Energy Transport and the Poynting Vector — Question 8

Question 8

Attenuated energy flux identifies where an absorbing slab is heated

A broad beam enters a slab at x=0 with intensity I0. Its measured forward intensity is $I(x)=I_0e^{-\alpha x}$, with positive alpha; neglect reflections and transverse energy flow. Steady deposited power per volume is q(x).

1. Use energy conservation in a thin slice to find q(x).

2. Find the fraction of power absorbed in thickness d and the thickness absorbing one-half of the incident power.

Question 8 - Solution

Attenuated energy flux identifies where an absorbing slab is heated

(a) Method and calculation

$$q(x)A\,dx=A[I(x)-I(x+dx)]\Rightarrow q=-dI/dx.$$$$\boxed{q(x)=\alpha I_0e^{-\alpha x}}\quad[\mathrm{W/m^3}].$$

(b) Interpretation and further calculation

$$\boxed{P_{\mathrm{abs}}/P_{\mathrm{in}}=1-e^{-\alpha d}}.$$$$1-e^{-\alpha d_{1/2}}=1/2\Rightarrow\boxed{d_{1/2}=\ln2/\alpha}.$$

Physical check

The attenuation coefficient describes intensity decay; a corresponding field-amplitude decay coefficient would be alpha/2.