Evaluation: Difference between revisions

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Mikael Häggström (talk | contribs)
Administrators, Editors, Image-reviewer, csmoderator, Interface administrators, Administrators, Upload Wizard campaign editors
7,148 edits
Mikael Häggström (talk | contribs)
Administrators, Editors, Image-reviewer, csmoderator, Interface administrators, Administrators, Upload Wizard campaign editors
7,148 edits
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<gallery mode=packed heights=220>
<gallery mode=packed heights=220>
File:Micrograph of a calibration slide, showing the diameter of the field of view.jpg|Micrograph of a calibration slide, showing the diameter of the field of view. On this slide, each square is 0.05 mm wide, and each line represents 0.01 mm, making the diameter of the field of view 0.55 mm in this case (and thus an area of 0.24 mm<sup>2</sup>).
File:Micrograph of a calibration slide, showing the diameter of the field of view.jpg|Micrograph of a calibration slide, showing the diameter of the field of view.  
</gallery>
</gallery>
On the imaged example, each square is 0.05 mm wide, and each line represents 0.01 mm, making the diameter of the field of view 0.55 mm in this case. Thus, you can calculate the area in mm<sup>2</sup> by Googling:
*0.55 x 0.55 x 0.79
:(equals approximately 0.24 mm<sup>2</sup>).


Sometimes "high power field" (HPF) is used for area, but it has a substantially different area for different microscopes, for example:
Sometimes "high power field" (HPF) is used for area, but it has a substantially different area for different microscopes, for example:
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