Reflection and Refraction of Electromagnetic waves

Consider a plane surface dividing a space into regions, 1 and 2, with different dielectric constants. If a plane wave with wave vector k and frequency \(\omega \) is incident on this surface from the side of region 1, part of this wave will reflect into region 1 and part will refract and transmit into region 2. If the refracted and reflected waves have vector k’ and k’’ respectively, then the electric field for the three waves are \(= (kr)e^{-ikr}\)

\begin{align*}E(r,t)= E_0e^i(\omega t - kr)\: (\text {Incident wave})\\ E'(r,t)= E_0'e^i(\omega' t - k'r)\: (\text {Refracted wave})\\E''(r,t)= E_0''e^i(\omega'' t - k''r)\: (\text {Reflected wave})\end{align*}

1. To prove that the frequency of the wave does not change under reflection and refraction:

   Let us use the boundary condition, the tangential component of the electric field vector remains continuous at the boundary of two media.
   \begin{align*} E_0e^i(\omega t - kr)+ E_0'e^i(\omega' t - k'r)&= E_0''e^i(\omega'' t - k''r)\\\text {or,}\: E e^{-ikr}. e^{i\omega t} + E' e^{-ik'r}. e^{i\omega' t}&= E'' e^{-ik''r}. e^{i\omega'' t}\:\dots (i)\end{align*}
   Differentiating both sides of the equation with respect to time we have,
   \begin{align*} i\omega E e^{-ikr}. e^{i\omega t} + i\omega 'E' e^{-ik'r}. e^{i\omega' t}&= i\omega ''E'' e^{-ik''r}. e^{i\omega'' t}\:\dots (ii)\end{align*}
   Substituting the value of E’’ from equation (i) in equation (ii), we get,
   \begin{align*} iE e^{-ikr}(\omega - \omega') e^{i\omega t} = iE' e^{-ik'r}(\omega'' - \omega') e^{i\omega' t}\dots (iii)\end{align*}
   This equation is satisfied for all values of t which is possible only if \(\omega = \omega ‘\).
   Similarly by eliminating E’ from equation (i) and (ii) and with the same argument, we have \(\omega = \omega ‘’\)
   $$\therefore \omega = \omega ' = \omega '' \dots (iv)$$

2. To prove that the incident, reflected and refracted rays lie in the same plane:

   Let us write the boundary condition equation (i) as
   $$a\: e^{-ikr} + b\: e^{-ik’r} = c\:e^{-ik'' r}\:\dots (v) $$
   where a, b and c are independent of r. Using the vector relation
   \begin{align*}r.\nabla e^{-ikr}&= \left (x\frac {\partial}{\partial x}+ y\frac {\partial}{\partial y} + z\frac {\partial}{\partial z}\right )e^{-ikr} \\ &= -(kr)e^{-ikr} \end{align*}
   Equation (v) now may be written as,
   \begin{align*}-ia(kr)e^{-ikr}- ib(kr)e^{-ik'r}=- ic(kr)e^{-ik''r} \dots (vi)\end{align*}
   Substituting the value of \(ce^{-ik’’r}\) from equation (v) in equation (vi), we have,
   \begin{align*}-ia[(kr) - (k''r)]e^{-ikr}=- ib[(k''r) - (k'r)]e^{-ik'r} \end{align*}
   This is valid for all vector r lying in the boundary plane which is possible only if \(k:r = k’r\). Similarly by eliminating \(be^{-k’r}\) from equation (v) and (vi) and with same argument, we have,
   \begin{align*}k.r &= k''r \\ \therefore kr &= k'r = k'' r \dots (vii)\end{align*}
   It shows that k, k’ and k’’ are coplanar i.e. incident, reflected and refracted rays lie in the same plane.

3. To find the relationship between the angles:

   Let us take the xy plane as the plane of incident, reflected and refracted rays. The y-axis lies along the boundary and the x-axis is perpendicular to it. If we choose the origin at same point on the negative y-axis, the direction of r will be along the positive y-axis. Therefore equation (vii) may be written as
   \begin{align*}kr\cos \theta =k'r\cos \theta' = k''r\cos\theta'' \dots (viii)\end{align*}
   The wave numbers k, k’ and k’’ in terms of velocities of the incident, refracted and reflected waves respectively as,
   \begin{align*}k= \frac {\omega}{\nu},\: k= \frac {\omega '}{\nu '} = \frac {\omega}{\nu'},\:k= \frac {\omega ''}{\nu ''} = \frac {\omega}{\nu''}\end{align*}
   Since the incident and reflected waves are propagated in the same medium, we may write \(v = v’’\) and \(k = k’’\)
   This along with equation (viii) gives
   $$\cos \theta = \cos \theta'' \:\: \text {or,}\: \theta = \theta ''\dots (xi)$$
   Hence, the angle of incident is equal to the angle of reflection, Equation (viii) can be written as
   $$\frac {1}{\nu}\cos \theta = \frac {1}{\nu '}\cos \theta'$$
   Since \(\cos\theta = \sin\theta _i\) and \(\cos\theta ‘= \sin\theta _t\), hence
   $$\frac {\sin\theta _i}{\sin \theta _t}\ = \frac {\nu}{\nu '}\:\dots (x)$$
   If n and n’ be the refractive indices of the first and second media respectively, then we have,
   $$\frac {\sin\theta _i}{\sin \theta _t}\ = \frac {\nu}{\nu '}= \frac {n'}{n}= 1^{n_{2}}\:\dots (x)$$
   Thus ratio of sine of the angle of incident is equal to the ratio of sine of the angle of refraction is equal to the ratio of the refractive index of the second medium with respect to the first i.e. Snell’s law.

Bibliography

P.B. Adhikari, Bhoj Raj Gautam, Lekha Nath Adhikari. A Textbook of Physics. kathmandu: Sukunda Pustak Bhawan, 2011.

Jha, V. K.; 'Lecture title'; Maxwell's Equation; St. Xavier's College, Kathmandu; 2016.