I've started studying basic electronics, and in this video I noticed he had a similar example as in my textbook: https://youtu.be/sTu3LwpF6XI?t=3m28s
Above is a screenshot if you don't bother to click on the link.
So, in essence.. when the switch is off, the LED to the right on. This means, there's a voltage drop across the second resistor and a voltage drop across the LED.
Now, by turning the switch on, we'll have voltage U across both first two paths. The first resistor should make sure not too much current is flowing through the base and thus, turning the transistor on. Now when the transistor is on, the voltage will be lowered to 0 after the second resistor.
Why the LED turns off at this point is because there's no (or at least not enough) current flowing through it (the potential is zero so there can't be). This should be due to the resistance of the LED itself. Or is there anything else going on here?