Post a New Question


posted by .

The Hubble Space Telescope (HST) can resolve objects that have a small
angular separation because there is no atmospheric distortion of the light. (i.e. The HST has excellent
resolving power.) Its 2.4-m-diameter primary mirror can collect light from distant galaxies that
formed early in the history of the universe. How far apart can two galaxies be from each other if they
are 10 billion light-years away from Earth and are barely resolved by the HST using visible light with
a wavelength of 400 nm? You may assume that the aperture diameter is the full 2.4 m diameter of the

okay so this is what I have tried so far:
I used a sin delta theta>/= 1.22 lambda nought
rearrange to get
delta theta>/= [inverse sin (1.22)(4.00*10^-9m)]/2.4m
so delta theta = 0.00001165

The only reason I used that formula is because it seemed like the best option. I'm not even sure if I'm using the right one.

I don't even know if I'm on the right track, and I have no idea what to do from here. Any help would be appreciated!

  • physics-telescope -

    You are on the right track but do not appear to be solving for the separation distance of the galaxies from each other. Call that x. That is what they are asking for.

    1.22 (lambda)/D = x/R

    where R is 10 million light years
    lambda = 400*10^-9 m
    D = 2.4 m

    Solve for x.
    If you leave both x and R in light years, you can solve for x in light years.
    x/R = 2.03*10^-7 (a dimensionless ratio)
    x = 203 light years (when the light was emitted 10 billion light years ago)

  • physics-telescope -

    My prof says that the answer should be of magnitude 10^19. I'm not sure what he's doing differently. Is there another way to solve this?

  • physics-telescope -

    I tried to get help from a friend, but her email just said to use tan theta = x/D. I'm not sure why she said to use tan theta, and when I do that-> tan 0.00001165*2.4m=x I get x= 0.000000487 which is obviously not right.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question