Angular magnification (definition)

Question

According to my physics book, "Angular magnification measures the change in the beam angle that converges in the image with respect to that of the beam emitted by the object and collected by the instrument". Can anyone explain this definition to me better?

Answer 1

I'm not sure about the wording in your physics book, but say you are looking at the moon. It subtends a certain angle at your eyes, that is, it has a certain angular extension in your field of view. If you look at the moon through a telescope, the diameter of the virtual image created by the scope is much smaller than the actual diameter of the moon. So how does this help? It helps because the angle subtended by the scope's moon image is greater than that subtended by the moon without the scope. The scope provides angular magnification and that makes the moon look bigger. Angular magnification is used when the image is virtual, as in telescopes, binoculars and microscopes.

Answer 2

It just means that your telescope makes the image have a larger apparent angular size,

$$ M_{\theta} = \frac{\tan\left(\theta_2\right) }{\tan\left( \theta_1\right)} \approx \frac{\theta_2}{\theta_1} $$

where $\theta_1$ is the angular size of the input beam and $\theta_2$ is the angular size of the exit beam. The trigonometric function appears from considering the focal distance and image height of the mean in the ray diagram, but can be approximated

In a little more detail.

In optical systems invariant is called the etendue. It a simplified form you can write this as,

$$ A_1 \sin^2\left(\theta_1\right) = A_2 \sin^2\left(\theta_2\right) $$

where $A_1$ is the input area, $A_2$ is the output area.

So this tells us that for a telescope to make the angular size larger (that is to magnify the target) it must do so by reducing the output area. Or stated another way, the input aperture must be very large.