Question

If the angular magnification of an astronomical telescope is 36 and the diameter of the objective is 75 mm, what is the minimum diameter of the eyepiece required to collect all the light entering the objective from a distant point source on the telescope axis?

Short answer

The minimum diameter of the eyepiece is 2.1 mm.

Step-by-step solution

Step 1: Given data

• Angular magnification= 36.
• Diameter of the objective is,$d=75\text{mm}=\text{0.075m}$.

Determining the concept

Find the magnification in terms of the diameter of the object and the eyepiece. Using that, calculate the minimum diameter of the eyepiece.

Formulae are as follows:

Magnification for the telescope is

$\begin{array}{rcl}M& =& \frac{Diameterofobject}{Diameteroftheeyepiece}\\ & =& \frac{D_o}{D_{ep}}\\ & & \end{array}$

Here, $D_o$is the diameter of the object and $D_{ep}$is the diameter of the eyepiece.

Determining the minimum diameter of the eyepiece required to collect all the light entering the objective from a distant point source on the telescope axis.

Let us find the magnification using the above diagram.

Let us find the ratio.

$\begin{array}{rcl}\frac{D_{ep}}{f_{ey}}& =& \frac{D_0}{f_{ob}+f_{ey}}\\ \frac{D_{ep}}{f_{ey}}& \approx & \frac{D_0}{f_{ob}}\\ D_{ep}& =& \frac{D_0}{f_{0b}}{f}_{ey}\\ & & \end{array}$

Angular magnification,

localid="1663080166275" $m_{\theta }=\frac{f_{0b}}{f_{ey}}$

$\therefore D_{ep}=\frac{D_0}{m_{\theta }}$

Using the given values,

$\begin{array}{rcl}{D}_{ep}& =& \frac{0.075}{36}\\ & =& 0.00208\text{m}\\ & =& 2.08\mathrm{mm}\\ & & ~2.1\mathrm{mm}\\ & & \end{array}$

Hence, the minimum diameter of the eyepiece is 2.1 mm.