A compound microscope uses two converging lenses working in sequence: the objective and the eyepiece.
The objective first forms a real, enlarged, inverted intermediate image of the small object. The eyepiece then acts like a magnifier for that intermediate image, producing a much larger apparent angular size for the observer.
Because of this two-stage process, microscope magnification is naturally written as a product.
In the standard introductory approximation, the objective magnification is
\[
m_o = \frac{L}{f_o},
\]
where \(L\) is the tube length and \(f_o\) is the focal length of the objective.
This formula comes from the fact that the objective works with the object placed very near its focal point, so it forms an enlarged real image far down the tube.
A shorter objective focal length gives a stronger enlargement.
The eyepiece contributes angular magnification, not ordinary linear magnification. If the final image is adjusted for a relaxed eye, so that it is effectively at infinity, then the eyepiece magnification is
\[
m_{e,\infty}=\frac{25}{f_e},
\]
where \(25\ \text{cm}\) is the conventional near-point distance and \(f_e\) is the eyepiece focal length in centimeters.
If instead the final image is adjusted to appear at the near point itself, the alternative formula is
\[
m_{e,25}=1+\frac{25}{f_e}.
\]
Since the microscope has both an objective stage and an eyepiece stage, the total magnification is the product
\[
m = m_o m_e.
\]
For the sample values \(f_o=0.5\ \text{cm}\), \(f_e=2\ \text{cm}\), and \(L=16\ \text{cm}\), the objective magnification is
\[
m_o=\frac{16}{0.5}=32.
\]
In relaxed-eye mode the eyepiece magnification is
\[
m_{e,\infty}=\frac{25}{2}=12.5,
\]
so the total magnification is
\[
m = 32 \times 12.5 = 400.
\]
This means the microscope gives a \(400\times\) magnification in the standard school approximation.
The final image of a compound microscope is typically inverted relative to the original object.
That happens because the objective already forms a real inverted intermediate image, and the eyepiece magnifies that image without restoring the original orientation.
In the sign language of magnification, the total signed magnification is therefore negative, even though the quoted microscope power is usually given as a positive magnitude such as \(400\times\).
The ray diagram reflects this two-stage story. First, the objective creates a real intermediate image inside the tube. Then the eyepiece uses that intermediate image as its object.
In relaxed-eye mode, the intermediate image lies essentially at the front focal plane of the eyepiece, so the outgoing rays leave the eyepiece parallel and the final image is treated as being at infinity.
In near-point mode, the intermediate image sits slightly inside the focal plane of the eyepiece, making the outgoing rays diverge so that their backward extensions meet at a virtual image about \(25\ \text{cm}\) away.
This calculator includes both eyepiece formulas because many textbooks switch between them. The relaxed-eye formula is usually simpler and is very common in school-level work, while the near-point version is often used when discussing maximum angular enlargement.
At higher levels, real microscope performance depends on much more than this simple product. Important advanced ideas include numerical aperture, diffraction-limited resolution, working distance, field of view, and aberration correction.
Even so, the basic product formula remains a useful starting point:
\[
m = \left(\frac{L}{f_o}\right)\left(\frac{25}{f_e}\right)
\]
or, in near-point mode,
\[
m = \left(\frac{L}{f_o}\right)\left(1+\frac{25}{f_e}\right).
\]
So the central lesson is simple: the objective does the first enlargement inside the tube, the eyepiece does the angular enlargement for the eye, and together they produce the large total power associated with a compound microscope.
