Chapter8 Experiment6: IntensityDistributions...

Chapter 8

Experiment 6: Intensity Distributionsof Diffraction Patterns

WARNINGThis experiment will employ Class III(a) lasers as a coherent, monochromatic lightsource. The student must read and understand the laser safety instructions onpage 91 before attending this week’s laboratory.

8.1 Introduction

In the electricity and magnetism laboratory, we observed the resonance of a RLC circuit.This is the frequency distribution of the circuit’s current response to an AC voltage stimulus.The Normal Distribution used in Statistics relates the probability of an observation to thevalue observed. Distributions are quite common in nature so it is fitting that we shouldstudy them.

In this lab you will observe and study, in one case, the intensity distribution of lightthat passes through a single slit and, in another case, a pair of identical slits; in bothobservations we will monitor the intensity as the angle between the original beam directionand the location of the observer is varied. In two previous laboratory exercises you haveobserved these distributions using your eyes as detectors. In this laboratory you will utilizea silicon PIN diode photodetector to get numerical values of intensity that can be plottedand analyzed graphically. Along the way, you will

1) learn about photodiode light intensity detectors,2) learn to use rotary optical encoders to monitor position,3) measure the intensity distribution of light diffracted by a single slit,4) measure the intensity distribution of light diffracted by double slits, and

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CHAPTER 8: EXPERIMENT 6

-7 -5 -3 -1 1 3 5 70

0.2

0.4

0.6

0.8

1

Intensity vs. Position

y (cm)

Figure 8.1: A graphical plot of one example intensity distribution for light transmittedthrough and diffracted by a single slit.

5) evaluate the theory of light as waves using this data.You should find that these quantitative observations have maxima and minima in preciselythe same places that we observed in the previous experiments.

The detailed theory of light transmitted through single slits is described in the appendixon page 106 and the double-slits’ theory are presented on page 108. You will be quizzed onthe material in that appendix as well as on the material in this chapter.

8.1.1 Intensity Distribution of Light Diffracted by Single Slits

We have learned in the fifth laboratory that the intensity distribution, I(θ), generated bya narrow slit of width, a, and infinite length, when illuminated with coherent light of wave

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length λ is given byI(θ)I(0) =

(sinαα

)2for α = πa

λsin θ, (8.1)

θ is the angle along which direction the intensity is observed, and I(0) is the intensity of theprincipal maximum. Figure 8.1 shows the expected ratio as calculated using Equation (8.1).Minima occur when the numerator is zero and the denominator is not; sin2 α = 0 or

α = ±mπ and m = 1, 2, 3, . . . (8.2)

Intensity maximum positions are given for values of α at which the derivative of Equa-tion (8.1) becomes zero. This leads to the equation tanα = α. An approximate solutionis

α = 0 and α = ±(m+ 1

2

)π, m = 1, 2, 3, . . . (8.3)

The ratio of intensity of a secondary maximum (order m) I(θ) to the principal maximumI(0) is

I(θ)I(0) ≈

sin2[(m+ 1

2

)π]

[(m+ 1

2

)π]2 = 1[(

m+ 12

)π]2 (8.4)

since sinα = ±1 at the approximate maxima.In the first part of this lab you will compare the intensity ratios calculated with Equa-

tion (8.4) to the measured ratio. Additionally, you will see how well Equation (8.1) describesthe intensity distribution details by graphing the measured intensity vs. the position of thedetector and by fitting the results to the Equation (8.1) model.

We will finish the first part by verifying that the slit width predicted by the graphicalanalysis is the same as that specified by the manufacturer.

8.1.2 Intensity Distribution of LightTransmitted Through Double-Slits

Historical AsideIn 1801 Thomas Young devised a classic experiment for demonstrating the wave natureof light similar to the one diagrammed in Figure 6.4. By passing a wide plane wavethrough two slits he effectively created two separate sources that could interfere witheach other. (A plane wave is a wave consisting of parallel wave-fronts all traveling inthe same direction.)

