APhO 2016, Експеримент — Задача 2. Reflection Phase Shift of Metal
Asian Physics Olympiad 2016 — Experiment Problem – E2: Reflection Phase Shift of Metal · 8 т.
Условие
Introduction
It is known that natural materials have refractive indexes () larger than that of vacuum (). However, meta-materials (artificial materials with nano-scale structures) can exhibit exotic phenomena, such as negative refraction which can be explained by considering negative effective refractive index. In general, the refractive index of a material, for instance metal, can also be a complex number when there is light absorption.
A simple method to study complex refractive index is by measuring the phase shift when a light beam is being reflected from the surface of a material. In optics, the phase shift due to reflection from a surface is important for many applications especially those in holographic measurements.
For ordinary glass, the reflection phase shift for normal incidence is (or in radian). However, can take different values for metals, depending on the absorptions of the metals. Phase measurement in optics is challenging and high precision is required. Many sophisticated techniques or methods have been developed for optical phase measurements. However, they are too complicated to be implemented for high school students. Here, we introduce a simple and inexpensive approach.
Objective
To study the phase shift of light reflected from Titanium (Ti) surface using Fabry-Perot laser interferometry.

List of apparatus
| [1] | Laser diode |
|---|---|
| [2] | Battery pack with batteries for the laser diode |
| [3] | White-color-capped screws for sample holder |
| [4] | Sample holder with two paper clips |
| [5] | Wires with banana connectors |
| [6] | Digital multi-meter |
| [7] | Sample box with one Titanium (Ti)-coated Fabry-Perot etalon with specific sample number |
| [8] | Flat-head screw driver |
| [9] | Optical platform with rotary disk, turntable detector holder, fixed angular scale, and laser diode holder |
| [10] | LED lamp |
Description of apparatus




Theory
Consider an ideal air-gap Fabry-Perot (FP) etalon as shown in Figure 10. The etalon consists of a top thick glass plate (with refractive index ) and a bottom sample plate (with refractive index ) sandwiching a thin air-gap ( micron) in-between.

We use a two-beam interference approximation to model the air-gap etalon. A light beam (beam 0) incident on the top glass plate, neglecting the reflection from the air-glass interface of the top glass, is partially reflected (beam 1) and partially transmitted (refracted) at the glass-air interface. The transmitted (refracted) beam then gets reflected at the air-sample interface and then transmitted (refracted) through the top glass plate, which is labeled as beam 2 shown in Figure 10. For beam 2, the reflection at the air-sample interface picks up an additional phase shift while there is no phase shift at other interfaces. The resulting intensity of reflected light for incident angle is the superposition of beams 1 and 2 given by:
where and are the intensities of beams 1 and 2, respectively, is the wavenumber and is the wavelength of the incident light. Equation (1) exhibits intensity peaks and troughs as a function of incident angle for fixed wavelength and air-gap spacing . In this experiment, we fix the polarization direction of the incident light and neglect polarization effects when the beam is reflected/refracted at the interfaces. Moreover, we use a Titanium (Ti)-coated glass plate as the bottom sample plate of the FP etalon.
Preparatory procedures/adjustments
- If you have chosen to do experiment E1 first, remember to remove the observation board and also the pin hole for E1 before assembling the apparatus for this experiment E2.
- CAUTION: DO NOT LOOK DIRECTLY AT THE LASER LIGHT OF THE LASER DIODE!!
- Install the laser diode into the circular hole of the laser holder as shown in Figure 11. Fasten the laser diode by using the yellow-color-capped screw. Make sure the body of the laser head is horizontal with the attached yellow label (see Figure 2 the laser diode with the yellow label) facing up (for setting the polarization of the laser light in the vertical direction). [Note: Do not over-tighten the (yellow-color-capped) screw of the laser holder; otherwise it will damage the laser diode!]
- Connect the battery pack to the laser diode as shown in Figure 11 and switch on the laser diode.
