PAMPHLET OF LASER CALCULATIONS
These sample problems will assist those doing Laser
HAZARD EVALUATIONS AND CLASSIFICATIONS
A Laser Hazard Analysis Software and loan copy of ANSI Standard Z136.1-2000 are
available in D.O.E.S. for your use. ANSI Standard Z136.1-2000 must be used to access
all the tables, factors and additional information and problems encountered.
Introduction
The cornea and lens focuses a laser beam on the retina and causes retinal or macula eye
damage. Exposure to the direct beam, even for a small collimated laser, results in an
image burn on the retina of roughly 25 um in diameter. The MPE’s, or maximum
permissible exposure levels, as corneal exposure, are set very low to offset the natural
focusing ability of the human eye. For retinal effects, the real hazard is related to the
amount of laser power or energy that enters the pupil and subsequently becomes focused
on the retina or macula.
Even though infrared (1.4 um to 1.0mm) and ultraviolet (0.180 um to 400 um) do not
present a defined retinal hazard, damage may still occur upon reaching a specific power
or energy level . The cornea and the lens of the eye do not possess any focusing ability
for these wavelengths, the MPE is therefore much larger.
For ultraviolet exposure, the photochemical effects will accumulate over a full day’s
exposure. For some wavelengths this continues into the subsequent days also.
MPE’s may be expressed as radiant exposure, J.cm-2 for exposures less than 10s.;
or as W.cm-2 for CW lasers with exposures lasting for several milliseconds.
For exposures exceeding 10s for extended visible laser sources, this will require concepts
involving radiance and integrated radiance calculations..
Reference: ANSI Z136.1-2000
The easy ones first!
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PAMPHLET OF LASER CALCULATIONS (Cont)
Sample MPE Calculations
1. With a single pulse near-infrared laser, find the MPE for a 1.064-m (Nd:YAG) laser
with a 7x10-4 s pulse. The formula is found by going to Table 5, p. 45:
9.0 Cc t 0.75 x 10 –3 J x cm-2
9.0 x (7x10-4 ) 0.75 x 10-3 J x cm-2 = 3.87 x 10-5 J x cm-2
3.87 x 10-5 J x cm-2 / 7 x 10-4s = 5.5 x 10-2 W x cm-2
Select this formula from the columns for the conditions shown:
MPE:H
MPE:E
2. Find the MPE for a single pulse visible laser with a 694.3 nm, ruby laser pulse of
7x10-4 s (0.7 ms). Select this equation for the conditions shown: 1.8 t0.75 x10-3 J.cm-2
MPE:H
1.8(10-3 )( 7x10-4 s) 0.75 J x cm-2 = 7.7 x 10-6 J x cm-2
Since E x t = H, the MPE may be expressed also as peak irradiance,
MPE:E = MPE:H/t = 7.7 x 10-6 J x cm-2 / 7x10-4 s = 1.1 x 10-2 W x cm-2
3. Find the maximum irradiance permitted for a 0.25 s exposure to a visible laser
(  = 400 nm to 700 nm.).
Well, the MPE for visible lasers from Table 5 for  = 400 nm to 700 nm. for
exposures of 18 s to 10 s is: MPE:H = 1.8 t 3/4 mJ x cm-2
So the MPE is 1.8 (0.25) 3/4 mJ x cm-2 = (1.8 x 0.345) mJ x cm-2 = 0.636 mJ x cm-2
For a single exposure, again the irradiance is: E = H/t H= radiant exposure t = pulse
duration
So….
0.636 mJ x cm-2 / 0.25 s = 2.55 mW cm-2 (2.55 mW per cm2)
The MPE may be expressed two different ways: As radiant exposure in J per cm2 or
as irradiance in W per cm2!
PAMPHLET OF LASER CALCULATIONS (Cont)
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PAMPHLET OF LASER CALCULATIONS (Cont)
4. A GaAs near infrared laser operates at room temperature at a peak =0.904 m .
What is the MPE for a 180 ns pulse? See Fig 4, page 65 and Fig 8, page 69. The MPE
for exposures of 1 ns to 18 s is 5 x 10-7 J x cm -2 . The correction factor for 0.9 m, CA ,
is 2.5 as read from Fig 8a. The product is 1.25 x 10-6 J x cm-2 .
