WON TECH WON COSJET ATR S, Руководство пользователя

Страница: 33 / 55

User Manual
WT-UM-08 (v4.4)
33/55
3. For the WON COSJET ATR S laser system, determine the intra-beam MPE for direct ocular
exposure to the radiation from a Nd:YAG laser(λ = 532 nm) operating at a frequency of F = 1 Hz with
a pulse width of t = 20 ns at Q-switching mode.
Wavelength = 532 nm;
Peak power per pulse P
p
= 25 MW;
Energy per pulse P
o
= 0.5 J;
Pulse repetition rate = 1 per second;
Exit aperture beam diameter = 10 mm;
Beam divergence angle = 50 mrad.
What is the effective NOHD on the basis of the single-pulse threshold (a) for exposure of the unaided
eye, and (b) when intra-beam viewing through 50 mm diameter optics is involved?
(Effects of beam attenuation or refractive focusing due to atmospheric transmission are neglected in
these calculations.)
The pulse width t
p
can be calculated from the condition
P
p
t
p
= P
o
by 25 × 10
6
× t
p
= 0.5
giving t
p
= 20 ns (i.e. 10
–9
< t
p
–5
s). The pulse repetition frequency F is 1/1 = 1 Hz.
In this example, it is assumed that α ≤ α
min
and for a small source C
6
= 1. If there is no intentional
viewing, the exposure duration to be used is 10 s; during this time, the number of pulses is
N = F
t = 1 Hz × 10 s = 10
The intra-beam MPE is taken as the most restrictive calculated from the application of 13.3.
Single-pulse assessment (condition 13.3a))
From table 6, the MPE for a single-pulse exposure from this laser is
H
MPE
= 5 × 10
–3
C
6
J·m
–2
where C
6
= 1, therefore
H
MPE, single
= 5 × 10
–3
J·m
–2
Average irradiance assessment (condition 13.3b))
From table 6, the MPE for the exposure duration of 10 s is
H
MPE
t
0.75
C
6
C
7
J·m
–2
where C
6
= 1 and C
7
= 1. There are 10 pulses in 10 s, therefore the average MPE per pulse is
H
PE, exposure
=
90 10
0.75
10
= 50.6 J m
-2
Multiple-pulse assessment (condition 13.3c))
The maximum exposure duration for which requirement c) should be applied is T2 in the wavelength
range 400 nm to 1400 nm, where T2 = 10 s for α ≤ α
min
. Therefore, the correction factor N
–1/4
= (10)
–
1/4
= 0.56 is used to calculate H
MPE
, train:
H
MPE, train
= H
MPE, single
N
–1/4
= 5 × 10
–3
× 0.56 = 2.8 × 10
–3
J·m
–2