Chapter 19.2, Problem 27SSC

To determine

The minimum separation between the stars such that the telescope will be able to resolve them.

Answer to Problem 27SSC

The minimum separation between the stars such that the telescope will be able to resolve them is $2.2\times {10}^{10}\text{m}$ .

Explanation of Solution

Given:

The wavelength of the light coming from the stars is $\lambda =550\text{nm}$ .

The diameter of the aperture of the telescope is $D=2.4\text{m}$ .

The distance between the system of stars and the Earth is $L_{\text{obj}}=8.44\text{light years}$ .

Formula used:

Consider two stars placed at a distance of $L_{\text{obj}}$ and the distance between the system of stars and the telescope on Earth is denoted by $x_{\text{obj}}$ .

The expression for the distance $x_{\text{obj}}$ is,

  $x_{\text{obj}}=\frac{1.22\lambda L_{\text{obj}}}{D}$

Here, $D$ is the diameter of the aperture, $\lambda$ is the wavelength of the light, and $L_{\text{obj}}$ is the distance between the Earth and the system of stars.

Calculation:

Consider the distance between the stars such that the telescope able to resolve them, is denoted by $x_{\text{obj}}$ .

The distance $\left(x_{\text{obj}}\right)$ of the stars from Earth is,

  $\begin{array}{l}{x}_{\text{obj}}=\frac{1.22\lambda L_{\text{obj}}}{D}\\ x_{\text{obj}}=\frac{1.22\left(550\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(8.44\text{light years}\times \frac{9.461\times {10}^{15}\text{m}}{1\text{light year}}\right)}{2.4\text{m}}\\ x_{\text{obj}}=\frac{1.22\left(550\times {10}^{-9}\text{m}\right)\left(7.98\times {10}^{16}\text{m}\right)}{2.4\text{m}}\\ x_{\text{obj}}=2.2\times {10}^{10}\text{m}\end{array}$

Conclusion:

Thus, the distance between the stars such that the telescope able to resolve them, is denoted by $2.2\times {10}^{10}\text{m}$ .