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

Depth,
`x=1.73* 10^(-5)\ m`

Step-by-step explanation:

It is given that,

Absorption coefficient,
`\alpha =4* 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,

`(I_o)/(2)=I_oe^(-\alpha x)`

`0.5=e^(-\alpha x)`

`ln(0.5)=-4* 10^4* x`

Since, ln (0.5) = -0.693

`x=(-0.693)/(-4* 10^4)`

x = 0.0000173 m

or

`x=1.73* 10^(-5)\ m`

So, the depth at which half the incident optical power has been absorbed in the material is
`1.73* 10^(-5)\ m`. Hence, this is the required solution.