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日期：2020-08-21 08:42

Experiment 6: Optical Reflection from a photovoltaic surface


0Aim: To investigate the effect of both the angle of incidence and polarization on the reflection of visible light from a commercial semi-conductor material (silicon).

Covid19: No experiments or videos are available. Only the data sets will be available.
Inside the box outside the box

Figure 1: Experimental set up for the optical reflection measurements.

THEORY:

Polarisation: A plane electromagnetic wave has both an electric field and magnetic field component perpendicular to the direction of propagation. The wave is said to be linearly polarised if the E field always lies in the same plane. If this is not the case then the wave is elliptically polarised, then the electric field has components in all directions perpendicular to the direction of propagation.

The reflection properties of a plane wave from a flat surface are different for the two polarisation cases:

1)E field in the plane of the surface (called TE polarisation), and
2)H field in the plane of the surface (called TM polarisation).

We can determine the plane of polarisation using a polarising window which allows only one plane of polarisation through. The ratio of the maximum to minimum electric field is called the ellipticity of the wave.

Fresnel’s Reflection: The reflection of plane electromagnetic radiation is described by the Fresnel equations. The reflection coefficient is simply the field strength of light reflected from the surface divided by the field strength of light falling on the surface. There are two different polarization cases. When the electric field is perpendicular to the plane of incidence, the interaction is described as the Transverse Electric (TE) case, and when the magnetic field of the incident radiation is perpendicular to the plane of incidence, the interaction is described as transverse magnetic (TM). Note that the plane of incidence is the plane containing both the incident and reflected wave vectors.

The TE reflection coefficient (also known as perpendicular polarization because the orientation of the electric field is parallel to the reflecting surface) is given by(8-58a)

where is the angle of incidence and is the angle of transmission. Both angles are measured relative to the normal to the surface. The wave impedance in air is (= 377?? refractive index = 1) and the wave impedance in the reflecting material is.

The TM reflection coefficient (also known as parallel polarization because the orientation of the electric field is perpendicular to the reflecting surface) is given by
where n is the refractive index of each medium. For TM polarization only, there is an angle (called the Brewster Angle, ), where the reflection coefficient is zero. This occurs when the angle of incidence is . This angle is given by
While this is exactly true for an insulating material (e.g. glass), when the reflecting material is a semi-conductor, the intrinsic impedance ? and refractive index n are complex numbers which are dependent on both the permittivity ? and the conductivity of the material ?.
In this experiment, we use both an LED light source and a laser diode light source, which we will assume has constant output intensity. A silicon semi-conductor detector provides a voltage output proportional to the light intensity. Note that it is necessary to orient the hardware to ensure the laser output strikes approximately the same position on the sensor surface.

The photodetector unit displays the light intensity in the units of “lux” (= lumens/m2). These units are wavelength dependent and match the sensitivity of the human eye. The conversion from lux to watts at the laser diode wavelength is

1 lux = 181 W/m2 (1)

PRE-LABORATORY QUESTIONS:

Q6.1 If the refractive index of glass is 1.8, calculate the relative permittivity, the intrinsic impedance and the Brewster angle for glass using equation 8.56.

EXPERIMENTAL PROCEDURE:

General Instructions:
This experiment is conducted in a light tight box which should be closed during measurements. Please do not touch the optical surfaces (polarisers, silicon cell, detector area). There are positioning devices to allow you to move the sample in and out of the laser beam and to rotate it (see Fig 2).


(a)Laser Polarisation
Replace the LED with the laser in the holder provided. Configure the optical rail with a 90 degree angle (reflection from the PV cell into the detector: see Figure 3) as follows:

Laser => Polarizer => Detector

Make sure the laser beam falls in the centre of the detector ball for every measurement. Note the colour of the laser in your report.

Place a polariser between the laser and the detector. With the box open and the polariser set at zero degrees, manually rotate the laser until the received signal is minimum (i.e. the red/green dot has lowest intensity). With the box shut, during each measurement, change the polariser angle in 5 degree steps through 180 degrees and note the received signal level.

The ellipticity angle of the laser is the ratio of the maximum to the minimum field strength (Ulaby p 341, equation 7.58). From your results calculate the ellipticity angle of the radiation from the laser. Is the radiation from the laser linearly polarised? Explain.


Figure 3: The optical components from left to right: laser diode, polarizer, PV cell reflection surface (housed in the white frame), polariser and optical detector. The PV cell output is measured outside the box by a voltmeter and the optical power meter.

