In the figure below the absorption coefficient as a function of the wavelength for several semiconductor materials is presented. Let's consider monochromatic light of photons with energy of Eph=1.55eV that incidents a film with thickness d. If we ignore possible reflection losses at the rear and front interfaces of the film, what thickness d (in μm) is required to achieve a light absorption of 90%?

UPDATE: Since it was very difficult to accurately read off the absorption coefficient values from the graph above, we have chosen to provide you with these values. The absorption coefficients for the different semiconductor materials at α(800nm) are:

αGaAs=2∗104cm−1
αInP=4∗104cm−1
αGe=6∗104cm−1
αSi=1∗103cm−1.

The thickness d (in μm) required to achieve a light absorption of 90% is:

1) For GaAs

2) For InP

3) For Ge

4) For Si

1.1513

0.5756
0.383764
23.026
;)

1.1513

0.5756
0.383764
23.026

I am also stuck at that question and can not solve it ... if you have the answer or at least some clue how can I solve it - please share!

users . encs. concordia. ca/~mojtaba/ ELEC425- assignment4- solutions . pdf

edit the spaces

Can you please just give out the answers.. thanx!!

please help in this q?

i am also stuck plz help

To calculate the thickness (d) required to achieve a light absorption of 90%, we need to use the absorption coefficient (α) for each semiconductor material.

The absorption coefficient represents the amount of light absorbed per unit length of the material. It is measured in units of cm^-1.

To calculate the required thickness, we can use the following formula:

d = -ln(1 - A) / α

where A is the desired light absorption (in this case, 0.9 or 90%), and ln represents the natural logarithm.

1) For GaAs:
Given α(GaAs) = 2 * 10^4 cm^-1, we can substitute the values into the formula:

d(GaAs) = -ln(1 - 0.9) / α(GaAs)
= -ln(0.1) / 2 * 10^4

Using a calculator, we find d(GaAs) ≈ 0.00162 cm or 16.2 μm

Therefore, the thickness required for GaAs is approximately 16.2 μm.

2) For InP:
Given α(InP) = 4 * 10^4 cm^-1, we can substitute the values into the formula:

d(InP) = -ln(1 - 0.9) / α(InP)
= -ln(0.1) / 4 * 10^4

Using a calculator, we find d(InP) ≈ 0.00081 cm or 8.1 μm

Therefore, the thickness required for InP is approximately 8.1 μm.

3) For Ge:
Given α(Ge) = 6 * 10^4 cm^-1, we can substitute the values into the formula:

d(Ge) = -ln(1 - 0.9) / α(Ge)
= -ln(0.1) / 6 * 10^4

Using a calculator, we find d(Ge) ≈ 0.00054 cm or 5.4 μm

Therefore, the thickness required for Ge is approximately 5.4 μm.

4) For Si:
Given α(Si) = 1 * 10^3 cm^-1, we can substitute the values into the formula:

d(Si) = -ln(1 - 0.9) / α(Si)
= -ln(0.1) / 1 * 10^3

Using a calculator, we find d(Si) ≈ 0.00921 cm or 92.1 μm

Therefore, the thickness required for Si is approximately 92.1 μm.

I think 1.51, not 1.15