5.2: Measure everything in anything

Last updated
Save as PDF

Page ID: 49693

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

It is easy to calculate the size of microscopic object. All what you need is a diameter of field (the whole visible circle) and number of objects which fit in this diameter. If, for example, the field diameter is 1 mm and 10 objects will fill this diameter, then the size of object is \(1\mbox{~mm}/10=0.1\) mm.

Today, we will use mostly \(400\times\) (or less) total magnification (\(40 \times 10 = 400\)):

The field diameter for this magnification is 0.43 mm.

If you use higher magnification, you should decrease the field size proportionally. If you use lower magnification, you should proportionally increase it. For example, if you use \(10\times\) objective lens instead of \(40\times\), the magnification is four times smaller, and the field diameter is four times bigger. So what is it? Calculate!

Now, when field diameter is known, you need to count how many objects of question (e.g., cells) will fit in the line crossing the center of the field. If 10 objects fit the line, and magnification is \(400\times\), then size of each object is \(0.43/10=0.043\) mm.

Very helpful animation movie “38 parrots” (your instructor could provide the link or simply show it) explains how to measure everything in anything. If you saw this movie, then imagine that your “boa” is a field diameter, and your “parrot” is an object of question.