Refer to Tables 5 and 6 to find the MPE. Select from Table 5, under “visible and near
infrared”, 0.700-1.05 m, 10-9 to 18 x 10-6 s, the
MPE:H = 5.0 x CA x CE x 10-7 J x cm-2
Source size, CE , is not given and is assumed to be 1.0
From Table 6, CA, for  = 0.7 to 1.05 m is calculated from:
CA = 102 ( -0.7 m) = 2.56
MPE:H = (2.56)(5 x 10-7) = 1.28 x 10-6 J x cm-2
5. A 4.0-mW laser operates at = 1.55 m with a beam diameter of 1.2 cm.
Find the MPE for a 10s exposure. Select from Table 5, the MPE for a 10s exposure
at 1550 nm is 1 J x cm-2. In terms of irradiance, this is then just:
MPE:H = 1 J x cm-2 / 10 s = 0.1W x cm-2
6. For an extremely short pulsed laser find the MPE for a single 100 fs pulse
(100 x 10-15 s) at 580 nm (0.58 m). Select from Table 5 for the conditions indicated:
MPE:H = 1.5 x 10-8 J x cm-2
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PAMPHLET OF LASER CALCULATIONS (Cont)
7. Find the MPE for a near infrared extremely short pulsed laser for a single pulse of
18 ps (1.8 x 10-11 s) at  = 1100 nm. Select the MPE from Table 5 for the above
conditions:
MPE:H = 27 Cc t 0.75 J x cm-2
where t is in seconds 10 ps to 1 ns
For 1100 nm, Cc = 1.0
MPE:H = 27 (1.0)(1.8 x 10-11) 0.75 J x cm-2
= 2.36 x 10-7 J x cm-2
8. Find the MPE for a visible repetitive-pulsed laser with a high pulse repetition rate
(PRF) for a 514.5 nm argon laser with a PRF of 10 Mhz and a pulse width t of 10 ns
(10-8 s). Let the exposure duration equal 0.25s.
When the PRF is g.t. 55 kHz, the average irradiance limitation from Rule 2 kicks in
and t becomes Tmax . So from Table 5,
MPE:H = 1.8 x t 0.75 x 10-3 J x cm-2
= 1.8 x (0.25s) 0.75 x 10-3 J x cm-2
= 6.36 x 10-4 J x cm-2
Since this MPE is the same as for a CW laser, the MPE expressed in terms of average
irradiance is:
MPE:E = MPE:H/ T = 6.36 x 10-4 J x cm-2 / 0.25s
= 2.55 x 10-3 W x cm-2
9. What is the MPE for a helium-cadium (He-Cd) laser at  = 3 25 nm when the worker
is in the lab only for the last 5 minutes of every hour, noting that an exposure can occur
at all locations within the laboratory?
Total exposure time = 7 min x 8 hours = 56 min. x 60 s / min = 3360 s
From Table 5, the MPE for these conditions is 1.0 J x cm-2 for exposures of 10s to 8 hrs
Note that the exposure is a limit and the time does not matter.
For the radiant exposure, then the time is a factor:
MPE:E = MPE:H/ t = 1 J x cm-2 / 3360 s = 0.3 mW x cm-2
Hold on to your calculators!
PAMPHLET OF LASER CALCULATIONS (Cont)
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PAMPHLET OF LASER CALCULATIONS (Cont)
Repetitive Pulse Laser Calculations
10. Find the MPE for a single pulse laser rangefinder at 1710 nm with a 15 ns pulse width. From
Table 5, the MPE for 1710 nm is 1 J x cm-2 for all durations of 1 ns to 10s.
What is the MPE for a train or group of pulses? The MPE group is expressed in average
irradiance. This is computed from the sum of the individual radiant exposures for all the pulses in
the group.
So, the MPE: E group = MPE:H/ T . There are a variety of ways to express the MPE for a group
of pulses.
First find the MPE for a person exposed to only one pulse of maximum energy
in the pulse train (MPE sp).
Second, find the combined energy of the entire pulse train (MPE:H group).
Third, to find the applicable MPE for an exposure to a repetitive pulse laser, the
wavelength, pulse repetition frequency (F), duration of a single pulse (t), duration of any
pulse groups (T), and the duration of a complete exposure must be known (T max).
