Problem 49652. Find the spot diameter from the intensity distribution matrix of single spot (circle)

Created by Sibi

Difficulty:Rate

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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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Last Solution submitted on Sep 01, 2023

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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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Problem 49652. Find the spot diameter from the intensity distribution matrix of single spot (circle)

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Problem 49652. Find the spot diameter from the intensity distribution matrix of single spot (circle)

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Last 200 Solutions

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