# Snell's law and law of reflection derivation -- confusing

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1. Nov 16, 2016

### zhouhao

• Member warned to use the homework template for posts in the homework sections of PF.

I am learning principle of optics written by Born&Wolf and confused with the method to derive Snell's laws and law of reflection.

In the textbook,$v_1$ is the speed of light in mediator 1 and $v_2$ is the one in mediator 2.
${\vec s}$ is the direction of light and ${\vec r}$ is the position.
As shown in Fig.1,
Incident light,Reflected light and Transmitted light are respectively recognized as plane waves and their behaivor at ${\vec r}$ corresponding to$F(t-\frac{{\vec r}\cdot{\vec s}^{(i)}}{v_1})$,$F(t-\frac{{\vec r}\cdot{\vec s}^{(r)}}{v_1})$ ,$F(t-\frac{{\vec r}\cdot{\vec s}^{(t)}}{v_2})$.
If ${\vec r}$ at interface, we have $t-\frac{{\vec r}\cdot{\vec s}^{(i)}}{v_1}=t-\frac{{\vec r}\cdot{\vec s}^{(r)}}{v_1}=t-\frac{{\vec r}\cdot{\vec s}^{(t)}}{v_2}$.Then we could get $\frac{s_x^{(i)}}{v_1}=\frac{s_x^{(r)}}{v_1}=\frac{s_x^{(t)}}{v_2}$ (Snell's law and Law of reflection included in).

My problem is :
1,Plane wave spreading over all space as shown in Fig.2,it is not like a stright ray(Does this means a light with width combined with many parallel rays behaves like plane wave?);

2,Where is the $F(t)$ with ${\vec r}=0$,I think if it is originating from somewhere of mediator 1,Reflection rays and Transmitting rays should be expressed as $F(t-\frac{{{\vec r}_A}\cdot{\vec s}^{(i)}}{v_1}-\frac{({\vec r}-{{\vec r}_A})\cdot{\vec s}^{(r)}}{v_1})$,$F(t-\frac{{{\vec r}_A}\cdot{\vec s}^{(i)}}{v_1}-\frac{({\vec r}-{{\vec r}_A})\cdot{\vec s}^{(t)}}{v_2})$,where A is the point of reflection and transmittion happening.I couldn't conduct the laws from my thoughts.So how does the authors think?

Last edited: Nov 16, 2016
2. Nov 17, 2016

### ehild

Yes, you can consider a ray as normal to the phase planes ( the planes where the phase $t-\frac{\vec s \cdot \vec r}{v}$ is constant).
The plane waves do not originate somewhere, they are of infinite extension. The function F is not the same for incident, reflected and transmitted waves. Easier to think of sinusoidal plane waves of given frequency. The electric and magnetic fields can be described as sum of plane waves. If the incident wave is $A_i \sin(\omega t-\vec r \cdot \vec s_i)$ , the reflected one is $A_r \sin(\omega t-\vec r \cdot \vec s_r+\phi_r)$ and the transmitted one is $A_t \sin(\omega t-\vec r \cdot \vec s_t+\phi_t)$, all with their own amplitudes and phase constants. In the medium, where there are both travelling and reflected waves, the field is the sum of them. The transmission and reflection happens at the whole interface between the two different media. Both the electric and the magnetic fields are of such form, but they are related according to Maxwell equations. Both the electric and magnetic fields are subject to the boundary condition, that the field components parallel with the interface at a certain time and place, have to be the same at both sides of the interface. From these condition, you can find the amplitudes and phase constants of the reflected and transmitted waves.