- It is desirable to have the laser diode to form a beam spot about 1 mm in diameter at a distance of 20 cm. To check the beam size, place a piece of paper at about 20 cm from the laser diode and observe the spot size of the laser beam emitted by the laser diode. Adjust the aperture of the laser diode using a flat-head screw driver to obtain a spot size about 1 mm in diameter if necessary.
- Connect the output terminals of the photo-detector pre-mounted on the detector holder to the multi-meter CAREFULLY using the electric wires as shown in Figure 12. The photo-detector is a photodiode connected in parallel with a resistor such that the current produced by the photodiode when illuminated with light can be recorded by measuring the voltage across the resistor.
- Adjust the position of the laser diode such that the laser beam shoots horizontally over the mark of the angular scale, crossing over the center of the rotary disk, and hits directly onto the detector positioned at the mark of the angular scale (as shown in Figure 13). You can turn on the multi-meter to monitor the intensity (in terms of voltage) of the laser beam to obtain an optimal condition. [Note: Remember to turn off the laser diode when you are done with this step!]
- Mount the sample holder to the rotary disk on the optical platform using the white-color-capped screws as shown in Figure 14.
- Then mount the Ti-coated etalon with the sample number (here #250) upright to the sample holder using the two paper clips provided such that the bottom glass plate of the sample is in contact with the wall surface of the holder as shown in Figure 15.
- Turn on the laser diode and adjust the vertical position of the etalon so that the laser beam shoots at the middle area of the top or the bottom Ti strip of the etalon and gets reflected to the photo-detector as shown in Figure 15. Then adjust the position of the photo-detector for maximum signal (in terms of the voltage measured by the multi-meter). Furthermore, you have to adjust the position of the sample holder using the two white-color-capped screws such that the Ti strip rotates with the same axis as the rotary disk, i.e. they are coaxial.
- Now rotate the sample holder together with the mounted Ti-coated etalon such that the laser beam makes an incident angle with respect to the normal of the sample surface as indicated by the angular scale. Then move the photo-detector to an angle of about and adjust the angular position of the photo-detector for obtaining a maximum signal from the multi-meter. After completing all the above steps/procedures, the setup is now ready for the experiment.





Note: All numerical values and answers should have three significant figures.
The Experiment




1 Measure the reflection intensity of the Ti-coated etalon in terms of voltage as a function of incident angle at fine enough angular steps, starting at such that as many intensity peaks as possible are recorded. Enter your data in Table E2_1. [1 т.]
2 In order to reduce errors resulted from the mis-alignment of the laser beam with respect to the sample holder, and also with respect to the angular scale of the optical platform, carry out similar measurements for incident angles on the other side of the mark of the angular scale by first turning the sample to face the opposite side of the mark of the angular scale and then moving the photo-detector accordingly for measuring intensity of the reflected laser beam. Enter your data in Table E2_2. [1 т.]
3 Plot the reflection intensity as a function of incident angle on a graph paper using the recorded data in Table E2_1 and label the graph as ‘Graph E2_1’. Plot a similar graph for the data in Table E2_2 on a different graph paper and label the graph as ‘Graph E2_2’. Show the variation of the reflection intensity by drawing a smooth curve fitting the data points in Graph E2_1 as well as in Graph E2_2. [0,9 т.]
4 Identify the peaks of the reflection intensity and label them with sequential peak numbers in Graphs E2_1 and E2_2, starting with the peak having the largest incident angle. [0,2 т.]
5 Derive the condition for the reflection intensity peaks in terms of , , , using Equation (1). Label this condition as Equation (2). [0,3 т.]
6 Choose a suitable function of , , as a new independent variable such that the reflection intensity peaks in Graphs E2_1 and E2_2 will be evenly spaced in a graph of reflection intensity vs . Then calculate the corresponding values of and insert them, as a new column, to Tables E2_1 and E2_2. [0,4 т.]