For a given direction, θ, the intensity observed at point P on the screen will be a

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x

Ry

Figure 8.2: A schematic diagram of our experiment showing how we can determine thescattering angle, θ, by measuring the distance from the optical axis, y, and the distancebetween the slits and the screen, x.

superposition of the light originating from every slit. Whether the two waves add or cancelat a particular point on the screen depends upon the relative phase between the waves atthat point. (The phase of a periodic function is ‘the stage it is at’ along its periodic cycle;the argument of the sine or cosine function is its phase.) The relative phase of the twowaves at the screen gradually increases as a function of direction causing periodic intensityminima and maxima called fringes (see Figure 6.2). The spacing of the fringes depends onthe distance between the two slits and the wavelength of the light.

Equation (8.1) gives the intensity versus direction for each of the two slits; however,interference occurs by the superposition of the two electric fields

E = E0sinαα

[sin(ωt− kx1) + sin(ωt− kx2)] (8.5)

where the two amplitudes are equal for a uniform plane wave incident upon two equal widthslits. As the observer moves around the mask with the two slits, x1 and x2 will vary. Welet the width of the two slits be a and the distance between the centers be d so that thenotation is consistent with previous developments. We place the origin of coordinates midwaybetween the slits so that x1 = x− d

2 sin θ and x2 = x+ d2 sin θ. We substitute this above into

Equation (8.5) and use a trigonometric identity sinA+ sinB = 2 sin 12(A+B) cos 1

2(A−B)to find

E = E0sinαα

[sin

(ωt− kx+ 1

2kd sin θ)

+ sin(ωt− kx− 1

2kd sin θ)]

= E0sinαα

[2 sin 1

2 (2ωt− 2kx) cos 12 (kd sin θ)

]= 2E0

sinαα

cos 12 (kd sin θ) sin (ωt− kx)

If we square this and average over a long time, the result becomes the intensity

I(θ) = I(0)(sinα

αcos β

)2= I(0)

(sinαα

)2cos2 β (8.6)

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where12kd sin θ = 1

2 · d ·2πλ

sin θ = πd

λsin θ = β. (8.7)

Figure 8.3: Graphical plots of light intensity for in-terference, diffraction, and the combined effects for lighttransmitted through matched double slits.

The average of sin2 ϕ is 12 and

the constants out front getabsorbed into I(0). From theresult in Equation (8.6) we seethat it is the simple product ofthe single slit result in Equa-tion (8.1) and the very nar-row double-slit result in Equa-tion (6.1). This combination isillustrated graphically in Fig-ure 8.3.

You will observe this in-tensity distribution in Sec-tion 8.2.2. We will take datausing the PIN photodetectorand treat the intensity versusdetector position graphically.We will see if the model ofEquation (8.6) fits the dataadequately and whether it pre-dicts the same a and d thatthe mask’s manufacturer spec-ified.

If the model satisfies theseaspects of your data, you will

have very strong evidence that Huygen’s principle that was used in each individual slit toachieve the observed diffraction is valid and that the superposition principle for electric andmagnetic fields is valid.

A similar analysis for N equally spaced identical slits yields

I(θ) = I(0)(sinα

α

)2 ( sinNβN sin β

)2

. (8.8)

This is very useful for predicting the detailed interaction of light with diffraction gratings.

8.2 The Apparatus

We will be using Pasco’s optical bench shown in Figure 8.4. Pasco’s laser diode modulefits their optical bench and will supply the coherent light for our experiment. The light

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will illuminate Pasco’s slit set shown in Figure 6.5 and Figure 7.3 at normal incidence.The resulting interference and diffraction patterns will then be detected by Pasco’s PINphotodetector. The photodetector can be moved across the optical axis and its position canbe monitored by their rotary encoder and rack-and-piñon gear.

