Does the optical transmittance of pulsed laser through a transparent material differ quantitatively from a continuous wave laser?

The absorption & transmittance of light through a transparent medium is described by the Beer-Lambert law. But the equation does not factor in the time parameter so I assume it describes continuous light rather than optical pulses. So the question is: Does the optical transmittance of pulsed laser light(assuming its fairly low power, << 1 Watt)for an arbitrary wavelength change quantitatively compared to a continuous beam of the same wavelength? Is so, is there a mathematical relationship between the pulse duration and transmittance?

• 1 W peak power? Or 1 W average power with an unspecified (possibly very low) duty cycle? – The Photon Apr 27 at 18:42
• In general ....no. If the power is similar for both then power out is the same minus whatever losses as a %. The losses in general are due to reflection and absorption, it's the absorption that can be non-linear at high power densities, for example the absorbing transitions can get saturated, unable to absorb photos at the same rate as at more normal powers. – PhysicsDave Apr 27 at 19:41

The propagation of an electric field through a linear medium is affected by the refractive index of the material, which is usually frequency dependent

$$\bar{n}\left(\omega\right)=n\left(\omega\right)+i\kappa\left(\omega\right)$$

For a monochromatic electric field then

$$E\left(x,t\right)=E_{0}e^{i\left(\bar{k}x-\omega t\right)}=E_{0}e^{-\frac{\alpha}{2} x}e^{i\left(kx-\omega t\right)}$$

where I used the relation $$\bar{k}=\frac{2\pi\bar{n}}{\lambda_{0}}=\underbrace{\frac{2\pi n}{\lambda_{0}}}_{k}+i\underbrace{\frac{2\pi\kappa}{\lambda_{0}}}_{\frac{\alpha}{2}}$$. If that's all the field you have, then you can immediately recover Beer-Lambert law $$I=I_{0}e^{-\alpha x}$$. However, in the general case your electric field at the interface with the material is made out of many frequencies, and you can use the Fourier decomposition to write

$$E\left(0,t\right)=\int E\left(\omega\right)e^{-i\omega t}{\rm d}\omega$$

Each of these components advances as a monochromatic light, and you get at the output

$$E\left(L,t\right)=\int E\left(\omega\right)e^{-\frac{\alpha\left(\omega\right)}{2} L}e^{i\left(k\left(\omega\right)L-\omega t\right)}{\rm d}\omega$$

The pulse duration is inversely proportional to its spectral width, and thus short pulses would get distorted and unevenly attenuated. To get more specific information, you will need the profile of $$\bar{n}\left(\omega\right)$$.