0
$\begingroup$

When attempting to trap a particle in a laser beam i was able to achieve the effect. enter image description here Pretty cool but now when i put of piece of plexiglass that is 1/8” thick i can no longer achieve the effect. The beam hits a focal lens when its diameter is around 6mm and focuses 50mm away. I put the piece if plexiglass 15mm after the focal lens. Now it is obvious that my 25mW laser will lose power but i wonder how, where, and how much? I imagine reflection and scattering accounts for most of it but there is more that plays a part. I want to know if theres a equation that takes a material, thickness, laser wavelength and power and will maybe give me my power on the other side of the material.

$\endgroup$
2
$\begingroup$

The equation you are looking for is

$$I(z) = (1-R) \cdot I_0 \cdot e^{-\alpha z}$$

where

$I(z)$: intensity of the laser beam after the distance $z$

$R$: reflectance of material

$I_0$: intensity of laser beam on the surface

$\alpha$: absorption coefficient (NOT the same as absorptance)

$z$: thickness of the material

When you plug the thickness of the plexiglas plate in as $z$, you get the intensity of the laser beam after it passes through the plate.

Note that the equation considers intensity instead of power, so you need to find the beam diameter on the surface of your plexiglas plate.

Further, you need to find the absorption coefficient of the material for the given wavelength. The absorption coefficient (or attenuation coefficient) is not a dimensionless number like the absorptance, but has the dimension $m^{-1}$, so they should not be mixed up.

$\endgroup$
2
  • $\begingroup$ Intensity vs. power doesn't really matter, since we assume uniform value of all coefficients across the plexi surface & volume. $\endgroup$ Nov 2 '18 at 17:01
  • $\begingroup$ @CarlWitthoft As long as we don't know the beam parameter product of the laser source, I wouldn't neglect the intensity change over 1/8", even if the power stays constant $\endgroup$ Nov 2 '18 at 17:41
2
$\begingroup$

There are a number of things to deal with. The plexi has a reflectivity of around 4% per surface (see, e.g. this spec sheet , but minimal absorption. So pure loss is 8% (but will increase dramatically if you tilt the plexi sheet)

However, if you've placed the sheet between the lens and the focal point, you have not only moved the focus (see "equivalent optical thickness") but almost certainly induced some coma or astigmatism, thus spoiling the power density in the focussed spot. If you can track down a ray-trace application or even a webpage-based one, you may be able to get an estimate of these effects.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.