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An optical microscope whose microscope objective is designed to form a real image at a specified finite distance. The intended image distance and optical aberration corrections belong together; mechanical tube length is a mounting specification, not automatically the image distance measured from a principal plane.^[ref]

Why can a small displacement produce a large image? For a converging thin lens, write $u_o=f_o+ϵ$ with $epsilon>0$. Substitution into the thin lens equation givesThe similar triangles construction supplies the minus sign. The example has $epsilon=2\mathrm{mm}$ and hence $m_o=-5$. Letting $epsilon$ decrease moves the real image farther away; it does not keep the microscope eyepiece in focus at a fixed position. Large transverse magnification therefore requires the correct conjugate geometry, not merely a short focal length.

An optical microscope whose microscope objective, with the specimen at its designed front focal plane, sends out a parallel bundle of light rays from each specimen point. Different specimen points generally give different bundle directions. A tube lens then forms the finite intermediate microscope image; the microscope eyepiece views it. This parallel region permits suitable optical accessories to be inserted. Finite-conjugate and infinity-corrected components must be matched to their intended system.^[ref]

In the ideal paraxial approximation, an object height $h_o$ at objective focal length $f_o$ gives bundle slope $-h_o/f_o$. A tube lens of focal length $f_t$ maps that slope to $h_i=-f_t{h}_o/f_o$, soThis is a two-assembly result. Substituting infinity into the single-objective transverse magnification $-v_o/u_o$ would not describe the finite recorded image. Infinity correction concerns the region after the objective; relaxed-eye viewing concerns the region after the eyepiece. They are distinct conditions.