Why the exponential part of Electromagnetic waves on both side of interface must be equal? Why the exponential part of Electromagnetic waves on both side of interface must be equal which implies equality of phases at the boundary at all the time to satisfy the boundary conditions. My text book and all the textbooks says the same thing without explaining the reason I am very confused. I am attaching a photo of my text book

 A: For simplicity, assume normal incidence at the point $r=0$, so that all the electric fields are parallel to the interface.
The E-field parallel to the interface is continuous (the same either side of the boundary).
$$ E_i \exp(i\omega_I t) + E_r \exp(i\omega_R t) = E_t\exp(i\omega_T t).$$
Now suppose that $\omega_I$, $\omega_R$ and $\omega_T$ were different.
Let's say that the equality works at $t=0$, so that $E_i + E_r = E_i$, which it has to be, since the sum of the E-field on each side of the boundary must be the same. We can then ask, at what other time is this true? And the answer is that it wouldn't be. There would always be some non-zero phase difference between the components at other values of $t$ (bar some special values if the frequencies were multiples of each other). The only way the relationship $E_i + E_r = E_t$ can be true for all $t$ is if the frequencies were the same.
The argument (and that's a self-contradictory pun!) doesn't change if you add the $\vec{k}\cdot \vec{r}$ parts, because you can always choose some point on the boundary where $\vec{k}\cdot \vec{r}=0$ and demand that the equality is true for all $t$ at that point. In a similar way (probably on the next page of your book) you can set $t=0$ and demand that the equality still holds for all $\vec{r}$ on the boundary - this leads to the law of reflection and Snell's law of refraction.
