= Microscope eyepiece
{title2=$M_e=D/f_e$}
The viewing <optical lens> assembly that acts as a magnifier of the <intermediate microscope image>. With that image at its front <focal plane>, the outgoing <light rays> from each point are parallel. In the <small-angle approximation>, the viewed <angle> is $h_i/f_e$, versus $h_i/D$ for the same image-sized object at reference distance $D$, giving $M_e=D/f_e$.<University Physics Volume 3 by Samuel J. Ling and others/2.8 Microscopes and Telescopes>[University Physics Volume 3, §2.8, relaxed viewing and Eq. 2.37]{ref}
A finite-distance <virtual image> requires a different eyepiece conjugate. In the ideal <thin lens> model, with the eye close to the eyepiece and the <virtual image> at distance $D$, the <angular magnification> becomes $1+D/f_e$. State the viewing condition before multiplying by the objective's <transverse magnification>; the relaxed-eye example above uses $D/f_e$.
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