Calibrate the Scale of an Image
The microscope display shows a scratch 30 pixels across.
Calibrate the Scale of an Image explained visually
The microscope display shows a scratch 30 pixels across. English captions are optional and start off. The complete transcript follows the written lesson.
What you will learn
The microscope display shows a scratch 30 pixels across.
A practical question
The microscope image shows a scratch thirty pixels wide. Is that three millimeters, a fraction of a millimeter, or something else? Pixels alone can't answer. We need a known physical length photographed under the same imaging geometry. That's the job of a calibration target.
The useful idea
Suppose ten millimeters on that target span one hundred pixels. Divide ten by one hundred and we get one tenth of a millimeter per pixel. Multiply by the scratch's thirty-pixel span: under the same setup, its estimated width is three millimeters. The calculation is simple; the conditions behind it are the part we must check.
Work through the example
Place the target and the fabric at comparable distances from the camera and use the same magnification. If someone adjusts the microscope zoom or moves the camera, the old scale may no longer describe the new image. A curved surface or changing depth can also make one scale inaccurate across the scene. Recalibrate or check the geometry before trusting another measurement.
Limits and next step
Save the target image, the scale calculation, and the original scratch image together. Then another inspector can trace how the physical number was obtained. Next, we'll study what an intensity histogram tells us about a faded image, and what it leaves out.
Worked example
If a known 10-millimeter span covers 100 pixels, 30 pixels correspond to about 3 millimeters under that same geometry.
Quick check: test your understanding
Try it yourself: In the situation described above, explain the method and one reason its output might be misleading.
Check the reasoning
A calibration target supplies a physical length per pixel for a fixed imaging setup. Changing magnification or camera distance invalidates an old scale.
What is the main idea behind calibrate the scale of an image?
A calibration target supplies a physical length per pixel for a fixed imaging setup.
How is it used in the example?
If a known 10-millimeter span covers 100 pixels, 30 pixels correspond to about 3 millimeters under that same geometry.
What limitation should be checked?
Changing magnification or camera distance invalidates an old scale.
Further reading
Digital Image Processing, 4th edition — Rafael C. Gonzalez and Richard E. Woods.
Original ToolsInVision teaching examples. Editorial approach · Suggest a correction
Accessible referenceLesson 20 transcript202 words · Video transcript · Matches the published narration and captionsOpen transcript
The microscope image shows a scratch thirty pixels wide. Is that three millimeters, a fraction of a millimeter, or something else? Pixels alone can't answer. We need a known physical length photographed under the same imaging geometry. That's the job of a calibration target.
Suppose ten millimeters on that target span one hundred pixels. Divide ten by one hundred and we get one tenth of a millimeter per pixel. Multiply by the scratch's thirty-pixel span: under the same setup, its estimated width is three millimeters. The calculation is simple; the conditions behind it are the part we must check.
Place the target and the fabric at comparable distances from the camera and use the same magnification. If someone adjusts the microscope zoom or moves the camera, the old scale may no longer describe the new image. A curved surface or changing depth can also make one scale inaccurate across the scene. Recalibrate or check the geometry before trusting another measurement.
Save the target image, the scale calculation, and the original scratch image together. Then another inspector can trace how the physical number was obtained. Next, we'll study what an intensity histogram tells us about a faded image, and what it leaves out.