I would wait for someone else to comment(I'm no EE either), but it looks like if this works then LS5 will get double the voltage and half the current as LS3(It won't be 5 and 10V though, since there's a voltage drop across the diodes).
A little careful thought will show that D2 and D3 must be reverse biased at all times and are thus redundant, you'll then see that both lamps get 10V (although one is a bit less due to the diode drops).
Basically if the batteries are in series they cannot be in parallel. You can arrange for switches to switch between the two configurations though.
You don't need any of the diodes. If you redraw the circuit to its simplest configuration (2 diodes removed as already suggested and the other 2 replaced by solid links) you will find that both batteries are in fact in series with one lamp across one battery and the other across both.
Indeed it is and that’s why in parenthesis I said “if you ignore the internal resistance of the battery”
Note that the original question related to batteries, hence my qualified (and I believe, correct) response.
I’m terribly sorry that I misinterpreted the original problem. I thought he was talking about a real circuit that he intended to build. I didn’t realize that he was talking about a mythical circuit with ideal batteries.