Posted by
Kevin Ryan on
URL: http://confocal-microscopy-list.275.s1.nabble.com/Question-about-Optical-Density-tp7584144p7584146.html
You have the basics correct. The optical density measurement is one of absorbance, which can be expressed as the log of transmittance:
OD = -log(Transmittance) = -log(Intensity of a blank / Intensity of the specimen)
A transmittance of 1.0 indicates no absorbance, and thus an OD of 0.0. From the ever handy WikiPedia (
https://en.wikipedia.org/wiki/Absorbance):Absorbance: −log10(Φet/Φei) Transmittance: Φet/Φei)
0 1
0.1 0.79
0.25 0.56
0.5 0.32
0.75 0.18
0.9 0.13
1 0.1
2 0.01
3 0.001
High OD values are, unfortunately, less certain - there's far less grayscale change involved in 2.0->2.1 OD than there is in 0.0->0.1 OD. A transmittance of 0.0 indicates an infinite OD as a result. This means that if you are quantifying OD you want _some_ transmittance in even the darkest portions of the sample, if at all possible.
Most imaging packages can apply an intensity calibration to convert grayscale to optical density/absorbance in transmitted light microscopy, with the OD scale indicating in a linear fashion the amount of material present. Note that in many packages (for example Image-Pro Plus/Premier, which I'm quite familiar with as our products) apply the current intensity calibration to _whatever_ is being measured, so that if an OD calibration has been applied to an image the IOD measurement is in OD terms, not the original gray levels. Check your software documentation on this to be certain that this is the case for your system.
Kevin Ryan
Senior Project Manager
Media Cybernetics, Inc.
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Good morning
I have a question about ImageJ measurement of optical density in photomicrographs obtained by digital cameras. As I understand the image captured by the camera CCD register the transmitted light. That should mean that the voltage (analogical signals) produced by the CCD is proportional to the brightness of the object. After the conversion of these signals to digital information, the brightness and gray shades of the object are translated into pixels values of gray that goes from 0 (black) to 255 (white).
If a pixel(a) has a gray value of 100 when compared with another pixel(b) that has a gray value of 200 we may conclude that the first one (a) is darker than the second one (b) and we may conclude that the object region corresponding to the first pixel with the gray value of 100 is also darker.
So, when we take a measurement of the mean gray level of an area of an object (translated as Optical Density – OD), be it by using ImageJ or Axiovision (Zeiss), we should conclude that the higher values correspond to a general bright object area and the lower values correspond to a darker area of the object. In other words the value of OD has an inverse correlation with the “darkness” of the area measured.
That reasoning may be extended to the Integrated Optical Density (IOD) that is calculated multiplying the pixel mean gray value of the area by the area of the region of the object where the measurement was taken (OD)Xarea = IOD, IOD is also the sum of the gray values of the area. That is, the lower the IOD number, the darker is the Area.
If we are measuring the product of an enzyme histochemical reaction or DAB deposition in a immunohistochemical DAB reaction, we may conclude that the lower the IOD, the darker is the area and consequently, the more quantity of reaction product is present the area. Of course, this relationship must be inverted if the image was originated from immunofluorescence.
Am I right? Is there something that I am not considering?
Francisco Blazquez
School of Veterinary Medicine
University of Sao Paulo