With the polariser oriented for maximum signal note the received signal power. Remove the polariser and again note the power level. What percentage of the optical power is lost through the polariser? Interpret your answer in terms of polarisation.

(b)Polarisation and Detector Sensitivity

Using the 90 degree set up in part (a), add the second polariser in front of the detector. The laser spot should still be located in the centre of the detector.

Laser => Polarizer 1 => Polarizer 2 => Detector

Set the laser and both polarisers for maximum detected power. Rotate the second polariser in 5 steps from 0 to 90, and note the output at each position. Linearise your data by plotting the square root of the detector output versus the sine/cosine of the angle of the second polariser, and so show that the relationship is linear. Write down the equation and the linear correlation coefficient. Why does the line not pass through zero?
Again calculate the percentage loss in power caused by the introduction of the second polariser.

(c)Reflection Experiment

Laser => Polarizer 1 => PV Sample => Polarizer 2 => Detector

Volt meter

Here we are attempting to reproduce part of Figure 8.19 in Ulaby for parallel polarisation and the equivalent plot for perpendicular polarisation. We cannot obtain measurements for angles of incidence of zero and 90 degrees, but a limited range in between. We are also measuring the “transmitted” light by recording the voltage out of the photovoltaic cell. This should be similar to the transmissivity in Figure 8-19 in Ulaby.

Set the two polarisers at 90 degrees and rotate the laser for maximum signal. Move the detector to the horizontal angle position at ? =90 (see Figure 1). Move the sample into the laser beam and rotate it so that the reflected spot is located at the centre of the detector. Note that the centre of the sample must be located at the axis of the rotating arm. Close the box and note the reading on the light detector and the voltmeter.

Repeat the procedure at each angle marked, noting the angle of the detector and the detector output. You might have to adjust the height of the detector and the lateral position of the PV cell.

Rotate the laser and BOTH the first and second polarisers to 0 degrees. You should still have maximum signal received. Again rotate through the horizontal angle ? =90 and record the light level as a function of angle.

Before plotting both graphs, convert your detector angle position ? to angle of incidence on the sample and convert the lux measurement to relative E field strength. Write down your conversion equation. Plot the detector output and voltmeter output versus angle of incidence with both data sets on the same graph paper using the same scale (4 lines). Comment of the PV output as a function of the angle of incidence.

Determine the Brewster Angle for the sample, and so determine the relative permittivity of the PV material.

POST-LABORATORY QUESTIONS

Q6.3 Use Matlab to calculate the complex Brewster Angle (in degrees) when the complex refractive index is 2.7-0.1j. How does this compare with the Brewster Angle when the refractive index is real and equal to 2.7? How do you interpret a complex reflective index?

Q6.4 From your measured data calculate the loss in optical power due to the single polariser aligned for maximum power transmission. Where do you think the energy has been lost?

Q6.5 Is the laser highly polarised?

Q6.6 Convert equations 8.58a and 8.66a to an equation which involves only the relative permittivities of air and the material, and the angle of incidence ?i. You must show the steps involved.

Part (a)
Laser power without polariser: 8800 lux


Laser Polarisation data:
Angle Lux
0 12
5 35
10 88
15 168
20 271
25 407
30 553
35 712
40 902
45 1076
50 1253
55 1422
60 1578
65 1718
70 1848
75 1942
80 2010
85 2050
90 2080

Part (b)
Laser with polariser – polarisation
Angle Lux
0 476
5 470
10 458
15 440
20 415
25 382
30 347
35 309
40 268
45 225
50 186
55 147
60 111
65 78
70 50
75 25
80 10
85 2
90 0

Part (c) Data
Angle of incidence data
theta = 90 angle(phi) Lux Voltage(mV)
10 188 47.85
15 142 58.59
20 97 67.4
25 68 74.13
30 45 78.33
35 28 81.46
40 17 83.6
45 9 84.92
50 4 85.89
55 1 86.49
60 0 86.77
65 0 86.9
70 0 86.94
75 0 86.9
80 1 86.75
85 2 86.56
90 3 86.37
theta=0 angle(phi) Lux Voltage(mV)
10 361 25.24
15 332 33.14
20 279 40.32
25 253 46.13
30 221 51.3
35 185 55.57
40 167 59.12
45 143 61.98
50 121 64.62
55 99 66.75
60 86 68.43
65 77 69.89
70 60 71.17
75 54 72.23
80 50 73.09
85 45 73.93
90 40 74.63