Thanks for the clarification. I was wondering the same things that Adam asked.
TX and RX mean transmit (or transmitter) (i.e. the LED) and receive (or receiver) (i.e. the phototransistor).
For that particular problem, a simple solution is to reduce the value of the emitter resistor.
Here's an explanation.
The phototransistor responds to light intensity by producing a roughly proportional photocurrent, or more correctly, by allowing a current to flow through it, from its collector to its emitter.
So if you connected a phototransistor across a power supply with no series resistor (don't actually do this because you could damage it if the light is bright enough), and measured the current through it, you would see the current vary roughly linearly in response to the light falling on the phototransistor.
Imagine that you subject the phototransistor to a steadily increasing light intensity. The current might start at 0.1 mA, increase smoothly through 1.0 mA, and eventually reach 10 mA.
The 555's trigger input responds to voltage. So you need some way to convert the current from the phototransistor into a voltage; hence the emitter resistor.
But the combination of the supply voltage and the value of the emitter resistance sets a limit for the current that can flow through the phototransistor, according to Ohm's Law: I = V / R. In your case, V isn't specified, so I'll assume it's 12V, and R is 8.2k. That limits the current to about 1.5 mA.
Now imagine that you subject the phototransistor to the same steadily increasing light intensity. Initially, say, 0.1 mA will flow. This produces a voltage drop of 0.82V across the resistor (from a rearrangement of Ohm's Law: V = I R). There will be 11.18V remaining across the phototransistor.
The current continues to increase, following the brightness of the light source. When it reaches 1 mA, the voltage across the emitter resistor will be 8.2V. There will be only 3.8V across the phototransistor. So you can see that as the phototransistor passes more current, more voltage is dropped across the emitter resistor, and less voltage remains across the phototransistor.
Now imagine that the illumination has increased to a level that should produce 1.4 mA current in the phototransistor. 1.4 mA through an 8.2k resistor produces 11.48V across the resistor, leaving only 0.52V across the phototransistor. The phototransistor is "running out of voltage"! It cannot create voltage from nothing; as the illumination continues to increase, the phototransistor will become "saturated" - that is, it "wants to" pass more current, but its collector-emitter current is being limited by the external components (the supply voltage and the emitter resistor) and cannot continue to increase in proportion to the "input" current (the photocurrent caused by the incident light).
So as the light intensity continues to increase, the phototransistor current flattens off into a horizontal line. A typical transistor saturation voltage is around 0.2V, leaving 11.8V across the 8k2 resistor; from I = V / R, the current will be about 1.44 mA. Even when the light intensity has increased to the point where the phototransistor "wants to" pass 10 mA of current, the actual current will only be 1.44 mA, about seven times less in this example.
When a transistor is saturated, it takes a while for it to "un-saturate", because solid state physics. Something called charge storage which I don't understand, and probably other factors too. So if the phototransistor sees a rapid drop in light intensity, it won't even start to respond until the light intensity has dropped by a factor of seven (in this example), and it will respond quite sluggishly.
So in this application, you really want to avoid, or at least minimise, saturating the phototransistor. This is done by changing the external circuit so the phototransistor can pass more current. Either increase the supply voltage, or reduce the emitter resistor - the resistor is the obvious choice.
Here's what I suggest. This may be a bit simplistic but it should be a start.
1. Make your LED brightness adjustable, or just switch-settable to full brightness and half brightness, for calibrating the phototransistor. For example, if your current limiting resistor is 220 ohms, connect another 220 ohm resistor in series with it, with a switch across it. Closing the switch puts only the first resistor in circuit, giving you full brightness. Opening the switch gives 440 ohms series resistance which will roughly halve the LED's brightness.
2. Determine the minimum emitter resistance using the phototransistor's data sheet. For example if the phototransistor data sheet says the maximum rated current is 100 mA, go for half that value, i.e. 50 mA, and calculate R using R = V / I where V is the power supply voltage and I is the current in amps.
3. Replace the emitter resistor with a preset potentiometer, aka "trimpot", with a resistance of, say, 1k, in series with a resistor with the value you calculated. This is just to prevent possible damage to the phototransistor.
4. Turn the trimpot to the maximum resistance end. Set the LED to half brightness. With your detector circuit set up normally with no obstructions in the light path, adjust the trimpot until the 555 circuit triggers.
5. Return the LED to full brightness.
That should do it. It calibrates the detector circuit so it triggers when the light intensity drops to half its normal value. Since the 555 triggers when pin 2 goes below 1/3 of the supply voltage, with the LED running at full brightness you should see around twice that voltage, i.e. 2/3 of the supply voltage, on the phototransistor's emitter. In that state, the phototransistor is not saturated (there is still 1/3 of the supply voltage across it) so it should respond very quickly to changes in light intensity.
Using a lower emitter resistance also reduces the effect of capacitance, which is another delaying factor. In general, operating a circuit at a higher current makes it faster. This is why desktop computers still have a performance advantage over laptops.
If you find that variations in temperature or other factors cause the circuit to trigger when it shouldn't, you can increase the emitter resistance. This may make the detector respond less quickly though.
. Kris if you are not a teacher you should be, your explanation cleared things up for me. I found a data sheet on the photo transistor and that gave me a starting point for my emitter resistor value. 