7 Construct Table E2_3 consisting of the peak number and the corresponding incident angle for the data in Tables E2_1 and E2_2. You may label them with the notations of LHS (left-hand side) and RHS (right-hand side), respectively, i.e. and . Then, for each matched reflection intensity peak, calculate the average values of and , and label the average as and and enter the results as new columns in Table E2_3.
[Note: The peak numbers in Tables E2_1 and E2_2 can be different. Make sure you match the corresponding peaks obtained from Graphs E2_1 and E2_2.] [0,6 т.]
8 Plot a graph on a separate graph paper to show the relationship between the peak number vs. . Label this graph as ‘Graph E2_3’. [0,6 т.]
9 Draw a straight line through all the data points in Graph E2_3 and then determine the slope and -intercept of this line. [Note: A graphical approach will also be accepted and error analysis is not required.] [0,4 т.]
10 Rewrite Equation (2) derived in Task 5 to give an expression of the interference order and the function obtained in Task 6. Label this equation as Equation (3a). From Equation (3a), write down the expression of in terms of , , and as Equation (3b). Hence, express the normalized reflection phase in terms of , , and as Equation (3c). Write down the range of . Then determine the values of for the peaks from the results obtained in Task 7 and insert them as a new column in Table E2_3. [1,2 т.]
11 Plot vs. on Graph E2_3, i.e. the same graph for the plot of peak number vs. . Draw a line through all data points for this plot. Hence determine the air-gap spacing of the etalon and also the reflection phase of the Ti. [Note: Again, graphical solution is acceptable and error analysis is not required.] [1,4 т.]
Отговори
9 Slope 12.0, y-intercept -5.45 (peak number vs. cos θaverage)
11 L = 3.903 μm; ϕs,n = -0.450, i.e. ϕs = -2.828 rad
Official sample values for etalon #15; the reflection phase is taken within (-2π, 0).
Решение
1
| (Deg) | Voltage (mV) | |
|---|---|---|
| 60.0 | 432 | 0.500 |
| 59.5 | 426 | 0.508 |
| 59.0 | 427 | 0.515 |
| 58.5 | 431 | 0.522 |
| 58.0 | 435 | 0.530 |
| 57.5 | 438 | 0.537 |
| 57.0 | 440 | 0.545 |
| 56.5 | 440 | 0.552 |
| 56.0 | 440 | 0.559 |
| 55.5 | 438 | 0.566 |
| 55.0 | 434 | 0.574 |
| 54.5 | 430 | 0.581 |
| 54.0 | 425 | 0.588 |
| 53.5 | 423 | 0.595 |
| 53.0 | 426 | 0.602 |
| 52.5 | 431 | 0.609 |
| 52.0 | 435 | 0.616 |
