I have seen angular magnification defined as "angle subtended by image"/"angle subtended by object, but that is very vague.

I have seen it defined as the ratio of the tangents of the same angles, which suggests that the first definition is a small angle approximation and the second is more accurate.

One thing I was trying to find out is exactly where the eye is supposed to be positioned during all of this? It certainly matters for the size of angle subtended by the object! Is it supposed to be mashed up against the eyepiece? At the far focal point of the eyepiece? Somewhere else?

Another question answer here on this site referred to "the angular size of the object if it were at the near point of the eye" but obviously if the Moon were at my near point I would be standing on it! So let's say there's some objective lens making a first image, and then the eyepiece is positioned so that the first image is sitting on the focal point of the eyepiece.

The first problem is, if the Moon is at "infinite distance" wouldn't the size of the first image be zero? All the light would gather right at the focal point! If the objective focal length were say 50 cm, how would I find the magnification, or angular magnification, of the first image, presuming that this is the first lens in a telescope say? Even if I know the Moon subtends an angle of $0.5$ degrees normally, where does that fit in?

Next, suppose somehow that the image has a height of $3$ cm, and the eyepiece focal length is $5$ cm. With the image sitting on the eyepiece focal point, the next image is at infinity.

Now, as for the "size of the object unaided", would that be trying to find the angular size of a $3$ cm tall object at a distance of $25$ cm, the standard near point? If so, that would be $\theta = \tan ^{-1} \big(\frac{3}{25}\big)$. Is that the correct answer or am I missing something else?

If the eye is supposed to be right up against the eyepiece, then the angle subtended by the image would be $\theta = \tan ^{-1} \big(\frac{3}{5}\big)$ and is that supposed to be the unaided angle instead?

I actually tutor students in this topic and generally can answer their typical questions, but someone asked me for more details and I haven't been able to find the nitpicks I need in online searching. I even studied an old optics textbook cover to cover and still am often caught feeling I am missing information just from the freshman problems.

Is there a good text that would explain the missing pieces? I'm trying to create tutorial videos on YouTube and I do not want to get any facts wrong. I want to learn this well enough that I am never stumped by a freshman physics problem again! Thanks for any suggestions.


1 Answer 1


I've also never found a text which explains any of this particularly well, so I'll try to give my own explanations after years of piecing bits together. I've come across four common magnification definitions and I give a brief description of each. There is a lot of text but I hope it helps.

Linear magnification

Even though you're asking about angular magnification, I'll include this for completeness. In this case the system produces an image in the image plane of an object in an object plane, both of which are not at infinity (note also that the image can be either real or virtual). The linear magnification is then simply defined as the linear size of the image (i.e. its physical length in the image plane) to the linear size of the object.

Here there is no need to talk about an eye anywhere.

Angular magnification of a telescope

Now the object is at infinity: the rays from any given point on it can be assumed to be parallel when they reach the entrance pupil of the system. However, the rays from a point on one side of the object will enter at a different angle (to the optic axis) to the rays from a point on the other side. This is where angular size comes in, defined as the angle subtended by the object at the entrance pupil, or equivalently the difference in angle of the rays from opposite sides of the object.

When such parallel rays from one side of the object pass through the system and leave the exit pupil, they will again be parallel (telescopes are designed to form an image at infinity of an object also at infinity), but at a different angle to the optic axis. Similarly, rays from the other side will be also be parallel and at a different angle. The angular size of the image is now defined as the difference between these new angles. Magnification is then simply the ratio of the angular size of the image to the angular size of the object.

If you want to discuss an eye here, it can technically be anywhere after the exit pupil as the rays are all parallel and you will see the same thing at any distance away. However, of course all the rays must actually enter your eye, so it is sensible to place your eye close to the exit pupil.

I hope this answers your question about the moon, something which would only ever be imaged with a telescope. The rays are not focussed onto a focal point but rather leave parallel as discussed above, and there is no need to discuss the near-point etc.

Angular magnification of a microscope

As in the case of linear magnification, the object is in an object plane which is not at infinity. However, the image is now formed at infinity. In the same way as for a telescope, rays from a given point on the object leave the exit pupil parallel and at a certain angle to the optic axis. The angular size of the image is defined as the difference between the angles of the rays leaving the exit pupil which originate from opposite sides of the object.

However, this time it doesn't make sense to talk about the angular size of the object from the point of view of the entrance pupil (which is generally very close to the object, compared to the telescope where it is at infinity), as a human eye would never be able to see it so close up. Rather, the best view the unaided human eye can get is when it is placed at the near point, so for a 'fair' comparison we work out the angular size of the object when it is placed at the near point with no imaging system. This is defined in terms of the tangent (which assumes one end of the object is on the optic axis, a convention many books use): if $h$ is the linear size of the object and $D$ the near point distance then the angular size of the object is $\alpha=\arctan(h/D)$ (but generally angles are small enough that $\alpha\approx h/D$ anyway). The magnification is now defined as the ratio of the angular size of the image to the angular size of the object at the near point.

Thus your example calculation for the "size of the object unaided" is correct, but your calculation for the angular size of the image is wrong; for that you'd need to know the specifics of the microscope to work out how points on the object are mapped to angles in the image.

Again, the eye can be anywhere after the exit pupil, but sensibly somewhere close such that all the rays actually go in.

Angular magnification of a magnifying glass

In my opinion this is the most annoying one. I'll consider the case where the magnifying glass is pressed right up to the eye (which is not how anyone ever uses one but seems to be the case covered by many optics books).

As with linear magnification, neither the object nor the image are at infinity, but (as long as the object is placed within the focal length of the magnifying glass) the image is always virtual. As the eye is placed right up to the glass, it simply sees a virtual image of linear size $H$ a distance $v$ away, the angular size of which is given by $\beta=\arctan(H/v)\approx H/v$. Just like for the microscope, for a fair comparison, we define the angular size of the object as that when it is at the near point with no imaging system, and the magnification is given by the ratio of the angular size of the image to the object as usual.

Just a final note. I've made a clear distinction between things being at infinity or not. However, something at infinity can still have its angular size defined in terms of linear size. It's just that its linear size is also infinite (of course in reality it's not but as far as the calculation is concerned) and so it is unhelpful to talk about it in terms of this.

  • $\begingroup$ Excellent answer, thank you. $\endgroup$ Dec 16, 2021 at 16:37

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