Wave reflections from mismatched impedances have inverted step waves for DC and inverted phases for AC. Just like waves in a pool. :)

added: Do you equate optically denser to higher relative permitivity to lower relative impedance? Think of the wavelet as a vector which can only reflect a range of inphase or opposite with null in balance of equal density.

" If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. This is due to changes in dielectric constant or other physical properties. https://books.google.ca/books?id=k1brJjXmXOQC&pg=PA43&lpg=PA43&dq=light+reflection+impedance+phase+inversion&source=bl&ots=G3qHMfPksC&sig=hwt5bC3GuiJ6OU3uI7n0XSmFjR4&hl=en&sa=X&ei=RS6rT6uXM4Wg8QT23Kka&ved=0CFkQ6AEwAQ#v=onepage&q=light%20reflection%20impedance%20phase%20inversion&f=false

Added: This illustration should answer your question intuitively with dark bands caused by out of phase or destructive reflection.

I propose removing this image or finding a suitable one. This diagram is part of Young's experiment which illustrates the phenomena of diffraction and interference, but I am afraid they do not illustrate reflection. No reflected wave is represented there. As you might see the two waves come from different sources. The reflected wave should have the same angle as the incident wave (both respect to the normal to the surface) which is not the case of the two last diagrams. The first one could be interpreted as a reflection without phase change, but that is more confusing than clarifying.

Wave reflections from mismatched impedances have inverted step waves for DC and inverted phases for AC. Just like waves in a pool. :)

added: Do you equate optically denser to higher relative permitivity or to lower relative impedance? Think of the wavelet as a vector which can only reflect a range of inphase or opposite with null in balance of equal density.

" If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. This is due to changes in dielectric constant or other physical properties. https://books.google.ca/books?id=k1brJjXmXOQC&pg=PA43&lpg=PA43&dq=light+reflection+impedance+phase+inversion&source=bl&ots=G3qHMfPksC&sig=hwt5bC3GuiJ6OU3uI7n0XSmFjR4&hl=en&sa=X&ei=RS6rT6uXM4Wg8QT23Kka&ved=0CFkQ6AEwAQ#v=onepage&q=light%20reflection%20impedance%20phase%20inversion&f=false

Added: This illustration should answer your question intuitively with dark bands caused by out of phase or destructive reflection.

I propose removing this image or finding a suitable one. This diagram is part of Young's experiment which illustrates the phenomena of diffraction and interference, but I am afraid they do not illustrate reflection. No reflected wave is represented there. As you might see the two waves come from different sources. The reflected wave should have the same angle as the incident wave (both respect to the normal to the surface) which is not the case of the two last diagrams. The first one could be interpreted as a reflection with phase change, but that is more confusing than clarifying.

Wave reflections from mismatched impedances have inverted step waves for DC and inverted phases for AC. Just like waves in a pool. :)

added: Do you equate optically denser to higher relative permitivity to lower relative impedance? Think of the wavelet as a vector which can only reflect a range of inphase or opposite with null in balance of equal density.

" If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. THis is due to changes in dielectric constant or other physical properties. http://goo.gl/vTwQq

Added: This illustration should answer your question intuitively with dark bands caused by out of phase or destructive reflection.

Wave reflections from mismatched impedances have inverted step waves for DC and inverted phases for AC. Just like waves in a pool. :)

added: Do you equate optically denser to higher relative permitivity to lower relative impedance? Think of the wavelet as a vector which can only reflect a range of inphase or opposite with null in balance of equal density.

" If terminal impedance is lower the reflection is inverted (-180deg) if higher it is in-phase, if equal, there is no reflection. This is due to changes in dielectric constant or other physical properties. https://books.google.ca/books?id=k1brJjXmXOQC&pg=PA43&lpg=PA43&dq=light+reflection+impedance+phase+inversion&source=bl&ots=G3qHMfPksC&sig=hwt5bC3GuiJ6OU3uI7n0XSmFjR4&hl=en&sa=X&ei=RS6rT6uXM4Wg8QT23Kka&ved=0CFkQ6AEwAQ#v=onepage&q=light%20reflection%20impedance%20phase%20inversion&f=false

Added: This illustration should answer your question intuitively with dark bands caused by out of phase or destructive reflection.

I propose removing this image or finding a suitable one. This diagram is part of Young's experiment which illustrates the phenomena of diffraction and interference, but I am afraid they do not illustrate reflection. No reflected wave is represented there. As you might see the two waves come from different sources. The reflected wave should have the same angle as the incident wave (both respect to the normal to the surface) which is not the case of the two last diagrams. The first one could be interpreted as a reflection without phase change, but that is more confusing than clarifying.