| 51.5 | 437 | 0.623 |
| 51.0 | 439 | 0.629 |
| 50.5 | 439 | 0.636 |
| 50.0 | 438 | 0.643 |
| 49.5 | 437 | 0.649 |
| 49.0 | 431 | 0.656 |
| 48.5 | 428 | 0.663 |
| 48.0 | 424 | 0.669 |
| 47.5 | 421 | 0.676 |
| 47.0 | 422 | 0.682 |
| 46.5 | 425 | 0.688 |
| 46.0 | 430 | 0.695 |
| 45.5 | 433 | 0.701 |
| 45.0 | 435 | 0.707 |
| 44.5 | 436 | 0.713 |
| 44.0 | 437 | 0.719 |
| 43.5 | 436 | 0.725 |
| 43.0 | 434 | 0.731 |
| 42.5 | 431 | 0.737 |
| 42.0 | 425 | 0.743 |
| 41.5 | 424 | 0.749 |
| 41.0 | 420 | 0.755 |
| 40.5 | 419 | 0.760 |
| 40.0 | 420 | 0.766 |
| 39.5 | 424 | 0.772 |
| 39.0 | 428 | 0.777 |
| 38.5 | 431 | 0.783 |
| 38.0 | 433 | 0.788 |
| 37.5 | 434 | 0.793 |
| 37.0 | 434 | 0.799 |
| 36.5 | 434 | 0.804 |
| 36.0 | 433 | 0.809 |
| 35.5 | 431 | 0.814 |
| 35.0 | 429 | 0.819 |
| 34.5 | 426 | 0.824 |
| 34.0 | 422 | 0.829 |
| 33.5 | 419 | 0.834 |
| 33.0 | 417 | 0.839 |
| 32.5 | 417 | 0.843 |
| 32.0 | 419 | 0.848 |
| 31.5 | 421 | 0.853 |
| 31.0 | 424 | 0.857 |
| 30.5 | 427 | 0.862 |
| 30.0 | 429 | 0.866 |
| 29.5 | 430 | 0.870 |
| 29.0 | 431 | 0.875 |
| 28.5 | 432 | 0.879 |
| 28.0 | 431 | 0.883 |
| 27.5 | 431 | 0.887 |
| 27.0 | 430 | 0.891 |
| 26.5 | 429 | 0.895 |
| 26.0 | 427 | 0.899 |
| 25.5 | 425 | 0.903 |
| 25.0 | 421 | 0.906 |
| 24.5 | 419 | 0.910 |
| 24.0 | 416 | 0.914 |
| 23.5 | 414 | 0.917 |
| 23.0 | 414 | 0.921 |
| 22.5 | 414 | 0.924 |
| 22.0 | 415 | 0.927 |
| 21.5 | 416 | 0.930 |
| 21.0 | 419 | 0.934 |
| 20.5 | 421 | 0.937 |
| 20.0 | 423 | 0.940 |
| 19.5 | 424 | 0.943 |
| 19.0 | 426 | 0.946 |
| 18.5 | 427 | 0.948 |
| 18.0 | 427 | 0.951 |
| 17.5 | 428 | 0.954 |
| 17.0 | 428 | 0.956 |
| 16.5 | 429 | 0.959 |
| 16.0 | 428 | 0.961 |
| 15.5 | 428 | 0.964 |
| 15.0 | 428 | 0.966 |
| 14.5 | 427 | 0.968 |
| 14.0 | 427 | 0.970 |
| 13.5 | 426 | 0.972 |
| 13.0 | 425 | 0.974 |
| 12.5 | 423 | 0.976 |
| 12.0 | 423 | 0.978 |
| 11.5 | 422 | 0.980 |
| 11.0 | 420 | 0.982 |
| 10.5 | 419 | 0.983 |
| 10.0 | 418 | 0.985 |
| 9.5 | 417 | 0.986 |
| 9.0 | 416 | 0.988 |
| 8.5 | 416 | 0.989 |
| 8.0 | 414 | 0.990 |
| 7.5 | 413 | 0.991 |
| 7.0 | 413 | 0.993 |
| 6.5 | 413 | 0.994 |
| 6.0 | 412 | 0.995 |
| 5.5 | 411 | 0.995 |
| 5.0 | 410 | 0.996 |
| 4.5 | - | - |
| 4.0 | - | - |
Reflection intensity against incident angle taken from to in intervals on one (LHS) side of the angular scale. The independent variable should be added to Table E2_1. (The Ti-coated etalon used is #15.)
