Let me take you through deriving the boundary conditions. I had a fantastic Prof, who explained this very well. We can derive them by evaluating some of <PERSON>'s equations at the interface.
Boundary Condition 1
First, let's assume there is a surface charge density $\rho_{sf}$. Let's begin with <PERSON>' Law: $$ \oint \vec{D}\cdot \mathrm{d}\vec{S} = Q_{encl} $$
We used the displacement field here, defined as $\vec{D} = \varepsilon \vec{E}$, and $Q_{encl}$ is the charge enclosed by a Gaussian box we're about to draw. Let's make it a cylinder around the interface, of height $h$.
In words, <PERSON>' law states that the net flux of an electric (displacement) field in a closed surface is directly proportional to the enclosed electric charge. Our cylinder is the closed surface - let's work the integral out!
First, there are no constraints on our choice of the Gaussian surface's height, so we can take the limit $h \rightarrow 0 $. 903
This means the surface integral for the curved surface goes to zero, and we're left with the ends:
$$ \int_{top} \vec{D}\cdot \mathrm{d}\vec{S} = (\vec{D}1\cdot \hat{n})A $$ $$ \int{bottom} \vec{D}\cdot \mathrm{d}\vec{S} = -(\vec{D}_2\cdot \hat{n})A $$
where A is the surface area of the top and bottom end of the cylinder, the subscript number indicates the medium and $\hat{n}$ is the surface normal. Note that for the bottom end, the integral is negative as the surface normal and the displacement field point in opposite directions.
Not forgetting about charge enclosed $Q_{end}$: in the limit of $h \rightarrow 0$, this becomes equal to the surface charge density multiplied by the area of the top/bottom surface! So, <PERSON>' law becomes:
$$ (\vec{D}1\cdot \hat{n} - \vec{D}_2\cdot \hat{n})A = A\rho{surf} $$
Cancelling $A$ and evaluating the dot products, we end up with the first boundary condition for the displacement field vector component perpendicular to the interface:
$$ \boxed{D_{1\perp} - D_{2\perp} = \rho_{surf}} $$
We could substitute $\vec{D} = \varepsilon \vec{E}$ here. Also, the presence of a surface charge will depend on the type of interface.
Boundary Condition 2
For the next boundary condition, let's evaluate the integral form of Faraday's Law:
$$ \oint \vec{E}\cdot\mathrm{d}\vec{l} = - \frac{\mathrm{d}}{\mathrm{d}t} \int \vec{B}\cdot\mathrm{d}\vec{S} $$
In words, this states that for any closed loop path, the sum of the length elements times the electric field in the direction of the length element is equal to the time derivative of the magnetic flux through the surface created by the closed loop path.
So, let's draw a rectangular loop of height $h$ and length $l$ around the interface. We can split the path integral of Faraday's Law up into a sum of 4 parts, for each side of the loop. Again, we can take the limit $h\rightarrow0$, which means we are only left with the path integrals parallel to the surface:
$$ \int_{top} \vec{E}\cdot\mathrm{d}\vec{l} = (\vec{E}1 \cdot \hat{n})l $$ $$ \int{bottom} \vec{E}\cdot\mathrm{d}\vec{l} = -(\vec{E}_2 \cdot \hat{n})l $$
where $\hat{n}$ is the unit vector in direction of the loop. Not forgetting about $\int \vec{B}\cdot\mathrm{d}\vec{S}$: in the limit of $h\rightarrow 0$, as the surface area of the enclosed path goes to 0, $\int \vec{B}\cdot\mathrm{d}\vec{S} = 0$. 707