I used 9V because the opamp drives the transistor so that the emitter goes to (the supply voltage minus 2V) which is not high enough (only 3V with a 5V supply) to turn on a white, blue or bright green 3.5V LED.I assume the resistors would have to be altered for the 9v?
Use Ohm's Law and simple artithmatic to calculate the resistor value:
1) Find out what is the minimum forward voltage of the LED from its datasheet.
2) The emitter goes to +9V with the switch so subtract the forward voltage and divide by how much current you want (20mA?).
An example is a 1.8V (minimum) red LED. The resistor is (9V - 1.8V)/20mA= 360 or 390 ohms.
Another example is a 3.2V (minimum) blue LED. The resistor is (9V - 3.2V)/20mA= 290 or 300 ohms.
You cannot connect two blue LEDs in series because their max forward voltage might be 3.5V then they won't have enough voltage across the resistor since the emitter is +7V when driven by the opamp.
You said you have a 3-positions switch: On-Off-On. In the off center position then the battery is connected to nuttin.1) (havent got that type of switch in my app but thats the rough idea!)