Can you please point me to the diode you're interested in and remind me how much optical power you want? I'm not following this thread very well. Then I'll try and answer each of your questions one by one.

EDIT:

I'll answer based on this diode on eBay:

405nm 50mw CW Violet/Blue Laser Diode LD SLD3232VF SONY New
1) Was i correct? i mean that 17 ohm calculation.

First, I would use the typical value, not the laser diode's maximum value. Second, the text in the eBay auction states the maximum current is

< 75mA (75mA=0.075A) however the datasheet provided by the link in the auction lists the maximum operating current at 65mA. From the datasheet the typical operating current of 55mA produces 50mW of optical power. Therefore the resistor value needed using an LM317 is 1.25V/0.065A

~19 ohms.

So yes, your calculation of 17 ohms using 75mA was done correctly - only I recommend that you use 55mA instead of 75mA for the reasons stated above.

2) I can see see voltage and current from your webpage. But can i calculate it somehow?

I am not aware of a model, calculation, program, etc. that computes optical power out as a function of input current. That doesn't mean one doesn't exist. Try

Google. I would guess that the manufacturer has extensive models based on their own internal research and development. Maybe try looking through a laser diode physics textbook. I've also noticed some diodes input current/output power relationship appears pretty linear so all you would need are a couple points to create your own equation/expression.

I use pictures taken from other people, e.g. DTR's website, or conduct a test myself.

3) I havent asked it yet but lets say im using potentiometer. How do i know which potentiometer should i use? Can i calculate it? 100 ohm is the best as i can see but why not 10 ohms or 1000 ohms?

As you probably already understand, using a potentiometer with an LM317 allows you to vary the current into the laser diode thus varying the output power. Be aware, however, potentiometers are 0 ohms on one end and whatever value it is on the other. You don't want to accidentally (or allow someone else to accidentally) allow too much current into the laser diode by allowing 0 ohms as it will destroy the diode. A way to fix a maximum current value but also allow variability is to use a fixed resistor and a pot in series with each other.

The first step is determine what range of input current you want. The datasheet doesn't give a minimum current but lists typical at 55mA and max at 65mA. For the sake of this calculation, let's pick a minimum current of 45mA. The fixed resistor determines the max current and is calculated to be 1.25V/.065A=19 ohms. The range of the pot is determined by the difference between the resistance values for minimum and maximum current. Calculating the resistor for minimum current we have 1.25V/0.045A=28 ohms. So 'ideally' you want a pot that goes from 0 ohms to (28-19)=9 ohms so you select a 10-ohm pot. Anything higher than a 10-ohm pot will simply produce less current thus producing less optical power. It's a design/requirements choice.

You also want to select the proper wattage for both the fixed resistor and the pot. Since power is (current) x (current) x (resistance), or (voltage) x (current), use the value that produce the largest power. In this case we have 0.065A going through a 19-ohm resistor which will dissipate 0.08 watts (or ~1/12th of a watt). The pot goes from 0 to 10-ohms but is in series with the fixed resistor so it's power dissipation is less than the fixed resistor. So a common choice for resistor wattage (fixed or pot) is 1/10th which meets your requirements. Some 'derate' their operational values which means choosing a value that is more conservative like 1/8th Watt or 1/4 Watt parts.

Let me know if you have any questions or find something that isn't calculated quite right, I'm off to dinner and will proof when I get back.