# Should the working distance of an infinity-corrected objective always be smaller than the focal length?

I'm building a homebrew imaging system with a 5X infinity-corrected apochromatic objective with the following relevant specs:

• Working distance: $$W = 45\,\mathrm{mm}$$
• Magnification: $$M = 5\times$$
• Tube lens focal length: $$L = 200\,{\mathrm{mm}}$$

From what I understand, infinity-corrected objectives are designed to work with the object in their focal plane (see e.g. this Zeiss tutorial) so that they produce in image at infinity. I would therefore expect the effective focal length of this compound lens system to be $$\geq 45\,\mathrm{mm}$$, otherwise an object at the working distance would be farther from the lens than the focal plane.

However, the focal length of an infinity-corrected objective can be determined from $$F = L/M$$, where $$M$$ is the objective's magnification spec and $$L$$ is the focal length of the microscope tube lens with which it is designed to be used (see e.g. Objective Tutorial on the Thorlabs website). Plugging in the appropriate quantities for my lens gives $$F = 200\,\mathrm{mm}/5\times = 40\,\mathrm{mm}$$, which is smaller than the working distance.

What am I missing here? Am I wrong that the working distance should always be smaller than the focal length, or am I misusing one of these specs?

Let's say I have an infinity-corrected objective. Now cascade it with a beam-shrinker (beam expander in reverse). If viewed as a black-box, this (objective + beam-shrinker) is an infinity-corrected objective with smaller exit pupil than the original one. Since exit pupil of an infinity-corrected objective is $$D = 2 NA \times EFL$$, my $$EFL$$ should shrink. So we can put whatever beam-shrinker until the new $$EFL$$ becomes shorter than the working distance.