum of the D-fields of the incident and reflected waves, multipled by $\sin \theta_i$ (recall that the D-field is perpendicular to the wave-vector). The normal component of the D-field in medium 2 is just the D-field of the transmitted wave multiplied by $\sin \theta_t$. Hence their ratio is given by (using $D = \epsilon_0 \epsilon_r E$ for a linear dielectric).
$$\frac{D_1\perp}{D_2\perp} =  \frac{\epsilon_{r,1}(E_i + E_r) \sin \theta_i}{\epsilon_{r,2} E_t \sin \theta_t} \ .$$
Snell's law tells us that $n_1 \sin \theta_i = n_2 \sin \theta_t$ and $\epsilon_{r,1}/\epsilon_{r,2} = n_1^2/n_2^2$. We can then use equations (1) and (2) to express all the fields in terms of $E_i$.
$$\frac{D_1\perp}{D_2\perp} = \left(\frac{n_1}{n_2}\right) \left(\frac{1 + (n_2 \cos \theta_i - n_1\cos \theta_t)/(n_1 \cos\theta_t + n_2\cos\theta_i)}{2n_1 \cos\theta_i/(n_1\cos\theta_t + n_2 \cos\theta_i)}\right)\ , $$
$$\frac{D_1\perp}{D_2\perp} =\left(\frac{n_1}{n_2}\right)\left(\frac{n_2}{n_1}\right)=1\ .$$
Thus the normal components of the D-field either side of the boundary are the same and there is no free charge density.
