The Hubble space telescope has a 2.4 meter primary mirror, and is in low-Earth o

ID: 1874705 • Letter: T

Question

The Hubble space telescope has a 2.4 meter primary mirror, and is in low-Earth

orbit at a distance from the center of the Earth of 6,917.5 km. (The Earth’s radius

is 6,371 km.) Imagine that it was turned around and used to take pictures of the

surface of the Earth (instead of using it to take pictures of the universe). If we

ignore atmospheric disturbances and the speed with which it is orbiting the Earth

and assume it is operating at the diffraction limit, what is the size of the smallest

feature it could resolve at a wavelength of 500 nm?

The angular resolution of telescope =1.22/a, =500nm, a= aperture =2.4 m gives 2.54X10^-7

Distance of Hubble from earth =6917.5-6371=546.5 km

Angle made by an object of width d (located at earth) on the Hubble= d/D=d/(546.5X10^3), which should be equal to resolution of hubble.

Equating the two gives d= 13.88 cm = size of object on earth which could be resolved.

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.