Marks: TOTAL = 1.0 points
0.6 points for two columns filled with values of and the voltage
0.2 points for including units
0.2 points for displaying the measurements with the correct number of significant figures
[-0.1 points for each unit missing
-0.1 points for displaying the incorrect number of significant figures for two or more variables
-0.1 points for angular step ]
2
| (Deg) | Voltage (mV) | |
|---|---|---|
| -60.0 | 426 | 0.500 |
| -59.5 | 432 | 0.508 |
| -59.0 | 437 | 0.515 |
| -58.5 | 439 | 0.522 |
| -58.0 | 441 | 0.530 |
| -57.5 | 441 | 0.537 |
| -57.0 | 440 | 0.545 |
| -56.5 | 438 | 0.552 |
| -56.0 | 434 | 0.559 |
| -55.5 | 426 | 0.566 |
| -55.0 | 421 | 0.574 |
| -54.5 | 421 | 0.581 |
| -54.0 | 427 | 0.588 |
| -53.5 | 432 | 0.595 |
| -53.0 | 435 | 0.602 |
| -52.5 | 438 | 0.609 |
| -52.0 | 439 | 0.616 |
| -51.5 | 439 | 0.623 |
| -51.0 | 437 | 0.629 |
| -50.5 | 434 | 0.636 |
| -50.0 | 429 | 0.643 |
| -49.5 | 422 | 0.649 |
| -49.0 | 418 | 0.656 |
| -48.5 | 420 | 0.663 |
| -48.0 | 425 | 0.669 |
| -47.5 | 431 | 0.676 |
| -47.0 | 434 | 0.682 |
| -46.5 | 436 | 0.688 |
| -46.0 | 436 | 0.695 |
| -45.5 | 436 | 0.701 |
| -45.0 | 436 | 0.707 |
| -44.5 | 433 | 0.713 |
| -44.0 | 431 | 0.719 |
| -43.5 | 426 | 0.725 |
| -43.0 | 421 | 0.731 |
| -42.5 | 413 | 0.737 |
| -42.0 | 417 | 0.743 |
| -41.5 | 420 | 0.749 |
| -41.0 | 423 | 0.755 |
| -40.5 | 428 | 0.760 |
| -40.0 | 431 | 0.766 |
| -39.5 | 431 | 0.772 |
| -39.0 | 433 | 0.777 |
| -38.5 | 432 | 0.783 |
| -38.0 | 433 | 0.788 |
| -37.5 | 431 | 0.793 |
| -37.0 | 422 | 0.799 |
| -36.5 | 425 | 0.804 |
| -36.0 | 417 | 0.809 |
| -35.5 | 416 | 0.814 |
| -35.0 | 414 | 0.819 |
| -34.5 | 414 | 0.824 |
| -34.0 | 419 | 0.829 |
| -33.5 | 421 | 0.834 |
| -33.0 | 425 | 0.839 |
| -32.5 | 427 | 0.843 |
| -32.0 | 428 | 0.848 |
| -31.5 | 430 | 0.853 |
| -31.0 | 431 | 0.857 |
| -30.5 | 432 | 0.862 |
| -30.0 | 431 | 0.866 |
| -29.5 | 430 | 0.870 |
| -29.0 | 428 | 0.875 |
| -28.5 | 427 | 0.879 |
| -28.0 | 424 | 0.883 |
| -27.5 | 421 | 0.887 |
| -27.0 | 418 | 0.891 |
| -26.5 | 415 | 0.895 |
| -26.0 | 413 | 0.899 |
| -25.5 | 413 | 0.903 |
| -25.0 | 413 | 0.906 |
| -24.5 | 416 | 0.910 |
| -24.0 | 418 | 0.914 |
| -23.5 | 420 | 0.917 |
| -23.0 | 422 | 0.921 |
| -22.5 | 424 | 0.924 |
| -22.0 | 426 | 0.927 |
| -21.5 | 427 | 0.930 |
| -21.0 | 428 | 0.934 |
| -20.5 | 428 | 0.937 |
| -20.0 | 429 | 0.940 |
| -19.5 | 429 | 0.943 |
| -19.0 | 428 | 0.946 |
| -18.5 | 429 | 0.948 |
| -18.0 | 428 | 0.951 |
| -17.5 | 427 | 0.954 |
| -17.0 | 426 | 0.956 |
| -16.5 | 425 | 0.959 |
| -16.0 | 423 | 0.961 |
| -15.5 | 422 | 0.964 |
| -15.0 | 419 | 0.966 |
| -14.5 | 418 | 0.968 |
| -14.0 | 416 | 0.970 |
| -13.5 | 414 | 0.972 |
| -13.0 | 414 | 0.974 |
| -12.5 | 413 | 0.976 |
| -12.0 | 412 | 0.978 |
| -11.5 | 411 | 0.980 |
| -11.0 | 411 | 0.982 |
| -10.5 | 411 | 0.983 |
| -10.0 | 411 | 0.985 |
| -9.5 | 412 | 0.986 |
| -9.0 | 412 | 0.988 |
| -8.5 | 413 | 0.989 |
| -8.0 | 414 | 0.990 |
| -7.5 | 414 | 0.991 |
| -7.0 | 415 | 0.993 |
| -6.5 | 416 | 0.994 |
| -6.0 | 416 | 0.995 |
| -5.5 | 417 | 0.995 |
| -5.0 | 418 | 0.996 |
| -4.5 | 426 | 0.997 |
| -4.0 | 425 | 0.998 |
Reflection intensity against incident angle taken from to in intervals on the (RHS) other side of the angular scale. The independent variable should be added to Table E2_2. (The Ti-coated etalon used is #15.)
