Chapter 25, Problem 51P

To determine

The minimum diameter of the earth based telescope’s aperture to obtain the image of the star and the planet system.

Answer to Problem 51P

Solution:

The minimum diameter of the telescope’s aperture required to obtain the image of the planet-star system is 17 cm.

Explanation of Solution

The nearest star to the Sun is at a distance s from the solar system and a planet orbits the star in an orbit of radius R. therefore, the distance between the star and the planet is R. The star- planet system subtends an angle $\theta$ to the telescope mounted on Earth.

Therefore,

$\theta =\frac{R}{s}$

The star- planet system will be resolved if the angular resolution $\theta$ of the telescope is equal to the angle subtended by the star- planet system to the objective of the telescope.

The angular resolution of a telescope is given by,

$\theta =\left(\frac{1.22\lambda }{D}\text{rad}\right)$

Equating both the expressions,

$\begin{array}{c}\theta =\left(\frac{1.22\lambda }{D}\text{rad}\right)=\frac{R}{s}\\ D=\frac{1.22\lambda s}{R}\end{array}$

Given:

The distance between the sun and the star $s=4\text{ ly}\times \frac{9.46\times {10}^{15}\text{m}}{1\text{ ly}}=3.784\times {10}^{16}\text{m}$

The radius of the orbit of the planet around the star $R=1.496\times {10}^{11}\text{m}$

The wavelength of light $\lambda =550\text{ nm}=5.50\times {10}^{-7}\text{m}$

Formula:

$D=\frac{1.22\lambda s}{R}$

Calculation:

Substitute the given values of $\lambda$, s and R in the expression and calculate the diameter.

$\begin{array}{c}D=\frac{1.22\lambda s}{R}\\ =\frac{\left(1.22\right)\left(5.50\times {10}^{-7}\text{m}\right)\left(3.784\times {10}^{16}\text{m}\right)}{1.496\times {10}^{11}\text{m}}\\ =0.1697\text{ m}\\ \text{=17 cm}\end{array}$