Microstrip Power Divider 
Attached is an image of a microstrip PCB I am working on. In the image the orange is the copper trace and the blue is the Rogers4350 substrate. Underneath the blue substrate is the ground plane. The imaged circuit is a power divider which takes the RF signal and divides it equally between two directions. Initially the microstrip line has 100 ohm impedance. At the T junction the microstrip line increases to 200 ohms (200ohms in parallel with 200 ohms is 100 ohms). From a circuit standpoint I can understand how and why the power flows from the 100 ohm line to the 200 ohm lines.
Where I am confused though is when I view the signal as an electromagnetic wave. I struggle to understand why the wave would change direction and flow into the 200 ohm lines as opposed to simply reflecting from the open in the microstrip? Does the wave diffract from the corners of the T junction? Is the fact that the wavelength of the EM wave is much larger than the T junction a relevant fact? My rational here being that due to a larger wavelength there is charge which is being forced to pile up at the T-junction. Eventually this charge must redistribute and flows towards the 200 ohm lines?
Simply put I am trying to better understand the mechanism by which the majority of the wavefront is able to squeeze from the wide 100 ohm trace into the two much thinner 200 ohms traces with less than -20 dB of reflected power.
Thank You
 A: Huygens' principle applies as much to RF waves propagating in waveguides as it does to optical waves passing through apertures. Each point on the wavefront can be treated as a source of spherical waves and the superposition of the waves radiating from all these sources will give the further propagation of the wave.
In this case, the wave encountering the "end" of the 100-ohm line segment radiates in all directions. In the "forward" direction it can't propagate further because the copper structure that guided it ends. But to the left and right it can propagate into the 200-ohm line segments. So that is what it does.
A: If you have not yet learned it then you will soon that discontinuities on a transmission line, such as sudden change in dimension or dielectric/magnetic properties, etc., correspond to a a network of lumped element circuits, specifically capacitors and inductors, and if the metallic/dielectric losses are also taken account then also including resistors. These lumped element circuit elements cause the (q)TEM wave (q-quasi: because it is not a pure TEM for microstrip) propagating along the $R_1=100\Omega$ transmission line would be reflected as you have expected it by the junction where it meets the two $R_2=200\Omega$ lines.
Notice that in the drawing those two high impedance lines are shaped so that they flare out going away from the junction. Because of the flare the wave impedance of the nominally $R_2=200\Omega$ line is actually higher than $R_2$ near the junction and then gradually after a couple or so wavelengths becomes $R_2$ towards the load. The purpose of that shaping is to tune out the mismatch caused by the junction (lumped element L/C circuit, losses are ignored), as much as possible. The longer is the flare the broader the tuning bandwidth is. If there is not enough room for the flare then tuning can also be accomplished by smaller reactive load(s) directly on the junction. In metallic waveguides screws and other metallic plates (iris) can be inserted to act as reactance(s). On a microstrip line you can put notches, slits, etc., to match the junction.
