I was studying about compound microscope here

I don't see why we multiply linear magnification of objective with angular magnification of eyepiece. Shouldn't it be both angular or both linear?

Can we call this magnification angular magnification of the system? I tried to find about angular magnification of a optical ststem on the internet but it took me to the realms of some matrices which I do not understand as I have never read about such.

  • $\begingroup$ I'm puzzled by the fact that the topic refers to telescope magnification, but the question is about microscope magnification. Telescope magnification is simply the focal length of the objective divided by the focal length of the eyepiece. $\endgroup$
    – Dr Chuck
    Mar 2 '14 at 17:14
  • $\begingroup$ @DrChuck Typo fixed. It is microscope as the link suggests. $\endgroup$
    – evil999man
    Mar 3 '14 at 4:13

Angular magnification is used to describe the magnification of an afocal system, typically a telescope like a Galilean telescope (which a modern microscope's eyepiece almost always is) or Keplerian Telescope. Indeed your question is a good one insofar that it underlines the meaninglessness of "magnification" (or at least calls into question how worthwhile this specification is for microscopes), especially in the modern, infinity conjugate microscope.

The "objective" is an infinity conjugate objective. It encodes positions on the object plane as tilts i.e. wavevectors of a pure plane wave at the output. In other words, a point source on the object plane begets the same output from an infinity conjugate objective as you see from a star in the night sky.

Infinity conjugate objective

so the whole field of view is encoded in a range of tilts as shown below.

Microscope System

The eyepiece, or occular, simply scales the tilt of the plane wave output from a point source so that the field of view fills a wider range of angles when incident on the eye. In the drawing above, the angular magnification of the occular telescope is $\frac{\theta_2}{\theta_1}$. The eye forms an image on the retina from the plane wave input.

So, what is the linear magnification of this system? By the strict, physicist's definition, if the eye's focal length is $f_e$, then it will be

$$M_a \frac{f_e}{f}$$

where $f$ is the effective focal length (distance from the object plane to the the objective's leftmost principal plane) and $M_a = \frac{\theta_2}{\theta_1}$ is the angular magnification of the occular telescope. This number depends on the particular person's eye size. So the strict linear magnification is actually poorly defined. For this reason, it is unlikely that this is the figure you get from multiplying the occular's angular magnification by the objective's magnification, even though this is what you are meant to do with these numbers! Indeed, the magnification of a true infinity conjugate objective, by the physicist's definition, is infinite! The objective's "magnification" as marked by the manufacturer is just a number that, when multiplied by the occular's (angular) magnification, is meant to give you an idea of how big something's angular subtense will be relative to how big this subtense would be without the microscope. This of course depends wholly on how far away from your eye this something is when you looked at it without the microscope. The "magnification" figure is chosen so that it is comparable with the linear magnifications that were meaningful for non-infinity conjugate objectives with, say 160mm tubelengths back in the old days (se my discussion of 160mm tube microscopes at the end). Therefore, the magnification stamped on the infinity conjugate objective is simply $160/f$, where $f$ is the objective's effective focal length, measured from the front principal plane: it is not a true linear magnification (which doesn't exist for the infinity conjugate objective). The only really meaningful specifications for a microscope then are its field of view $F$ and its resolution $\delta$, reciprocally related to its numerical aperture $\eta$ through $\delta \approx 0.7 \lambda/\eta$. You can think of the magnification on an infinity conjugate objective as a "magnification relative to a 1x objective by the same manufacturer", i.e. an Mx objective will yield a magnification of $M/N$ times what an Nx objective by the same manufacturer will yield.


What is meaningful as a linear magnification is if the manufacturer gives you a measuring reticule as I show in the photo above. Wontedly this will be in millimetres, and you divide any measurement from this reticule by the system magnification, as worked out by the objective's "magnification" multiplied by the occular's angular magnifications. But you need to be careful if you use different manufacturer's objectives: a 4x objective from an Olympus microscope will yield different linear magnifications when it is used on an Olympus or Zeiss microscopes. It's often best to calibrate your setup with an electron microscopy calibration grid (commonly with a $12.5{\rm \mu m}$ grid period) and see what the true distance is that corresponds to your reticule intervals.

Linear magnification was meaningful in the days when infinity conjugate lenses weren't used and instead the objective formed a real image which was viewed by the eyepiece. This real image was a "tube length" distance from the back principal plane of the objective. The tubelength was typically 160mm (for Zeiss and Olympus microscopes). The magnification was then $160 / f$, where $f$ is the objective's effective focal length, measured from the front principal plane. Then the eyepiece's magnification was a linear magnification calculated using the manufacturer's model of the "ideal eye".

See my answer here for more discussion of the infinity conjugate microscope system.


the answer is at the bottom of the page you linked to, go to Linear and angular magnification

read the content, but the brief answer is that the first lens projects from air to air, and in this case you use linear magnification. the second lens projects from air into your eye (some kind of liquid?), so you use angular magnification

  • $\begingroup$ Can we call this angular magnification? $\endgroup$
    – evil999man
    Mar 3 '14 at 4:19
  • $\begingroup$ @Awesome, they're certainly calling it such. the term sounds reasonable to me $\endgroup$ Mar 3 '14 at 4:26
  • $\begingroup$ Why not the term where we multiply both angular magnification? Or is it just $defined$ to be called this way? $\endgroup$
    – evil999man
    Mar 3 '14 at 4:28
  • $\begingroup$ @Awesome, i'm not sure i understand the question. for any lens you can compute its angular or linear magnification. you use the lense's linear magnification until the eyepiece for which you use angular, because its image gets inside the eye $\endgroup$ Mar 3 '14 at 5:04
  • $\begingroup$ I want to know why do we do so? $\endgroup$
    – evil999man
    Mar 3 '14 at 5:22

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