Pasco’s rotary encoder, CI-6538, reports the angle of its shaft and pulley rotation inradians (rad). We would like to determine the detector’s perpendicular location, y, in meters(m). After examining the rack-and-piñon gears carefully, we have determined that the piñongear has 24 teeth. The rack is threaded at 3 teeth/cm. As the encoder makes one revolution,the computer will report 2π rad and the carriage will move the distance of 24 teeth. We canthen find the distance, y, that the carriage will move to be

y(ϕ) =24 teeth2π rad3 teeth0.01 m

× ϕ = 0.04mπ rad ϕ. (8.9)

Additionally, both the slit width α and the separation β contain z = πλ

sin θ ≈ πyλx

and wecan choose to separate our slits from our screen by 1.000 m; so we can let Ga3 calculate acolumn for z ≈ πy

λx= (0.04 m)πϕ

(1.000 m)πλ = 0.04 ϕ670×10−6 mm using

0.04 * “Phi” / (670 * 10^-6)

(You need to define “z” somewhere in your Procedures or Data so your reader will know themeaning of your x-axis: z = π

λsin θ.) The units of z are “1/mm”; we will let this be our

horizontal axis and the light intensity be our vertical axis when we plot our data.

Figure 8.4: A photograph of the apparatus used to measure the intensity distribution ofour light patterns. The x and y axes are indicated as are the diode laser light source, theslit pattern wheel, and the rotary encoder used to move the detector along the y-axis and tomeasure its motion.

8.2.1 Light Intensity Distribution for Single Slit

Configure Pasco’s Capstone program to gather your data. A suitable setup file can bedownloaded from the lab’s website at

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http://groups.physics.northwestern.edu/lab/intensity.html

Measure the intensity distribution of the diffraction pattern generated by the narrowestvertical slit on the detector shape wheel. Set your sample slits ‘exactly’ 1.000m from yourdetector shape wheel. Turn on your laser diode and verify that the reflection from thesample pattern has the same horizontal (y-direction) position as the laser aperture. Thisassures that the laser and slit pattern are very nearly perpendicular. It might be necessaryto fold up some scrap paper and to wedge it between the pattern wheel holder and theoptical bench track to obtain correct performance. Move the detector along the rack to oneside of the diffraction pattern where the pattern is very dim but visible in low light. Click“Monitor” at the bottom left and very, VERY slowly turn the rotary encoder wheel atconstant speed until the detector passes through the diffraction pattern. As you turn thewheel, the computer should plot your diffraction pattern on the graph. Click “Stop” at thebottom left. If the distribution is asymmetric, try a different single slit, remove or blocklight sources that illuminate only one side of the optical bench, and/or ask your TA to checkand to correct the apparatus’ alignment. If the peak intensity is less than 10%, increase thedetector’s gain and try again. If necessary, use a wider slit to get the central peak heightabove 10%.

Activate the Smart Tool ( ) on the data graph and measure the height of the centralpeak, the heights of 3-4 other peaks, and the lowest ‘background’ intensity. Subtract thebackground from all of the peak heights and divide the corrected peak heights by thecorrected central peak height. Now use Equation (8.4) to predict these ratios for comparison.Does it seem likely from this that our distribution function might be correct?

Copy and paste your data from Capstone’s table into Vernier Software’s Ga3 graphicalanalysis program. A suitable configuration file for Ga3 is also available for download fromthe website. Plot the intensity on the y-axis and “z” on the x-axis.

Try to fit your data to the theoretical model in Equation (8.1). It would be wise to saveyour data. Draw a box around your data points, Analyze/Curve Fit..., and select the “singleslit” model from the bottom of the list. Be sure the model agrees with

bg + I0 * (sin (a * (x-x0)) / (a * (x-x0)))^2

It is necessary to fit manually until the model is quite close to the data points. “bg” is thevertical offset due to background room light and detector leakage due to thermal energy, “I0”is the height of the central maximum, “x0” is the location of the central maximum on thex-axis, and “a” is the width a of the single slit. Once the fit is fairly close, enable ‘automaticfit’ and ‘try fit’ to allow the computer to tune the parameters better. It might be convenientto ‘OK’ after the manual fit so that your manual parameters are not lost to the auto-fitter.Once the fit is optimized, ‘OK’ to exit the fitter.

