Question

The nearest known exoplanets (planets beyond the solar system) are around 20 light-years away. What would have to be the minimum diameter of an optical telescope to resolve a Jupiter-sized planet at that distance using light of wavelength 600 nm? (Express your answer to two significant figures.)

Answer 1

To solve the problem, it is necessary to apply the concepts related to the diffraction given in circular spaces. By definition it is expressed as

$\Delta \theta = 1.22\frac{\lambda}{d}$

Where,

\lambda = Wavelength

d = Optical Diameter

$\theta =$ Angular resolution

In turn you can calculate the angle through the diameter and the arc length, that is,

$\Delta \theta = \frac{x}{D}$

Where,

x = The length of the arc

D = Distance

From known data we know that Jupiter's diameter is,

$x_J = 1.43*10^8m$

$D = 20*9.4608*10^{15}$

$\lambda = 600*10^{-9}m$

Replacing we have that,

$\frac{x}{D} = 1.22\frac{\lambda}{d}$

$\frac{1.43*10^8}{20*9.4608*10^{15} } = 1.22\frac{600*10^{-9}}{d}$

Re-arrange to find d,

$d = 968.5m = 0.968Km$

Therefore the minimum diameter of an optical telescope to resolve a Jupiter-sized planet is 0.968Km.