Marks: TOTAL = 1.0 points
0.6 points for two columns filled with values of and the voltage
0.2 points for including units
0.2 points for displaying the measurements with the correct number of significant figures
[-0.1 points for each unit missing
-0.1 points for displaying the incorrect number of significant figures for two or more variables
-0.1 points for angular step ]
3

Graphs E2_1 and E2_2 show the relationship between the intensity and for the positive (LHS) and negative (RHS) incident angles respectively. The peak numbers are also labeled for all graphs.
Marks: TOTAL = 0.9 points
0.4 points for displaying the data points (for both graphs)
0.1 points for displaying the units (for both graphs)
0.1 points for displaying the axis label (for both graphs)
0.1 points for displaying the axis ticks label (for both graphs)
0.2 points for smooth curve (for both graphs)
[deduct half of the points if the above items are not shown in both graphs]
4
Refer to Graphs E2_1 and E2_2.
Marks: TOTAL = 0.2 points
0.2 points for labeling the peaks with appropriate peak numbers (for both graphs)
[-0.1 points for wrong peak number order]
5
From Equation (1), the peaks correspond to constructive interference where the total phase difference of the two beams is equal to multiples of , i.e.
where = 1, 2, etc… is the interference order and is the corresponding incident angle for peak reflection intensity.
Thus a plot of the interference order vs. will give a straight line with a slope related to the air-gap spacing and an intercept related to the reflection phase . Since the interference order changes sequentially with the peak number, a plot of the peak number vs. will also give a straight line with the same slope as the plot for the interference order vs. . However, the y-intercept is now shifted along the -axis with respect to the plot for the interference order vs. .
Thus should be chosen as the independent variable such that the intensity peaks will be evenly spaced in a plot of reflection intensity vs. . Furthermore, a plot of peak number vs. will give a straight line that can be used to obtain the air-gap spacing of the Ti-coated etalon and also the reflection phase of the Ti.
Marks: TOTAL = 0.3 points
0.2 points for deriving the correct equation (i.e. Eq.(2))
0.1 points for equals to integer number
6
Independent variable
Refer to Tables E2_1 and E2_2
Marks: TOTAL = 0.4 points
0.2 points for correct independent
0.2 points for working out the numbers of in Tables E2_1 and E2_2
7
Ideally, the locations of the peaks for (RHS) negative incident angles should be the same as for (LHS) positive incident angles for perfect optical alignment. Since there could be an offset in the locations of the peaks due to mis-alignment in the optics, the peak locations are better determined by averaging the negative and positive incident angles. After pairing, one would get the following table:
| Peak number (LHS) | (Deg) | Peak number (RHS) | (Deg) | (Deg) | ||
|---|---|---|---|---|---|---|
| 6 | 16.50 | 6 | -19.25 | 17.875 | 0.952 | 11 |
| 5 | 28.50 | 5 | -30.75 | 29.625 | 0.869 | 10 |
| 4 | 36.50 | 4 | -38.50 | 37.500 | 0.793 | 9 |
| 3 | 44.00 | 3 | -45.75 | 44.875 | 0.709 | 8 |
| 2 | 51.25 | 2 | -51.75 | 51.500 | 0.623 | 7 |
| 1 | 56.50 | 1 | -57.75 | 57.125 | 0.543 | 6 |
refers to the peak location obtained from Table E2_1.
refers to the peak location obtained from Table E2_2.
is the average of and .
