The Magnification is a combination of all of the focal lengths of the picture you have shown above. A real image is created by the objective and tube lens. This creates an image of what you have at the object plane that is magnified by:
$M = \frac {f_{tube lens}}{f_{objective}}$
So, if you were to measure the size of the image, it would be M times larger than the object that is placed to the left of the objective lens.
One common confusion is that many microscope objectives actually create an image all by themselves without a tube lens. There are several standards including objectives that create images 160 mm and 170 mm away from the microscope objective. In your diagram, it implies an infinity corrected objective lens. This means that the image created by the microscope objective is infinitely far from the objective lens. This might lead you to believe that since the light is collimated from the microscope objective, you can place the tube lens anywhere you want. That is not technically correct because of two factors: vignetting and the optical design of the tube lens.
Vignetting means that the light escapes the size of the lens. In your diagram, this would happen if the tube lens is too small. Many infinity corrected objectives are designed for tube lenses that are 180 mm from the objective lens. If the tube lens is not placed at a distance close to 180 mm, you can have vignetting or performance from optical aberrations may cause the image to degrade.
Now, take the final step to the eye of the observer. This is the eyepiece. Your eye prefers (is relaxed) when looking at infinity. Therefore, the eyepiece is typically designed to project the image created by the objective-tube lens pair to infinity. Your diagram actually shows the image at 25 cm instead of at infinity. For this case, the eyepiece is placed at nearly one focal length away from real image (image plane 3 in your diagram).
The final magnification is $ M_{total} = M \times \frac{25 cm}{f_{eyepiece}} $
There are additional considerations including:
- Working distance of the objective (distance between objective and object)
- Eye relief (distance between eyepiece and observer's eye)
- Pupil or eyebox (sometimes you look inside a microscope and it is black until you line up your eye with the microscope)
- Illumination (most objects require some external light source to illuminate them so you can see them!)
One last consideration. If you just want to put the image onto a camera. In which case, you don't need the eyepiece!
I suspect that you are getting a focused image of objects that are normally out of focus because of the point like nature of the light source. The spatial coherence of the light means that you can effectively "focus" on these things - and you are adding an amount of magnification because this is like a "pinhole shadowgraph" - the glint being closer to the object (speck of dust, random cell) than that object is to the retina.
Best Answer
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.
so the whole field of view is encoded in a range of tilts as shown below.
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.