Select the MPE/pulse which indicates the greatest hazard for any single pulse during the exposure
from testing of the following three rules!
1. Single pulse limit. The MPE is limited by the MPE sp for any single pulse during the
exposure (assuming exposure to only one pulse).
2. Average- power limit. The MPE is limited to the MPE for the duration of all pulse trains,
T, divided by the number of pulses, n, during T, for all exposure durations up to T max.
3. Repetitive Pulse limit*. The MPE is limited to the MPE sp multiplied by the correction
factor Cp, i.e. = n –0.25, where n is the number of pulses that occur during the exposure
duration T max.
The critical frequency  is the PRF above which the MPE from Rule 2 yields the smallest MPE.
For the retinal hazard region, the critical frequency is generally 55 kHz for
 = 0.4 to 1.05 m and 22 kHz for wavelengths between 1.05 and 1.4 m.
* For rule 3, pulses that occur within t min , are considered a single pulse and the MPEsp ,
is based on a minimum pulse width of t min, Rule 3 does not apply for wavelengths less than
0.4 m, except for the thermal limit expressed as 0.56 x t 0.25 J x cm-2 .++

When sub nano-second pulses are involved, the critical frequency could be higher. When the
wavelength is outside the retinal hazard region, or the duration of the pulse train exceeds 10s, the
critical frequency is less.
++ Exposure durations exceeding 10s, only those pulses contained within a time equal to T 2, are
considered when computing Cp. The value of T2 is 10s for small sources. For extended sources
see equation B88 in B7.2 for computation of T2.
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PAMPHLET OF LASER CALCULATIONS (Cont)
11. For a near infrared repetitive pulsed laser, find the MPE for a 905 nm GaAs
laser with a 100 ns pulse width (1 x 10-7) and a PRF of 1 kHz. The PRF is less than the
critical frequency, so all three rules must be tested.
At = 905 nm, there will not be a natural aversion response as would be for a visible
wavelength laser, assume a 10-s exposure duration for Tmax. The total number of pulses
(n), in a 10 s exposure interval is the product of the exposure duration (T) and the PRF (F).
So…. n = F x T = 1 x 104 pulses and from Fig 13, page 80, the reduction factor n –0.25
is found to be 0.1. Now from Table 6, p.47 or Fig 8a, page 68, the  correction factor is
2.57 at 905 nm.
The MPE/pulse is the lowest and most conservative from testing the three rules. The
MPE from Table 5 for a single 1 ns pulse is :
Rule 1: Single Pulse Limit:
MPEsp = 0.5 CA x 10-6 J x cm-2 = 1.29 x 10-6 J x cm-2
Rule 2: Average Power Limit: The MPE for a 10s exposure is :
MPE:H group = 1.8 x 10-3 CA t 0.75 J x cm-2 = 2.6 x 10-6 J x cm-2
104 pulses
Rule 3: Repetitive Pulse Limit: MPE/pulse = n-0.25 MPEsp = 10,000-0.25 x
1.29 x 10-6 J x cm-2 = 1.3 x 10-7 J x cm-2
Rule 3 yields the lowest value and is selected as the MPE/pulse.
Now the cumulative exposure MPE for the entire pulse train is found:
MPE:H group = T x F x MPE/pulse = (10s) (103 Hz) (1.3 x 10-7 J x cm-2)
= 1.3 x 10-3 J x cm-2
In terms of average irradiance this is:
MPE:E = MPE:H group = 1.3 x 10-7 J x cm-2 = 1.3 x 10-4 W x cm-2
T
10s
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PAMPHLET OF LASER CALCULATIONS (Cont)
12. Find the lowest MPE for a 10 s exposure when using a XeCl excimer laser at
308 nm that emits 18 ns pulses at a pulse repetition frequency (PRF) of 190 hz.
MPEs for the Ultraviolet region is based on a dual set of limits for both photochemical
and thermal effects.
MPE at 308 nm = 40 mJ x cm-2 …………. for exposures of 1 ns to 30 ks
(based on photochemical effects of eye and skin)
MPE of 0.56 t 0.25 cant be exceeded because of thermal effects!
(this is the same MPE used for middle and far infrared exceeding a few ns)
The three rules must be applied. However, both thermal and photochemical effects
must be determined for each rule.
