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1.Why it is necessary that phase of incident, reflected and refracted wave must equal at the interface of two medium to satisfy the boundary conditions at the interface? 2. According to boundary conditions fields left to the interface must join fields right to interface why we just add amplitudes of fields E(I) + E (R) = E(T) here E(I), E (R) and E(T) are amplitudes of incident,reflected and transmitted wave enter image description here

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  • $\begingroup$ It's hard to answer without knowing your level. Are you familiar the way to derive conditions on fields at boundaries using Maxwell's equations? $\endgroup$ Commented Nov 28, 2020 at 20:39
  • $\begingroup$ You may want to look into the Fresnel equations. $\endgroup$
    – my2cts
    Commented Nov 28, 2020 at 21:12
  • $\begingroup$ Prof.Andrew Steane my particular doubt is if according to boundary conditions fields left to the interface must join fields right to interface why we just add amplitudes of fields E(I) + E (R) = E(T) here E(I), E (R) and E(T) are amplitudes of incident,reflected and transmitted wave $\endgroup$ Commented Nov 29, 2020 at 4:54

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It's not necessary that the phase should be equal, as in the case of $v_2<v_1$, the reflected and incident wave is out of phase.

What is important that the field must satisfy the electrodynamic boundary conditions : $$(1) \ \ \epsilon_1E_1^\perp=\epsilon_2E^\perp_2, \ \ \ \ \ (2) \ \ \ \ E^\parallel_1=E_2^\parallel$$ $$(3) \ \ \ B_1^\perp=B_2^\perp, \ \ \ \ \ \ (4) \ \ \ \ \ \frac{1}{\mu_1}B^\parallel_1=\frac{1}{\mu_2}B^\parallel_2$$ These equations relate the electric and magnetic fields just to the left and just to the right of the interface between two linear media.

The rest follows from here.


Additional Answer On Edited Question:

It's not necessary to use Amplitude, You can, if want to, use the fields. Suppose like as in Griffith, the fields are : $$\mathbf{E}_I(z,t)=E_{0I}e^{i(k_1z-\omega t)}\hat{x}$$ $$\mathbf{E}_R(z,t)=E_{0R}e^{i(-k_1z-\omega t)}\hat{x}$$ $$\mathbf{E}_T(z,t)=E_{0T}e^{i(k_1z-\omega t)}\hat{x}$$

The interface is at $z=0$ so $$\mathbf{E}_I(0,t)+\mathbf{E}_R(0,t)=\mathbf{E}_T(0,t)$$

$$\Rightarrow E_{0I}e^{i(k_1\cdot 0-\omega t)}+E_{0R}e^{i(-k_1\cdot 0-\omega t)}=E_{0T}e^{i(k_1\cdot 0-\omega t)}$$ $$E_{0I}+E_{0R}=E_{0T}$$ That's the same thing as to add amplitudes.

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  • $\begingroup$ Yes I am aware of this point I think I just asked wrong questions pls review my edited question $\endgroup$ Commented Nov 29, 2020 at 4:57
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    $\begingroup$ See the edit and I don't think you have permission to add new questions through the edit, It's only for clarifying meaning or mistake. You can ask a separate questions. $\endgroup$ Commented Nov 29, 2020 at 5:14
  • $\begingroup$ Reflection occurs at a point and that can choose to be the origin. There is no loss of generality. $\endgroup$ Commented Nov 29, 2020 at 5:32
  • $\begingroup$ What happens in the case of oblique incident. Is this result still valid and if yes what about values of x and y they may not be zero ?This where the doubt of equality of phases came in my mind because this the only way exponential factor on both side cancel out and give the result u just derived $\endgroup$ Commented Nov 29, 2020 at 5:34
  • $\begingroup$ As I said, The reflection point occurs at a point in the reflection plane. You can choose it to be origin so that $(x,y,z)=(0,0,0)$. $\endgroup$ Commented Nov 29, 2020 at 5:39

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