11
$\begingroup$

I want to measure the size of the optical slit from the microscope image.

Here is the image:

single slit

The dimension of image data is $1300\times1030$ and the real pixel size is $6.7$ micron.

But the problem is the slit in this image is tilted.

How can I measure width of the slit from this image?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$
    – Michael E2
    May 5, 2016 at 14:44
  • 2
    $\begingroup$ Rotate the image. $\endgroup$
    – David G. Stork
    May 5, 2016 at 15:20
  • 6
    $\begingroup$ mathematica.stackexchange.com/questions/86396/… $\endgroup$
    – Karsten7
    May 5, 2016 at 16:20
  • $\begingroup$ Although the answers linked by @Karsten can also answer this question, I think this one's still distinct because it doesn't focus on the variation in slit width. This simplification makes it possible to get results with a simpler approach, which I point out in my answer. It would also apply to the other question, but would require more work to characterize the variations. $\endgroup$
    – Jens
    May 5, 2016 at 19:41

2 Answers 2

Reset to default
22
$\begingroup$

Extract lines at the edge of the object:

img = Import["https://i.stack.imgur.com/pT8aP.jpg"]
lines = ImageLines[EdgeDetect[FillingTransform[Binarize[img]]]];
HighlightImage[img, {Thick, Yellow, Line /@ lines}]

Mathematica graphics

From here you can rotate the image if you get the angle:

θ = Mean[ArcTan @@@ Subtract @@@ lines]
(* 1.67222 *)

ImageTransformation[img, RotationTransform[\[Theta] - Pi/2], 
 Padding -> 0, PlotRange -> All]

Mathematica graphics

and from here proceed how you would have if it wasn't tilted.

On the other hand, you can try the following as well. Here I take random points from one of the lines and find the minimum distance to the other line. I then take the mean of all of these minimum distances.

pts = RandomPoint[Line[lines[[2]]], 500];
Mean[RegionDistance[Line[lines[[1]]]] /@ pts]
(* 83.175 *)

in microns:

Mean[RegionDistance[Line[lines[[1]]]] /@ pts]*Quantity[6.7, "Microns"]
(* Quantity[556.521, "Microns"] *)
Improve this answer
$\endgroup$
17
$\begingroup$

Here is an approach that requires no manual interaction:

g = Import["https://i.stack.imgur.com/pT8aP.jpg"];

shape = Position[ImageData[Binarize[g]], 1, {2}];
pc = PrincipalComponents[N@shape];
lm1 = Internal`ListMin[pc][[All, 2]];
lm2 = -Internal`ListMin[-pc][[All, 2]];

slitWidth = Mean[lm2] - Mean[lm1]

(* ==> 83.6331 *)

Here, I'm using the undocumented function ListMin from Daniel's answer here after first doing the image rotation automatically using PrincipalComponents as I also did here.

To see how this works, look at the points collected in PrincipalComponents:

ListPlot[pc, AspectRatio -> Automatic, PlotRange -> All]

components

The points making up the slit have been aligned so that we now only have to determine a measure of the vertical extent. This is what I do with ListMin.

Of course, you can now convert to physical units by multiplying the pixel count with the micron ratio:

6.7 slitWidth

(* ==> 560.342 *)
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.