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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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.
The example had an error, now it's fixed.
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