# Divergence of light rays - parallel approximation

https://www.handprint.com/ASTRO/ae1.html

the author gives this calculation to show light rays from very distant objects are essentially parallel:

In astronomical applications, light sources are so distant that the concentric wavefronts become a series of equally spaced parallel planes across the width of any practical telescope aperture. To illustrate: across the aperture of a 1 meter (39.4") telescope, light rays from a single point on the Moon, the closest astronomical object at 384,403 kilometers, diverge from parallel by no more than 1/384403000 of a radian or 0.0000026 millimeter, which is 0.0047 or 1/200 a wavelength of "green" light. Since the fabrication limits of the highest quality astronomical optics are around λ/20 wave, or 10 times larger than the wavefront divergence, optical calculations can assume perfectly flat and parallel wavefronts from a distant light source.

I don't follow the steps in this reasoning. Is he using a small angle approximation? I know it's very elementary, but could somebody explain it to me?

Is he using a small angle approximation?

Yes!

Consider this triangle:

The bottom side which I labeled "384403000 m" as the Earth-Moon distance, and "1 m" is the diameter of the aperture. The angle $$\theta$$ is the angular separation of two rays, in your words, the divergence of two light rays. By trig function:

$$\tan\theta = \frac{1}{384403000}.$$

Because the fraction is so small, when one expand the $$\tan \theta$$ function by Taylor series, all the quantities with high orders will be practically seen as 0, which gives $$\tan \theta \approx \theta$$, in your words, the small angle approximation. So now we reduce $$\tan \theta$$ to only $$\theta$$.

Note $$\theta$$ is also extremely small, so we can approximate $$\theta \approx 0$$. Remember the unit of the angle is radian! Not degree!

That is the answer of your question, $$\theta$$ is just too small for astronomers to consider about it.