Problem 49652. Find the spot diameter from the intensity distribution matrix of single spot (circle)
Intensity distribution will be same as gaussian dsitribution and check for gaussian beam distribution
Distance between each index will be equal to 1 mm. Grid will be n*n matrix of distribution . n will be odd number . If n=3 center of the spot will be (2,2)
Find spot diameter in mm.
Hint : w(z) is the radius of the laser beam where the irradiance is 1/e2 (13.5%) of Intensity
example
x 0.0000 0.0045 0.0335 0.0045 0.0000
0.0045 1.8316 13.5335 1.8316 0.0045
0.0335 13.5335 100.0000 13.5335 0.0335
0.0045 1.8316 13.5335 1.8316 0.0045
0.0000 0.0045 0.0335 0.0045 0.0000]
I=100 (max intensity);
I(13.5 %)=13.5
radius=1 mm;Spot diameter=2 mm
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3 Comments
Tim
on 30 Dec 2020
In your example, the distance from 100 to 13.53 is sqrt(2), so the diameter should be 2*sqrt(2). The test suite values also do not seem to make sense.
Sibi
on 5 Jan 2021
The example had an error, now it's fixed.
Dyuman Joshi
on 13 Aug 2021
Test 7, max(x(:))=152; 13.5% of 152 = 20.52, which is not present in the matrix. Same goes for the rest of the cases.
I feel like I am missing something.
@Sibi
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