Marks: TOTAL = 0.6 points

0.2 points for identifying the peaks and the corresponding LHS/RHS incident angles
0.2 points for matching of peaks
0.2 points for calculating the average of the independent variable
[-0.1 points for each unit missing
-0.1 points for displaying the incorrect number of significant figures except for peak and interference numbers
-0.1 points for mis-matching peak number]
8
Peak number vs.
The slope is 12.0
The -intercept is -5.45
Interference order vs.
The slope is 12.0
The -intercept is -0.45
By plotting peak number against and drawing a line through the data points, one could get the slope and the -intercept, as shown in Graph E2_3. The same principle applies to plotting the interference order against .
(Graphical solutions for the slope and intercept will be accepted.)
Note: Plotting these two graphs separately will also be acceptable as shown in Graph E2_3a and Graph E2_3b i.e.


Marks: TOTAL = 0.6 points
0.3 points for displaying the data points
0.1 points for displaying the units
0.1 points for displaying the axis label
0.1 points for displaying the axis ticks label
[-0.2 points for plotting vs. peak number or interference order (i.e. - axis are reserved)]
9
Refer to Graph E2_3
Peak number vs.
The slope is 12.0
The -intercept is -5.45
Marks: TOTAL = 0.4 points
0.2 points for fitting a straight line
0.1 points for deriving the value of the slope
0.1 points for deriving the value of the -intercept
10
Equation (2) in Tasks 5+6 can be rewritten in a simpler form for each constructive interference order at incident angle as below:
In Equation (3a), one can take the integer part of as the interference order, i.e.
Then the decimal part of is related to the reflection phase by
A simpler method to find the interference order is to add to the peak number directly the absolute value of the integer part of the -intercept obtained from the plot of peak number vs. as shown in Graph E2_3, i.e. = peak number + |integer part of -intercept in Graph E2_3.
Then, the -intercept of a plot of vs. will give the normalized reflection phase directly.
is now defined within (-1, 0), corresponding to chosen with (, 0).
Refer to Table E2_3 for .
Marks: TOTAL = 1.2 points
0.3 points for deriving the correct equation (i.e. Eq. (3a))
0.3 points for getting the expression for (i.e. Eq. (3b))
0.2 points for getting the expression for normalized reflection phase (i.e. Eq. (3c))
0.2 points for defining the range for the reflection phase
0.2 points for interference order added to Table E2_3.
11
Refer to Graph E2_3:
From Graph E2_3, the slope for vs, is 12.01.
Since slope of the line is equal to , one can then obtain the value of as
The -intercept is the normalized reflection phase
Marks: Total = 1.4 points
0.4 points for displaying the data points
0.3 points for fitting a straight line
0.4 points for deriving the value of the air-gap spacing
0.3 points for deriving the value of the -intercept
[-0.1 points for outside (-0.9, -0.1)]
Appendix:
Path difference calculation for an ideal air-gap etalon:

Path difference for beams 1 and 2 is equal to:
This is the path difference used in Equation (1).
It is also acceptable to calculate the path difference directly using Equation (1), but will only be given half of the points as writing down Eq. (4a).
A mis-alignment of angle between the laser beam and the angular scale corresponds to a correction of for the incident angle taken directly from the angular scale. Thus the incident angle is now . Hence the corrected path difference between beams 1 and 2 with a mis-alignment of angle is:
Thus the correction of the path difference for a mis-alignment of angle as compared to the ideal case of perfect alignment is:
Hence, the error for the reflection phase due to mis-alignment is , or when normalized by .
Now one can use , , and to get the error for the reflection phase due to a mis-alignment by using Equation (4c):
The value of by using Equation (4c) is:
Now, the phase error is is , or

Оригинал в Архива: exp2.pdf · официални решения: exp2_sol.pdf