Chapter 25, Problem 53P

To determine

The focal length of the eyepiece required to construct a telescope which can resolve features 6.5 km across the Moon and the resolution limit set by the size of the objective lens due to diffraction.

Answer to Problem 53P

Solution:

The focal length of the eyepiece is found to be 8.5 cm and the angular resolution limit of the objective is found to be $6.2\times {10}^{-6}\text{rad}$ with the spatial resolution limit as 2.4 km.:

Explanation of Solution

In a telescope, the resolving powers of the objective and the eyepiece must be equal and it should be equal to the resolving power of the eye.

The objective resolves two points at a distance $d_0$ located on Moon, which is at a distance R from the Earth. Then,

${\theta }_0=\frac{d_0}{R}$

If the angular resolution of the objective is ${\theta }_0$, then, the resolving power of the objective is given by,

$R{P}_o={\theta }_0{f}_0$

Similarly, the angular resolution and the resolving power of the eye, which can resolve two objects separated by a distance $d_e$ and located at a distance $R_e$ can be written as,

${\theta }_e=\frac{d_e}{R_e}$ and

$R{P}_e={\theta }_e{f}_e$

Since $R{P}_e=R{P}_o$, therefore,

$\begin{array}{c}{\theta }_e{f}_e={\theta }_0{f}_0\\ \frac{f_e}{f_0}=\frac{{\theta }_0}{{\theta }_e}\end{array}$

Using the expressions for ${\theta }_0$ and ${\theta }_e$ in the expression,the focal length of the eyepiece can be calculated.

$\begin{array}{c}\frac{f_e}{f_0}=\frac{{\theta }_0}{{\theta }_e}\\ =\frac{d_0/R}{d_e/R_e}\\ f_e=\left(\frac{d_0/R}{d_e/R_e}\right)f_0\end{array}$

The angular resolution limit set by the objective lens is given by,

${\theta }_0=\frac{1.22\lambda }{D}$

The spatial resolution is given by,

$\Delta l={\theta }_0\times R$

Given:

The distance that can be resolved by the telescope on the Moon $d_0=6.5\text{ km}$

The distance between Earth and Moon $R=384000\text{ km}$

Focal length of the objective $f_0=2.0\text{ m}$

Diameter of the objective’s aperture $D=11.0\text{ cm}$

The distance that can be resolved by the eye $d_e=0.10\text{ mm}$

The distance at which the eye resolves the objects $R_e=25\text{ cm}=250\text{ mm}$

Wavelength of light used $\lambda =560\text{ nm}=5.6\times {10}^{-7}\text{m}$

Formula:

The focal length of the eyepiece is calculated using the expression,

$f_e=\left(\frac{d_0/R}{d_e/R_e}\right)f_0$

The angular resolution limit of the objective is calculated using the expression,

${\theta }_0=\frac{1.22\lambda }{D}$

The spatial resolution is given by,

$\Delta l={\theta }_0\times R$

Calculation:

Using the given quantities in the expression for focal length of the eyepiece, the following calculation is done.

$\begin{array}{c}{f}_e=\left(\frac{d_0/R}{d_e/R_e}\right)f_0\\ =\left[\frac{\left(6.5\text{ km}\right)/384000\text{ km}}{\left(0.10\text{ mm}\right)/\left(250\text{ mm}\right)}\right]\left(2.0\text{ m}\right)\\ =0.0846\text{ m}\\ =8.5\text{ cm}\end{array}$

The resolution limit set by the objective is calculated as follows:

$\begin{array}{c}{\theta }_0=\frac{1.22\lambda }{D}\\ =\frac{\left(1.22\right)\left(5.6\times {10}^{-7}\text{m}\right)}{11.0\text{ cm}}\\ =6.2\times {10}^{-6}\text{rad}\end{array}$

The spatial resolution is calculated as follows:

$\begin{array}{c}\Delta l={\theta }_0\times R\\ =\left(6.2\times {10}^{-6}\text{rad}\right)\left(384000\text{ km}\right)\\ =2.4\text{ km}\end{array}$