3. Nov 23, 2016

### zhouhao

Thank you.I got your idea and understood the way to deduce Snell's law and Reflection law in the Born&Wolf's book.(It was shown below).There is still two problem.
Incident wave:$A_i \sin(\omega t-\frac{\vec r \cdot \vec s_i}{{\lambda}_i}+\phi_i)=A_i \sin(\omega t-\frac{(\vec r-\vec r_o) \cdot \vec s_i}{{\lambda}_i}+\frac{\vec r_o \cdot \vec s_i}{{\lambda}_i}+\phi_i)$
Reflected wave:$A_r \sin(\omega t-\frac{\vec r \cdot \vec s_r}{{\lambda}_i}+\phi_r)=A_r\sin(\omega t-\frac{(\vec r-\vec r_o) \cdot \vec s_r}{{\lambda}_i}+\frac{\vec r_o \cdot \vec s_r}{{\lambda}_i}+\phi_r)$
Transmitted wave:$A_t \sin(\omega t-\frac{\vec r \cdot \vec s_t}{{\lambda}_i}+\phi_t)=A_t \sin(\omega t-\frac{(\vec r-\vec r_o) \cdot \vec s_t}{{\lambda}_t}+\frac{\vec r_o \cdot \vec s_t}{{\lambda}_t}+\phi_t)$

Choose appropriate $\vec r_o$ to make $\frac{\vec r_o \cdot \vec s_i}{{\lambda}_i}+\phi_i=\frac{\vec r_o \cdot \vec s_r}{{\lambda}_i}+\phi_r=\frac{\vec r_o \cdot \vec s_t}{{\lambda}_t}+\phi_t=0$ and let ${\vec r_h}=\vec r-\vec r_o$

Incident wave,reflected wave and transmitted wave could be expressed as
$A_i \sin(\omega t-\frac{\vec r_h \cdot \vec s_i}{{\lambda}_i})$,$A_r \sin(\omega t-\frac{\vec r_h \cdot \vec s_r}{{\lambda}_i})$,$A_t \sin(\omega t-\frac{\vec r_h \cdot \vec s_t}{{\lambda}_t})$.

At the interface(Fig.1),the three waves' phases are equal:$\frac{\vec r_h \cdot \vec s_i}{{\lambda}_i}=\frac{\vec r_h \cdot \vec s_r}{{\lambda}_i}=\frac{\vec r_h \cdot \vec s_t}{{\lambda}_t} {(1)}$

If we define $\vec r_h=(x,z)$,$\vec s_i=(\sin{\theta_i},\cos{\theta_i})$$\vec s_r=(\sin{\theta_r},-\cos{\theta_r})$,$\vec s_t=(\sin{\theta_t},\cos{\theta_t})$,
then through equation (1) and the interface condition(fig.1):$x$ could be varied while $z$ was fixed , we got
$\theta_i=\theta_r$,$\frac{\sin{\theta_i}}{\sin{\theta_t}}=\frac{\lambda_i}{\lambda_t}$

Problem:
1,Is there a wave function to express a light existing during finite time and finite space?
2,Why the three waves' phase at interface is equal?

4. Nov 23, 2016

### ehild

You can not cancel all phase constant with a single "appropriate" ro.
,
You do not need to play with that "appropriate ro" From the original waveforms, you also get that the phase change at reflection can be zero or pi, and zero at refraction. These are valid for transparent materials with real refractive indices.
1. there is, but it is not a plane wave.
2. They are not all equal, but are determined by the boundary conditions at the interface. The change of the phases with time and place is the same. Therefore the frequency is the same for the incident, reflected and transmitted waves, and also sinθ/λ are the same.

Using complex wavefunctions of form $Ae^{i(ωt-\vec k \cdot \vec r) }$, the phase constants are included into the amplitudes, and it is more clear that the phase $(ωt-\vec k \cdot \vec r)$ most be the same for all the incident, reflected and transmitted waves, in order that the boundary conditions be valid all time and all places on the interface.

5. Dec 13, 2016

### zhouhao

Yes,there is no appropriate $\vec r_o$.I have made a mistake.Thank you for clarifying.
"phase$(wt-\vec k \cdot \vec r)$ must be same for all the incident,reflected and transmitted waves",