What kind of telescope could see planets in other galaxies?

[Maximum7]

I was reading the description for Retro Sci Fi Tales #9 and it mentions in the description that their is a science fair in Paris in a fictitious 1800’s universe with a telescope that can see the surfaces of planets in other galaxies. Now Neil deGrasse Tyson said using the sun itself as a lens could allow us to see the atmospheres of exoplanets in our home galaxy. I however would be curious to take it up a notch and see exoplanets in other galaxies; essentially an intergalactic telescope. Speculatist and pseudo-futurist Isaac Arthur mentioned on his channel that a lens made of dark matter could warp spacetime and allow this but he did not really go into detail and someone on Yahoo Answers said the dark matter lens would have to be the size of a galaxy. That’s no good. Way too big. I’m trying to think of a Sci-fiy way to explain a telescope (perhaps a large mountainside one) that can see intergalacticly. I haven’t established the upper limits of its range but I’m sure it cannot see all galaxies in the universe; just the ones near the Milky Way like Andromeda. What pseudo-science grounded in some real believable science can I use to make this telescope for my Sci fi story?

[swansont]

You can apply the Rayleigh criterion to find out. Size is going to be an important factor. 

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/Raylei.html

 

[MigL]

Pretty well constrained by the Physics ...

Angular resolution = 1.22 * waelength / aperture diameter

( the Rayleigh criterion that Swansont mentioned )

This is a sci-fi story,; make something up.

[Curious layman]

these replies don't give much away. 

you've got to know your shit to work out how big your telescope's going to be from that link Swansont.

Quote

If all parts of an imaging system are considered to be perfect, then the resolution of any imaging process will be limited by diffraction. Considering the single slit expression above, then when the wavelength is equal to the slit width, the angle for the first diffraction minimum is 90°. This means that the wave is spread all the way to the plane of the slit and will not contain resolvable information about the source of the wave. This leads to the simplified statement that the limit of resolution of any imaging process is going to be on the order of the wavelength of the wave used to image it.

It's a sci-fI book, not a thesis. 😂 

[MigL]

You can easily calculate the aperture diameter needed, for visible light, to get the desired resolution.
The Rayleigh criterion is a simple formula.

You don't need to know how it works; just plug in approximate numbers.

Edited March 9, 2021 by MigL

[Curious layman]

this? 

What does the first bit, up to the equals sign, mean? The 0 r  and the triangle over the d specifically. Please.

 

is it the angle of the light? The 0 r

that thing over the d, a hypotenuse?

 

Edited March 9, 2021 by Curious layman

[swansont]

54 minutes ago, Curious layman said:

these replies don't give much away. 

you've got to know your shit to work out how big your telescope's going to be from that link Swansont.

It's a sci-fI book, not a thesis. 😂 

You need to know trigonometry. 

46 minutes ago, Curious layman said:

this? 

What does the first bit, up to the equals sign, mean? The 0 r  and the triangle over the d specifically. Please.

 

is it the angle of the light? The 0 r

that thing over the d, a hypotenuse?

 

That’s explained at the link. The numbers are given and identified as wavelength (numerator) and iris diameter (denominator)