Unfortunately, Ga3’s fitter does not estimate the uncertainty in ’a’ correctly. Since we stillneed an uncertainty estimate, we will generate one manually. Analyze/Curve Fit. . . again andenter the parameters from the fit above. To the right of the ’a’ parameter are three controls:increment/decrement immediately adjacent and change amount (∆) slightly farther. Click

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∆ and set the amount to ’0.001’mm. Now click the increment control until the model curvepeaks and valleys no longer align with the data. Write down this upper limit for ’aul’. Nowclick the decrement control until ’a’ passes the best fit value and the peaks and valleys onceagain no longer align with your data. Write down this lower limit ’all’. Your best estimatefor a’s uncertainty is likely about

δa ≈ aul − all

6 . (8.10)

8.2.2 Light Intensity Distribution for Double Slits

Now, observe the various intensity distributions for the available double-slit pattern masks.Choose one for detailed analysis and repeat the procedure in Section 8.2.1 above except nowthe fit function must describe the distribution for the “double-slit” sample pattern,

bg + I0 * (sin (a * (x-a0)) * cos (d * (x-d0)) / (a * (x-a0)))^2

In this case “a0” is the location of the envelope’s central maximum, “d0” is the horizontaloffset of the narrow fringes, and “d” is the distance d between the slits. The remainingparameters are the same as for the single slit model. Begin with d0=a0 at the distribution’scenter. Adjust d0 as little as possible from this value to align the fast oscillations with yourdata. (You can use the ∆ and increment/decrement controls to adjust d0 if you prefer.)

Ga3’s fitter especially does not like to fit the double-slit data automatically. It might bebest simply to enter Pasco’s specified slit separation, d, and slit width, a, into the fitter andthen to observe how well the model fits the data. Click ’OK’ to preserve these parameters onyour graph. As an option you might then Analyze/Curve Fit. . . again enter the parametersagain and adjust a and d to improve the alignment between the envelope peaks and valleysand between the fringe peaks and valleys, respectively. Once the fit is VERY good you mightturn ‘Automatic’ back on and ’Try Fit’; but you might want to note the new parametersfirst. . .

You also need to obtain reasonable uncertainties δa and δd.

8.3 Analysis

Our consideration of light waves interfering and diffracting led us to Equation (8.1) andEquation (8.6). Try to imagine inventing these models without using wave interference;how well do you imagine your model would fit your data? Keep your answers to thesequestions in mind when you are deciding what your conclusions should be later. Beforewe do that, however, our beautiful model spat out slit dimensions that we can compareto measurements from the manufacturer obtained via other means. . . probably using amicroscope and micrometer scale. So a natural question is whether these two measurementsyield the same values for these respective dimensions. Surely by now we are all gettingfamiliar with the statistics needed to determine the answers to these questions.

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Calculate the difference between the manufacturer specified single-slit width and themeasurement delivered by your model fitting

∆ = Fitted - Specified.

Calculate the error in this difference using

σ =√

(δFitted)2 + (δSpecified)2.

For each, is ∆ < σ? Is ∆ < 2σ? Is ∆ > 3σ? Recall that ∆ > σ about 1:3 and ∆ > 2σabout 1:22 just from random fluctuations; these probabilities are too large for us to assertthat the compared two numbers are not equal. But ∆ > 3σ only 1:370 due only to randomfluctuations, so differences this large suggest that we conclude that the two numbers areindeed different; statistics cannot tell us why, however. Repeat this quantitative analysis forthe single slit width, a1, the double-slit width, a2, and the double-slit separation, d2. Alsoanalyze the measured peak height ratios with those predicted by Equation (8.4).

