Shining a laser directly into the lens of a camera is probably a bad idea.

By scattering the laser beam and then taking advantage of the inverse square law, we can get any attenuation we want.

A Lambert surface scatters light in a predictable manner. Paper can make a good Lambertian surface. link We had some paper that matched the description of what was used in the link.

We imagine a hemisphere centered on the point at which the laser beam strikes the paper. If we know the radiant flux density (watts per square meter) on the surface of the sphere directly above the paper, we can predict what it should be anywhere else on the surface of the hemisphere.

E' = E cosθWhere:

E is the radiant flux density on the surface of the hemisphereThe process will go as follows:

exactly above the middle of our Lambertian surface.

E' is the radiant flux density at the camera lens.

θ is the angle between the reference point and the camera lens

- Calculate the power received by the camera.
- Calculate the power at the reference point on our imaginary hemisphere.
- Figure out how the power on the reference point relates to the laser power.
- Calculate the laser power it took to create the power at the reference point.

The aperture was left wide open. The analog gain was set to unity. The exposure time was 1/1000 second.

Full Well Capacity (FWC) = 8180 #taken from Strolls with my Dog

The data from the camera was surprising so a total of four images was taken with the paper being rotated between each image.

The surprise is how much the energy seems to be concentrated in the middle of the laser beam.

Based on an initial guess about the width of the laser beam, the statistical calculations were based on 1 mm cubes.

With the equipment set up as it is, 1 mm at the center of the graduated cylinder is about 14 pixels at the camera.

fwc = 8180. #full well capacityThe result is:

mdn = 255. #max digital number

f = open("ave.txt", 'r')

total = 0

l = f.readline()

while l != "":

pos = l.find(" ")

num = int(l[pos:])

total = total + num

l = f.readline()

#that's for an array 15 x 1

#to get a 1 mm square area multiply by 206/15

total = total * 206./15.

f.close()

print(total)

#the red pixels are 1/4 of the total pixels

#to get the correct energy landing on the 1 mm square,

#multiply the electrons by four

electrons = 4 * int(fwc * total / mdn)

print("electrons", electrons)

h = 6.6e-34

c = 3e8

lmbda = 640e-9

ee = h*c/lmbda #electron energy (joules)

print("electron energy", ee, "joules")

eas = electrons * ee #energy at sensor (joules) per 1 mm square

print("energy at sensor", eas, "joules")

6743.066666666667

electrons 865228

electron energy 3.09375e-19 joules

energy at sensor 2.676799125e-13 joules

The hemisphere was modeled as a stack of rings of varying radius. The inner surface of each ring was 1 mm wide so the inner surface area of each ring would be 1 mm x 2 * pi * r where r is in mm.

The total power illuminating each ring would be the ring area x cosθ. In the following code, angle α was used to calcualte each ring radius so the power illuminating the ring would be cos(90° - α).

#a set of stacked rings approximating a hemisphereThe result was:

#the power flux at the top of the hemisphere is set to 1

from math import pi

from math import cos

from math import acos

from math import sin

from math import asin

pMax = 1.

r = 300. #mm

arcLen = pi*r/2. #quarter circle arc length

arcAng = pi/2 #radians in quarter circle

inc = 1. #1 mm approximate square

incAng = arcAng * inc / arcLen

totalPwr = 0

ringAng = incAng/2

while ringAng < arcAng:

ringRadius = r - (r * sin(ringAng))

print(ringAng, ringRadius)

ringCirc = 2*pi*ringRadius

ringPwr = inc * ringCirc * cos(arcAng - ringAng)

totalPwr = totalPwr + ringPwr

ringAng = ringAng + incAng

print("total power", totalPwr)

total power 121354.7So, if we multiply the power illuminating the reference point by the ratio, rat = 121355, we get the total power illuminating the inside surface of the hemisphere.

The base, b = 280 mm.

The angle α = 0.367 radians = 21°

The angle θ = π/2 - α = 1.2 radians = 69°

From above, energy at sensor, es = 2.676799125e-13 joules

The energy at the reference point, er = es x cos(69°)

The reflectance of the paper was taken as, refl = 0.8

The laser energy, le = er x rat / refl = 1.13e-7 joules

The exposure time was et = 0.001 second.

The laser power was, lp = le / et = 0.11 mw.