I've seen people talking about the angular resolution of the HST is, if using Rayleigh's criterion, equal to: $$\theta = 1.220 \frac{\lambda}{D}$$ My question is, since the diameter of the HST $D$ (2.4m) is wayyy larger than the wavelength $\lambda$, how on earth would diffraction happen? Wouldn't the light just go through without any effects? Thanks.

  • $\begingroup$ Indeed, the effects are almost negligible. Calculating $\theta$ yields a very small number. $\endgroup$
    – kristjan
    Feb 16, 2015 at 21:59
  • $\begingroup$ Hi @kristjan, I think the smaller $\theta$ is the better, as it stands for the ability of the telescope to differentiate objects to each other. $\endgroup$
    – Lampard
    Feb 17, 2015 at 9:37

1 Answer 1


It all depends on what you want to resolve and how far away it is. For the Andromeda galaxy at a distance of about 2.5 million light years, HST will resolve visible light objects about 0.6 light years apart. If we take our separation from Proxima Centauri (4.3 ly) as typical of separations in the spiral (certainly it's less in the center of the galaxy), the HST will easily resolve individual stars, but not planets in solar systems (too small). In fact, the latest (2014) HST images of Andromeda show individual stars. So the diffraction already limits us to stars.

If we push to galaxies at 25 million ly, the resolution drops to 6 ly and we can't resolve separate stars. That limits us identifying specific single stars that go into supernovas. If we're watching [EDIT here] UV or gamma, it's better because of the shorter wavelengths (smaller $\theta$ is better resolution, and supernovas have very interesting UV and gamma profiles. It's nice to have stellar spectra both before and after the supernova, but if we can't resolve the star it hurts the overall wavelength range analysis. [End edit]

Most recently HST has looked at galaxies about 13 billion ly away. In 500 nm light, that would be a resolution of 3300 ly. That makes it difficult (but not impossible) to determine any shape or structure of the galaxy.

I wouldn't say the light goes through "without any effects."

Added: The reason there is diffraction is the circular aperture always interrupts the EM wave. This partial sampling of the wave is what creates the diffraction effects. It's always there, but we usually don't notice it because other effects will blur an image first (atmospheric, optical aberration, etc.) While the diffraction is small, it is noticeable for images with small angular size. HST's huge aperture located outside the atmosphere allows it to have an impressive distance reach before Rayleigh diffraction is noticable.

  • $\begingroup$ Hi @Bill, thanks for your reply. I guess in my question I was not clear about my point. I know how well HST works but what I'm asking here is what makes it work so well. According to textbooks diffraction happens (at least only noticable) when a wave going through a slit whose width is comparable to its wavelength, but since the diameter of HST is way larger than the light wave length, how would diffraction happen and how is the HST making use of the phenomenon to interpret the atlases? Thanks. $\endgroup$
    – Lampard
    Feb 17, 2015 at 10:34
  • $\begingroup$ @Lampard : I added a section to try to address your question more directly. I don't know about interpreting the atlases. $\endgroup$
    – Bill N
    Feb 17, 2015 at 23:40

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