Can you think of anything in the experiments that might reasonably affect the measure-ments that we have not already included in our expected measurement errors? Discuss eachof these using complete sentences and paragraphs. In your view is it more likely that thetheory that has survived for a hundred years of professional tests is false or that you havemade some mistake or overlooked some significant sources of error that might explain yournonzero difference(s)?

8.4 Conclusions

We are now ready to state simply and briefly what our data says about light. We havealready detailed the reasons why and should not repeat those reasons here; Conclusionsare merely a series of results that our data supports or contradicts clearly stated. Alwayscommunicate with complete sentences to introduce valid equations (names like Newton’ssecond law or Equation (2) works best). We must state our conclusions clearly so that ourreader can understand them completely without reading anything else in our report.

Finally, report any measured quantities that we or other experimenters might find usefulin future work. If we expect that we might re-use our slit patterns, we might report theirmeasured dimensions. In each case communicate with complete sentences and include theunits and errors for the measurements. What improvements might you suggest?

8.4.1 References

Sections of this write up were taken from:

1. D. Halliday & R. Resnick, Physics Part 2, John Wiley & Sons.

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2. R. Blum and D.E. Roller, Physics Volume 2, Electricity Magnetism and Light; Holden-Day.

3. E. Hecht/A. Zajac, Optics, Addison-Wesley Publishing Company.

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8.5 APPENDIX

8.5.1 The Semiconductor Diode Laser

Semiconductors are materials that can be pretty good electrical conductors in some situationsand pretty good insulators in other situations. The most widely used semiconductor materialis silicon (Si), but carbon (diamond) and germanium (Ge) have similar properties. Eachsilicon atom in a solid crystal is bonded to four nearest neighbor silicon atoms. Since siliconis in column IVA of the periodic table of the elements, this means that every atom is verycontent with its bonding structure; every atom shares one electron covalently with each ofits four nearest neighbors and effectively has a full valence shell of eight electrons. All of theelectrons in the crystal are already busy holding the crystal together so that no electrons arefree to move about and to conduct electrical current. In solid state physics we say that anenergy gap has developed between the valence band and the conduction band, the valenceband is completely filled, and the conduction band is completely empty. Pure silicon is avery good electrical insulator at low temperature and moderate electric fields. Pure silicon isimpossible to obtain, however, and no object has infinite volume so the surface atoms haveno nearest neighbors with which to share electrons.

If we imagine substituting a column IIIA atom like gallium (Ga) or boron (B) for one ofthe atoms in the silicon crystal, we see that boron has only three electrons to share with itsfour nearest neighbors. One of the boron atom’s neighbors wants another electron that boroncannot give. It turns out that this “hole” (h+) can be filled by any one of the four sharedelectrons near this hole moving into this position, but this leaves behind another hole. Thishole, in turn, can be filled by any of the four electrons near its position. If we consider thatthe positively charged atomic nuclei are not moving along with their electrons, we will seethat the volume around the boron atom has become negatively charged and that the volumearound the hole has become positively charged. The negatively charged boron atom is fixedin place by the crystal lattice, but the positively charged hole is now moving around inside thematerial and conducting electrical charge. The addition of type IIIA elements to a siliconlattice turns it into a conductor of positive charge. We call this “p-type” semiconductormaterial. In solid state physics we say that the substitution of boron has added a hole tothe valence band. This is illustrated in Figure 8.5(a).

If we imagine substituting a type VA atom like phosphorous (P) or arsenic (As) for oneof the atoms in the silicon crystal, we see that arsenic has five electrons to share with itsfour nearest neighbors. This leaves an extra electron (e−) that is not needed to hold thecrystal together. This extra electron, then, is free to move around inside the solid and totake its charge along with it. The arsenic nucleus, however, is fixed in place by the fourbonds to the lattice and when we removed one of its electrons the volume around the arseniccore became positively charged. Similarly, the mobile volume around the extra electron isnegatively charged. This charge motion is an electrical current. The addition of type VAatoms to a silicon crystal turns it into a conductor of negative charge. We call this “n-type”semiconductor material. Figure 8.5(b) illustrates this situation. In solid state physics we saythat the substitution of arsenic has added electrons to the conduction band.

