# Focal length vs working distance in an infinity corrected objective

Consider the following microscope objective (infinity corrected):

Magnification $$= 50 \times$$

Glass Thickness $$= 3.5$$mm

NA $$= 0.50$$

WD $$= 13.89$$mm

Focal Length $$= 4$$mm

I'm confused about the difference between $$f$$ (focal length) and $$WD$$ (working distance).

Also, what does $$M=50 \times$$ mean in the context of an infinity corrected objective? The magnification should be determined by whatever tube lens I use right (i.e. $$M=\frac{f_{tube}}{f_{objective}}$$)?

Can someone explain this?

You are correct that for a single lens the working distance would be the focal length. For compound lenses, like microscope objectives, you have to look at the entire optical system to figure out the working distance. The short answer to your question is that the focal length and working distance are not what you expect because the objective is a compound lens. The magnification for microscope objectives is confusing because to calculate the magnification you must know something about the tube lens that is used along with the objective.

I'm going to go through an example, for quantities I don't define explicitly here, please see Figures 9 and 12 in refrence 1.

Suppose you have two focusing lenses (f1, and f2) separated by a distance d were the first lens is a distance W from the object you are trying to image. Microscope objectives have to fit in a specific space so they have a required parfocal distance PD = d + W (I am assuming thin lenses).

The ray tracing optics for this lens system would be: $$\begin{equation} M= \left( \begin{array}{cc} M_{11} & M_{12} \\ M_{21} & M_{22} \\ \end{array} \right) = \left( \begin{array}{cc} 1 & 0 \\ -\frac{1}{\text{f2}} & 1 \\ \end{array} \right).\left( \begin{array}{cc} 1 & PD-W \\ 0 & 1 \\ \end{array} \right).\left( \begin{array}{cc} 1 & 0 \\ -\frac{1}{\text{f1}} & 1 \\ \end{array} \right).\left( \begin{array}{cc} 1 & W \\ 0 & 1 \\ \end{array} \right) \end{equation}$$

Actual performing the math we get: $$\begin{equation} M= \left( \begin{array}{cc} 1-\frac{\text{PD}-W}{\text{f1}} & \text{PD}+\left(1-\frac{\text{PD}-W}{\text{f1}}\right) W-W \\ -\frac{1-\frac{\text{PD}-W}{\text{f2}}}{\text{f1}}-\frac{1}{\text{f2}} & -\frac{\text{PD}-W}{\text{f2}}+\left(-\frac{1-\frac{\text{PD}-W}{\text{f2}}}{\text{f1}}-\frac{1}{\text{f2}}\right) W+1 \\ \end{array} \right) \end{equation}$$

In order for this objective to image properly we set the requirement that M12 = 0. When we are happy that our objective correctly images we can figure out the effective focal length by calculating the M21 element.

Let's work an example. Suppose that PD = 45 mm, f1 = 11 mm, f2 = 16.5 mm. I mostly made these numbers up, but PD for microscope objectives is 45 - 60 mm. This is on purpose to make objectives interchangeable.

From the requirement that M12 = 0 we can calculate W: $$\begin{equation} M_{12} = W \left(1-\frac{\text{PD}-W}{\text{f1}}\right)+\text{PD}-W = 0 \\ \implies W = 19.15 mm, \end{equation}$$ so the working distance is 19.15 mm.

Next we want to calculate the focal distance of this lens, which we can get from the M21 element: $$\begin{equation} M_{21} = -(1/f2) - (1 - (PD - W)/f2)/f1 = -0.0091 mm^{-1}, \end{equation}$$ but focal lengths are usually written like $$M_{21} = -1/f_{o}$$ so the focal length of our compound lens system is $$f_{o}$$ = 110 mm. So very clearly, compound lenses do not behave like single lenses.

What about the magnification? To answer that question we have to look at the M matrix with all the numbers plugged in: $$\begin{equation} M= \left( \begin{array}{cc} -1.35037 & 0. \\ -0.00906831 & -0.740536 \\ \end{array} \right). \end{equation}$$

From inspection you'd expect the magnification of this lens to be 1.35 (the minus sign just means the image is upside down). This is where your confusion comes in: magnification for microscope objectives require knowledge of the tube lens. From this example of someone trying to sell you a microscope objective you can see that it clearly states that:

All stated magnifications are based on a tube lens focal length of 200mm.


So for our example above using a 200 mm tube lens the magnification would be $$M=f_{tube}/f_{objective} = 200 / 110 = 1.82$$. Scrolling down the page of reference 2 you can see that the FL=100 mm lens has a magnification of 2.

• Ok, that really cleared things up for me, makes sense! Thanks a lot for the great answer! – CPE Nov 2 '18 at 12:20
• Thanks. Would it be possible to accept it? – Chad Sexington Nov 5 '18 at 3:08