An electromagnetic wave of unit amplitude is incident on one side of a boundary between two media, and the amplitudes of the reflected and transmitted waves are \(r\) and \(t\), respectively. If the rays are all reversed in direction, then \(r\) and \(t\) should combine to give just the original unit amplitude. Use this argument to show that \(r^{\prime}=-r\) and \(r^{2}+t t^{\prime}=1\), where \(r^{\prime}\) and \(t^{\prime}\) are the values of \(r\) and \(t\) for incidence on the other side of the boundary. Show that the Fresnel coefficients obey these relations.

Consider an electromagnetic wave of unit amplitude incident on the boundary between two media. The wave is partially reflected with amplitude \(r\) and partially transmitted with amplitude \(t\).

Now, when the rays are reversed in direction, the transmitted wave comes back with an amplitude of \(t'\) and the reflected wave comes back with an amplitude of \(r'\). They must combine to give just the original unit amplitude.

Given that the original unit amplitude should be obtained when the reflected wave \(r\) combines with the transmitted wave from the other side \(t'\), we can write \(r + t' = 1\) Now, consider the amplitude of the transmitted wave on the other side \(t\) and the reflected wave on that side \(r'\). We have \(t - r' = 1\) Taking the difference between the two equations, we get: \(r + t' - (t - r') = r + r' = 0\) Thus, \(r^{\prime} = -r\) To find the relationship between the powers, we take the square of the first equation and add it to the square of the second equation: \((r + t')^2 + (t - r')^2 = 1 + 1\) The relation between the powers becomes: \(r^2 + t t' = 1\)

The Fresnel coefficients describe the relation between the amplitudes of the reflected and transmitted waves. They depend on the incident angle and refractive index of the media involved. The Fresnel coefficients for amplitude \(r\) and \(t\) are given by: \(r = \frac {n_{1}\cos\theta_{1} - n_{2}\cos\theta_{2}}{n_{1}\cos\theta_{1} + n_{2}\cos\theta_{2}}\), \(t = \frac {2n_{1}\cos\theta_{1}}{n_{1}\cos\theta_{1} + n_{2}\cos\theta_{2}}\) For the reverse direction, we have: \(r' = \frac {n_{2}\cos\theta_{2} - n_{1}\cos\theta_{1}}{n_{2}\cos\theta_{2} + n_{1}\cos\theta_{1}}\), \(t' = \frac {2n_{2}\cos\theta_{2}}{n_{2}\cos\theta_{2} + n_{1}\cos\theta_{1}}\) On comparing \(r'\) and \(r\), we find \(r^{\prime} = -r\). Similarly, we can find: \(r^2 + t t' = \left(\frac {n_{1}\cos\theta_{1} - n_{2}\cos\theta_{2}}{n_{1}\cos\theta_{1} + n_{2}\cos\theta_{2}}\right)^{2}+\frac {2n_{1}\cos\theta_{1}}{n_{1}\cos\theta_{1} + n_{2}\cos\theta_{2}}\cdot\frac {2n_{2}\cos\theta_{2}}{n_{2}\cos\theta_{2} + n_{1}\cos\theta_{1}} = 1\) Thus, the Fresnel coefficients obey the given relations.