Astronomers have discovered a planetary system orbiting a star, which is at a distance of 5.4 x 10^19 m from the earth. One planet is believed to be located at a distance of 2.0 x 10^10 m from the star. Using visible light with a vacuum wavelength of 556 nm, what is the minimum necessary aperture diameter that a telescope must have so that it can resolve the planet and the star?© BrainMass Inc. brainmass.com December 24, 2021, 6:33 pm ad1c9bdddf
SOLUTION This solution is FREE courtesy of BrainMass!
The resolving power of a telescope is its ability to show distinctly the images of two distant objects lying close by. It is measured by the angle (alpha) subtended at the objective of the telescope by those two distant objects whose images are just seen as separate through the telescope. Resolving power alpha is given by the following expression :
Alpha = 1.22 x Wave length of light/Diameter of the objective (Wave length and diameter must be in the same units and alpha is in radians) .......(1)
In the given problem:
Distance between the star and the planet = 2 x 10^10 m
Distance of the star from earth = 5.4 x 10^19 m
As the distance between the planet and star is very small as compared to the distance of the star from earth, we can consider the line joining the planet and the star as an arc of a circle of radius 5.4 x 10^19 m. As we know:
Angle subtended at the centre of the circle (in radians) = arc/radius = 2 x 10^10/5.4 x 10^19 = 0.37 x 10^-9 radians .......(2)
As the telescope just shows the star and the planet as separate objects, the angle given by (2) is the resolving power alpha.
Alpha = 0.37 x 10^-9 radians
Using equation (1) and putting values of alpha and wave length we get :
0.37 x 10^-9 = 1.22 x 556 x 10^-9/D
Or D = 1833 m
(Note : A telescope with such a large aperture diameter is not practically possible)© BrainMass Inc. brainmass.com December 24, 2021, 6:33 pm ad1c9bdddf>