# When viewed from a telescope, does an object get more magnified if its angular diameter is increased?

and if yes, how can the diameter be increased? Lets say I want to view saturn from a small telescope. Increasing the angular diameter will give a better magnification if the answer to the above question is yes. Then how can I increase the angular diameter?

• The angular diameter of the object (Saturn)? Or of the observed image? The magnification is related to how much bigger the angular diameter of the image is relative to the object. You can increase this by increasing the magnification :) i.e., the strength of your telescope. If you mean the angular diameter of your object, magnification can change if the properties of your lens system is not ideal. If you mean the angular diameter of your primary aperture, it just gives you more light gathering power, but not necessarily magnification. – lionelbrits Nov 10 '13 at 18:45

## 2 Answers

For optical systems there are 2 classes of magnification:

1. lateral magnification, and

2. angular magnification.

Lateral magnification is a ratio of the actual size of the image (sometimes called $y'$) to size of the object ($y$). If the image is inverted, the image size is considered to be negative: $$m=\dfrac{y'}{y}.$$ In other words, a magnification of 2 would mean that the image of a 5 m object would be 10 m. This type of magnification usually occurs for projection of images on screens, such as a movie or video projector.

Angular magnification is the ratio of the angular size of the image to the angular size of the object: $$\mathcal{M}=\dfrac{\Theta '}{\Theta}.$$ Angular size is itself a ratio of the lateral size of the image or object to the distance from the observer to the image or object: $$\Theta=\dfrac{y}{r_\textrm{obj}}\hspace{1in}\Theta=\dfrac{y'}{r_\textrm{img}'}$$ For example, the angular size of the moon viewed from Earth is about $\Theta=0.50^o$. If a telescope produces an image $y' =$ 10 mm high at a distance of 30 mm then $\Theta '=0.333\text{ rad} = 19^o$ for an angular magnification of $\mathcal{M}=38$. On the other hand, the lateral magnification is $$m=\dfrac{10}{3369882000}=3\times10^{-9}.$$

Angular magnification gives the sense that the distance has been reduced by a factor of $\dfrac{1}{\mathcal{M}}$.

Telescopes are designed to provide angular magnification, not lateral magnification. The basic method of changing the angular magnification is to change the eyepiece, and for basic telescopes designed for an eyeball near-point focus $$\mathcal{M}=\dfrac{f_\textrm{objective}}{f_\textrm{eyepiece}}$$ where the $f$s are the focal lengths of the objective and the eyepiece. So if the objective mirror or lens of the telescope has a focal length of 500 mm, a 50 mm eyepiece will yield an angular magnification of 10. A 20 mm eyepiece would yield angular magnification of 25.

It may seem that one could increase the angular magnification almost indefinitely but larger angular magnifications can result in some problems such as excessive vibration distortion, vignetting, and spherical or chromatic aberration. High quality short focal length eyepieces are difficult to make and are very expensive.

Get a higher-magnification eyepiece. Increasing the magnification is equivalent to increasing the apparent angular diameter of the object. Practically, this means getting an eyepiece with a shorter focal length (usually marked on the barrel). Keep in mind that, since objects that aren't point sources have a set brightness per unit angular area, things will seem dimmer if you magnify more.