Chapter 19, Problem 52A

To determine

The large object that can be resolved by the telescope.

Answer to Problem 52A

  $x_{obj}=2.6\times {10}^{-2}\text{ m}$

Explanation of Solution

Given:

The given figure is shown below.

  

Figure 1

Diameter of the telescope is $D=2.4\text{ m}$

Distance between the Earth and the orbit of a telescope is $L_{obj}=1.0\times {10}^5\text{ m}$

Wavelength of light is $\lambda =515\text{ nm = 5}\text{.15}\times {10}^{-7}\text{ m}$

Formula used:

The large object that can be resolved by the telescope can be determined using the formula,

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

  $\Rightarrow x_{obj}=\frac{1.22\lambda }{D}{L}_{obj}$ $\left(1\right)$

Where, $x_{obj}$ is the size of the object that can be resolved by the telescope,

  $L_{obj}$ is the distance between the object and the telescope,

  $D$ is the diameter of the telescope

  $\lambda$ is the wavelength of light

Calculation:

Substituting the numerical values in equation $\left(1\right)$ ,

  $x_{obj}=\frac{1.22\left(\text{5}\text{.15}\times {10}^{-7}\text{ m}\right)\left(1.0\times {10}^5\text{ m}\right)}{2.4\text{ m}}$

  $=2.6\times {10}^{-2}\text{ m}$

Conclusion:

The large object that can be resolved by the telescope is $2.6\times {10}^{-2}\text{ m}$ across.