Question

The absorption coefficient is 4X10^4 cm-1 and the surface reflectivity os 0.1 for a silicon wafer illuminated with a monochromatic light having a hv of 3 eV. Calcuate the depth at which half the incident optical power has been absorbed in the material

Answer 1

Answer:

Depth, $x=1.73\times 10^{-5}\ m$

Explanation:

It is given that,

Absorption coefficient, $\alpha =4\times 10^4\ cm^{-1}$

Surface reflectivity, T = 0.1

Energy of light, E = 3 eV

We need to find the depth at which half the incident optical power has been absorbed in the material. The relationship between the absorption coefficient and intensity of light is given by :

$I=I_oe^{-\alpha x}$

Since, the incident power is becoming half. So,

$\dfrac{I_o}{2}=I_oe^{-\alpha x}$

$0.5=e^{-\alpha x}$

$ln(0.5)=-4\times 10^4\times x$

Since, ln (0.5) = -0.693

$x=\dfrac{-0.693}{-4\times 10^4}$

x = 0.0000173 m

or

$x=1.73\times 10^{-5}\ m$

So, the depth at which half the incident optical power has been absorbed in the material is $1.73\times 10^{-5}\ m$. Hence, this is the required solution.