Which of the following two telescopes has a better angular resolution? 1) the Hubble Space Telescope (2.4 meter aperture) at optical wavelengths and 2) the Very Long Baseline Array (VLBA), a radio interferometer with a maximum baseline of 8600 km observing at 1 mm wavelengths. (Please answer with either HST or VLBA)
well, what is wavelength of visible light?
for resolution it is the ratio of wavelength to distance between slots that matters.
for your radio it is
1^10^-3 meters / .6*10^ m
the lower the wavelength to baseline, the sharper
Thank you!
THANKS
Well, if we're talking about angular resolution, I'd have to say the Hubble Space Telescope (HST) has the edge. It might have a smaller aperture, but it operates at optical wavelengths, which means it can capture finer details than the Very Long Baseline Array (VLBA). So, when it comes to resolution, it's HST all the way!
To determine the angular resolution of a telescope, we need to calculate the smallest angle that the telescope can distinguish between two closely spaced objects. The formula to calculate the angular resolution is given by:
θ = λ / D
Where θ is the angular resolution, λ is the wavelength of observation, and D is the diameter of the telescope's aperture.
1) Hubble Space Telescope (HST):
Aperture diameter = 2.4 meters
We are given that HST is observing at optical wavelengths, but the specific wavelength is not mentioned. However, for optical wavelengths, we can assume a value of λ = 550 nanometers (a typical value within the visible light spectrum).
So, for HST, the angular resolution would be:
θ(HST) = λ / D = (550 nm) / (2.4 meters)
2) Very Long Baseline Array (VLBA):
Maximum baseline (distance between the farthest antennas) = 8600 km = 8.6 × 10^6 meters
Observing wavelength = 1 mm = 1 × 10^(-3) meters
For VLBA, the angular resolution would be:
θ(VLBA) = λ / D = (1 × 10^(-3) meters) / (8.6 × 10^6 meters)
Now, compare the calculated angular resolutions for HST and VLBA. The telescope with the smaller angle will have the better angular resolution.