Hi...
As you know, the output of the 4011 oscillator is a square wave.
This means that the duty cycle of the IR LED current is equal to 50% when Dout is on.
I downloaded the datasheet of LD271. On page 5, for pulses about 15us (as in your case) and a duty cycle = 50%, the permissible IF (forward current) is a bit higher than 300mA.
Based on your original circuit, you can test it as follows:
(1) Discoonect Dout from pin1 of 4011.
(2) Add a 10K resistor between pin1 and Vcc (12V).
(3) Run the circuit, you will get continuous pulses.
(4) by a DC voltmeter, measure the voltage (Vr) across R7 (47 Ohm)
(5) The LED peak current is equal to 2*Vr/47
(6) Now the power dissipated on R7 is around 2W.
(7) To increase the current you can add one or more resistors in parallel with R7, so I_LED = 2*Vr_new / R_equivalent
Note if you have another LED try to add in series with the first one. The current will be reduced but you can decrease the value of R7. In this case the dissipated power is less since some of the previous one is taken by the added IR LED. You may need to experiment the new distance range after using 2 LEDs... perhaps at some points they cancel each other... while at other points their signals are added.
Kerim