Numerical aperture (NA) of an optical fiber The Numerical Aperture (NA) (for fiber optics) is usually used to denote the acceptance cone for a multi-mode fiber. 
Does NA also describe the expansion of light emitted from the end of a fiber?
I have a 1 mm core 0.22 NA PMMA fiber, I would like to collimate the light once it's reached a 1.5" diameter. 
$$
n\sin(\theta)=NA \\
\theta = \arcsin(\frac{.22}{1.4914}) \\
\frac{.5 \cdot 38 [mm]}{ \tan(\theta)}=\left \{ \text{focal length} \right \}
$$
So, I should need a ~130 mm focusing lens?
edited in response to answer below
Is n the index of refraction of the fiber core (1.4914), or free space (1)?
 A: Yes, but.
The quality of the collimated beam may not be fantastic.  And it depends on how the light is launched into the fiber.  If the incident light has a narrow angular dispersion, and the fiber is short and/or of very good quality with few bends, then the light coming out will have a low angular dispersion.  But if the incident light is converging at an angle close to the NA of the fiber, then your analysis should be ok.
A: I'm sorry, but your formulas do not seem correct.
First, the NA is defined in free space, not inside the fiber.  So, NA = 0.22 means that is the NA in air.  Therefore, the angle is $sin(\theta) = 0.22 \rightarrow \theta = 12.7 deg$. [no need to divide by the index of refraction]
Next, NA refers to the half angle, not the full angle (see:  http://en.wikipedia.org/wiki/Numerical_aperture ).  So: $ tan(\theta) = \frac{D/2}{f} = \frac{38 mm/2}{f}$ which gives you $f = \frac{17.5 mm}{tan(\theta)} = 77.6 mm$ .
Finally, please appreciate that collimation is identical to imaging at infinity.  What this means in practice is two-fold


*

*If you look on a wall that is much more than 77 mm from your lens, you'll see an image of the fiber tip (defects included!)

*The divergence angle of the collimated light is non-zero.  Therefore, it does not stay the same diameter as it propagates.  As soon as the light exits the lens, it will begin to diverge at a rate of:  $\theta \approx \frac{D_{fiber}/2}{f} = \frac{0.5 mm}{77.6 mm} = 6.4 mrad = 0.37 deg $

