Photocoupler interface
The forward current is the absolute maximum rating.  Just about any current will suffice to power the LED.  Minimize the current for your application, generally 5 to 10 mA is sufficient for firing the LED.

The reverse voltage should be limited to prevent damage to the device.  Although I don't see any reverse protection in the symbol, just keep the maximum reverse voltage below what the manufacturer recommends.

You do not have to ensure 50mA forward current on the emitter, it's just an LED.  The resistor is the same as in a normal LED, size it to provide your desired current flow with your desired supply voltage.  The diagram on the data sheet doesn't say what the supply voltage is.

The only thing that can change the emitter intensity is current through it.  The current through it depends on the supply voltage and the series resistor (or other current limiting means).  Why would you want to change the intensity anyway? That device is a simple optocoupler.  It is meant to switch on and off, not convey an analog signal.   Actually from the packaging of that device, it is most likely meant to be used with an encoder wheel with gaps to alternately pass and interrupt the emitter light.

The maximum open collector voltage appears to be 30 VDC.
Choose your load again, based on your collector supply to ensure reliable and clean switching of the output.  Almost any value will work as long as you don't exceed the output transistors current rating of 20mA.  The lower the load resistor however, the longer the output rise time will be.