Hang on! Here we go again!
Rule 1: Single Pulse Limit.
The thermal MPE limit is ………
MPEsp = 0.56 x (18 x 10-9)0.25 Jx cm-2 = 0.56 x 1.16 x 10-2 J x cm-2
= 6.50 mJ x cm-2
The photochemical MPE limit is ….
MPEphoto sp = same as for entire exposure i.e. 40 mJ x cm-2
Therefore, for Rule 1: The thermal MPE predominates and is the most conservative!
Rule 1:
MPE sp = 6.50 mJ x cm-2 (thermal)
Rule 2: Average Power Limit. The MPE (thermal effects) = 0.56 t0.25 Jx cm-2
Where t becomes the total duration of the exposure, Tmax, at 10s.
Rule 2: MPE = 0.56 (10)0.25 J x cm-2 = 0.56 x 1.78 J x cm-2 = 1.0 J cm-2 (thermal)
For 10s of exposure, the number of pulses are 190 x 10 = 1900 pulses
The thermal MPE for each pulse = 1.0 J x cm-2 /1900 pulses = 0.526 mJ x cm-2
The photochemical MPE limit is …..
MPEphoto = the accumulated exposure over the entire 10 s = 40 mJ x cm-2
MPE/pulse (Rule 2) = 40 mJ x cm-2 = 21 J x cm-2
1900 pulses
The photochemical MPE, 21 J x cm-2, is much less than the thermal MPE (Rule 2),
1.0 J cm-2, so it is selected for Rule 2.
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PAMPHLET OF LASER CALCULATIONS (Cont)
Rule 3: Repetitive Pulse Limit. To find the MPE for Rule 3, apply the repetitive
pulse correction factor to the thermal MPE sp for a single pulse but not to the
photochemical limit of 40 mJ x cm-2.
MPE sp = 6.50 mJ x cm-2 (thermal for 1900 pulses of exposure)
CP= n-0.25 = 0.15
MPE/pulse (Rule 3) = 6.50 mJ x cm-2 x 0.15 = 0.975 mJ x cm-2
The final MPE selected is the lowest from all the Rules, Rule 2, 21 J x cm-2.
13. Find the MPE of a 0.6943 m Q-switched laser with three 200 ps pulses 100 ns
apart? This is a visible wavelength (643 nm), where the pulse train isn’t more than 0.25s.
Rule 1: Single Pulse Limit: Table 5 (B2), single pulse:
MPEsp = 2.7 x t 0.75 J x cm-2
= 1.44 x 10-7 J x cm-2
Rule 2: Average Power Limit. The Pulse train of 200 ns is less than tmin of
18 s (B12), So the ……(B3 Table 5, p 45)
MPE:H group = 5 x 10-7 J x cm-2 = 1.67 x 10-7 J x cm-2
3 pulses
Rule 3: Repetitive Pulse Limit. Where T is less than tmin , all the pulses are
considered the same as 1 pulse, so sum the energies of all three pulses. CP is 1.0.
The MPE for tmin is 5 x 10-7 J x cm-2 and the …..
MPE/pulse = based on Rule 3 is the same as for Rule 2!
= 1.67 x 10-7 J x cm-2
Select Rule 1: MPE/pulse = 1.44 x 10-7 J x cm-2
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PAMPHLET OF LASER CALCULATIONS (Cont)
14. Find the MPE at 20 cm from the laser exit port for a GaAs diode laser with a 1.5 cm
circular beam at the exit port and operating at 0.904 m with a PRF of 2.53 kHz. The
laser is focused into a fiber optic cable by placing the tip of the cable at the focal point of
a short focal length lens. The collimating lens is 2.0 cm in diameter. The source size is a
constant 2.8 mrad within a distance of 667 cm, then the exit port limits the source size to
2 cm for all viewing distances that follows.
For this laser use the small source MPE: MPEsp (Rule 3):
(MPE/pulse)small = MPEsp x Cp J x cm-2
Next, find the number of pulses determining Cp from T2, which is based on source size.
The equation for T2 is taken from Table 6, p. 47.
