1. Diode current = bulb current = slightly less than 3A in this case
2. The internal resistance of each diode is about 1V /3A = 0.33Ω
3. The voltage applied to the light bulb = approx 10V
At rated current, silicon diodes drop approx 1V each--rather than consider the elusive resistance (that changes with load), it is far easer to estimate voltage drop as the silicon diode tends to have a fixed voltage drop. At light loads, this voltage drop is about 0.65V.
Two series diodes are required in order to obtain sufficient voltage to forward bias the LED inside the opto-coupler.
I can envision a superior circuit that would use a TL080 op amp and a low value series resistor that would drop perhaps 0.1V and apply 11.9V to the bulb thus making it much brighter. Dim light bulbs bother me...