bjt - 'Leftover' voltage when using constant current

I came across this circuit,

^{simulate this circuit – Schematic created using CircuitLab}

Which states that the current through the LED is equal to 0.5*R2/1000. I'm not 100% sure how to come to that conclusion, but that is only a part of my question (a part that I am still happy to have answered! :))

My main concern is as follows: Let's say I choose R2 = 40 ohms and the current through the LED is therefore 20 mA. And let's also say that, from the datasheet, the forward voltage of the LED is 3 V. From Ohm's Law the voltage across the resistor is equal to 0.02 * 40 = 0.8 V. What happens to the remaining 1.2 V from the constant 5V DC voltage source?

I'm reluctant to test this circuit without knowing the answer in case it causes excessive heating or damage to my components.

Answer

The remaining voltage will be dropped across Q1. Or, the voltage across Q1 will vary to ensure that 20 mA flows through the LED and R2.