T2 = 10 x 10 (-1.5)/98.5 = 10 x 10 (2.8-1.5)/98.5
= 10.30 s
The MPEsp is 1.28 x 10-6 j x cm-2 as noted in example 4. The number of pulses emitted
in 10.30 s is 26059. The Cp value is 0.0787. Now the MPE becomes:
MPE/pulse extended =1.28 x 10-6 j x cm-2 x Cp where Cp = (26059) –0.25
= 1.28 x 10-6 J x cm-2 x 0.0787
= 1.007 x 10-7 J x cm-2
The extended source MPE is found from the angular source size and the small-source
MPE. When the source subtends and angle greater than 1.5 mrad, the small source MPE
is multiplied by CE. Since the evaluation distance is less than 667 cm, the source
subtends an angle of 3 mrad for this laser and:
CE = /min = 2.8 mrad / 1.5 mrad = 1.86 (B14)
The corresponding extended source MPE becomes:
(MPE/pulse) extended = (MPE/pulse)small x CE
= 1.86 x 1.007 x 10-7 J x cm-2
= 1.87 x 10-7 J x cm-2
)
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PAMPHLET OF LASER CALCULATIONS (Cont)
15. Determine the emergent beam diameter necessary to preclude a skin hazard for a
1.0 watt He-Ne Laser operating at 0.543 m.
The MPE for visible wavelength skin exposures greater than 10 s is found in Table 7, p. 48.
MPE skin = 0.2 x CA w x cm-2 (with a limiting aperture of 3.5 mm)
For a visible laser, CA = 1.0, so the ….
MPE skin = 0.2 x 1.0 = 0.2 w x cm-2
Where the output is greater than 0.5 W, the beam would have to be large enough to
reduce the irradiance to below 200 mW x cm-2.
MPE skin = E0 = 4 /a2 and assume a is larger than 3.5 mm so…..
a=
4/MPE
4(1)/(3.2)
=
= 2.52 cm
The beam diameter must be greater than 2.5 cm to preclude a skin hazard.
16. Find the separation distance required for an open beam 200 watt Class IV
ND:YAG laser etching system with a beam divergence of 3.00 mrad and an exit beam
diameter of 0.5 cm.
Assume an eight hour long worst case exposure situation and a TL of 50 W x cm-2
Then ……..
Ds = 1/3.0 x 10-3
4 x 200 -(0.5)2
 x 50
where TL= barrier threshold
limit value in w x cm-2
= 7.33 m
So, placement of this barrier closer than 7.33 meters toward the laser would result in a
hole being burned through the barrier!
17. What is the worst case Optical Density (OD) required for a He-Ne laser operating at
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PAMPHLET OF LASER CALCULATIONS (Cont)
a wavelength of 0.600 m with a 1 s pulse of 10 mJ with a 4 mm diameter beam?
The worst case optical density required for this laser is calculated by averaging the
radiant energy over the limiting aperture.
This radiant energy is given as 10 mJ. The limiting aperture is 7 mm since the laser
operates at 0.500 m. The radiant exposure averaged over 7 mm is…..
H = 1 x 102 = 26.0 mJ x cm-2
0.385
Now note from Table 5, the MPE for a 1 s visible wavelength single pulse is:
MPE = 0.5 J x cm-2 So, the minimum worst case OD is…..
D = log H/MPE = log
2.6 x 10-2
5 x 10-7
= 4.72 O.D.
18. Find D, the laser beam barrier penetration distance, for a 2000 watt CO2 laser with a
4 -inch focal length lens in place. The beam size is 1 inch at the lens.
Assume a worst case long term 8-hr exposure and a TL of 50 W x cm-2
/
D = 0 b
4
= 4/1
TL
= 0.285 m
PAMPHLET OF LASER CALCULATIONS (Cont)
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4 x 2000/3.14 x 50
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PAMPHLET OF LASER CALCULATIONS (Cont)
19. Find the power passing through the limiting aperture of 7 mm for a He-Ne, 4 mW
laser where the beam diameter specified at 1/e2 of the peak irradiance points is 1.8 cm.
The beam diameter is given at the 1/e2 points, but for laser calculations, we use 1/e!
So………
1/e = 1.8 cm
2 = 1.27 cm
Now, the power that passes through an aperture of diameter Df is:
d = 0 1-e –{df/dl}2
= 4 mW 1-e
–{0.7cm/1.27cm}2
= 1.05 mW
20. Does a He-Ne laser at 632.8 nm with a 1 mm exit beam diameter and a specified
maximum output power of 1.25 mW exceed the MPE for a 0.25s exposure near the exit?
MPE for a visible laser for 0.25s is 2.55 x 10-3 W x cm-2
Near the laser exit port, the increase in beam diameter may be ignored here.
For the retinal hazard region, Df is 7 mm. This represents a fully dilated pupil in
daylight. The irradiance of the laser is:
E=
1.27 
= 1.27 (1.25 x 10-3) = 3.24 x 10-3 W x cm-2
2
[ max (a,Df) ]
( 0.7)2
So….. 3.24 x 10-3 W x cm-2 is greater than 2.55 W x cm-2
This laser does exceed the MPE.
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PAMPHLET OF LASER CALCULATIONS (Cont)
21. What is the laser class for a single pulsed Q-switched ruby laser with a
manufacturer’s listed peak power output of 25 mW for 32 ns?. The laser rod diameter
is 5/8 inch.
First find the output energy per pulse using this formula:
Q=xt
where  = laser peak power
7
-8
= (2.5 x 10 W)(3.2 x 10 s)
= 0.8 J
Now select from Table 6, CA , the correction factor = 1.0 for 694.3 nm (0.694 m)
The Class 3b limit formula is: 0.03 x CA = 30mJ
This laser is 27 times this limit!
( 0.8J/0.03J = 26.66) So it is a Class 4 laser.
22. Classify the following laser: A dye laser with a peak = 0.580 m. Power output is
10 mJ for a 5 mm beam for a 1 s burst.
With a pulse width less than 18 s, the Table 5 MPE selected is:
MPE = 5 x 10-7 J x cm-2
again, CA =1.0 at 590 nm (Table 6)
The Class 1 AEL is defined as the MPE x Area of limiting aperture from Table 8!
At 590 nm, the limiting aperture is:
7 mm
So……
Df2
aperture area= 0.385 cm2
J x cm-2
AEL = MPE x
4
= 5 x 10-7 J x cm-2 x 0.385 cm2
= 1.9 10-7 J
The Class 3a AEL is 5 times the Class 1 AEL or 9.6 x 10-7 J and the Class 3b limit is
previously noted as : 0.03 x CA = 30mJ/pulse.
The 10 mJ output lies between the limits of 0.96 J and 30 mJ. Therefore, it’s a
Class 3b laser.
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PAMPHLET OF LASER CALCULATIONS (Cont)
23. Find the minimum energy of a Q-switched single pulsed laser operating at = 750 nm
with an exit diameter of 1.0 cm that will produce a diffuse reflection hazard.
From Table 3, find the energy that will not produce a hazardous diffuse reflection.
CA = 1.26 for a laser at this wavelength.
So….. The maximum energy that will not produce a hazardous diffuse reflection is:
22 x 1.26 mJ …
at a 20 cm viewing distance!
At a 100 cm viewing distance, the:
minimum energy = 110 x 1.26 mJ = 140 mJ
For a 10 m viewing distance, the incident beam energy greater than:
1.6 x 1.26 = 2.0 J
diameter spot.
…… will produce a hazardous diffuse reflection for a 1 cm
24. For an infrared CW Nd:YAG laser, operating at 35 W, with a beam diameter at the output
(at 1/e) of 1.0 cm, find the minimum OD required to protect against accidental intra-beam
viewing. =1064 nm.
This is an infrared laser, so assume a worst case exposure of 10s for accidental viewing.
Irradiance MPE under these conditions is 5 mW/cm2 (5.0 Cc x 10-3 w/cm2, Table 5a, Cc=1.0 )
DL = 1 cm and Df = 0.7 cm
a2 =1.0 because DL is g.t. Df
d
2
First, lets find the power through the aperture: n = 1 – e – a2 = 0.387
So the power going through the 0.7 cm aperture is: 13.55 watts
E avg = 13.55 / 0.385 cm2 = 35.2 W/cm2 and the MPE = 0.005 w/ cm2
So… based on average irradiance, the OD= log10 (Eavg/MPE) =
log10 (35.2/ 0.005) = log10(7040)
OD = 3.85
Do select an OD slightly higher!
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