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Figure 8.5: Two illustrations of how doping silicon can generate a conductor. In (a) aboron (B) atom replaces a silicon (Si) atom and creates an electron vacancy or hole in thevalence band. In (b) an arsenic (As) atom replaces a Si atom and leaves an extra free electronin the conduction band.

It is also possible to form semiconductors using molecules such as GaAs or InP insteadof atoms to build up the crystal. These binary III-V semiconductors are generally dopedby replacing their atoms with silicon; replacing the column III atoms with silicon yieldsconduction electrons and replacing the column V atoms with silicon yields valence band holes.Ternary semiconductors like GaAlAs also have properties different than other materials; wecan exploit specific properties of the various materials to build up the particular set ofproperties we need for a particular device to have. Finally, there are organic semiconductorsthat usually exploit a molecular substrate instead of a crystal lattice to obtain semiconductiveproperties. This allows organic semiconductors to be very flexible instead of brittle likesilicon.

If we imagine building a semiconductor device by beginning with a p-type substrate andadding layer after layer of n-type material, the result is a pn-junction. The converse processof beginning with a n-type substrate and adding p-type layers also generates a pn-junction.The second law of thermodynamics would suggest that the extra electrons diffuse from then-type material to fill the entire volume of the crystal uniformly with electrons. Similarly,the holes should diffuse from the p-type material and fill the crystal uniformly with holes.The crystal does, in fact, try to obey the second law of thermodynamics; however, electronsand holes are each electrically charged so that a very large electric field develops in thevicinity of the junction. As electrons moved into the junction, they left behind a positivecharge; similarly, holes left behind a negative charge. This makes a large electric field pointfrom the n-type material toward the p-type material as shown in Figure 8.6. Ordinarily,this electric field would result in a large current but most of the electrons that enter thejunction recombine with the holes that enter the junction and become a silicon bond onceagain. Sometimes a hole and an electron can form a metastable “exciton” wherein theyorbit each other like the electron and proton in the hydrogen atom. Once all of the chargesare thus immobilized in the junction region, the junction becomes a very good insulatorthat cannot conduct electric current; the junction region has no free mobile charges so wecall it an “intrinsic” semiconductor. In photodetectors we intentionally add pure (intrinsic)

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semiconductor material between the p and n materials to increase the likelihood of absorbinga photon in this volume; this forms a p-type - intrinsic - n-type or pin-junction.

Figure 8.6: An illustration of charges flowing across a pn-junction. Most of the time theelectrons and holes simply recombine once again to become an immobile, neutral silicon bond.Sometimes an electron and hole form a metastable exciton where they orbit each other likethe charges in a hydrogen atom. In (c) these processes are shown from the band theory ofsolids perspective.

If we attach the positive terminal of a battery to the p-type material and the negativeterminal to the n-type material, current will flow. In fact, so much current will flow that ifwe do not somehow limit the current, the device will quickly heat up and burn out. As thebattery helps cancel the electric field in the junction region, the electrons and holes enter thejunction as a battery current and recombine. Each time an electron-hole pair recombinesthe energy of attraction between them is released as a photon. The energy of the photon isequal to the band gap energy Eg of the semiconductor, Eg = hf = hc

λ. A pn-junction is a

light emitting diode (LED) but the light is outside the human visible range (400-700 nm) formost diodes. Gallium arsenide (GaAs) diodes emit the bright red light used in calculatordisplays in the 1970s.

Over the years we have developed ways to trap the photons emitted by some pn-junctionsby making the surfaces around the junction region highly reflective. By containing thelight in this manner, we greatly increase the rate that these trapped photons stimulatephoton emissions and pair recombination. As we may have learned earlier in studying theHeNe laser, these stimulated emissions have exactly the same energy (wavelength), direction(momentum), and phase as the photon that stimulated the emission; the two photons arecoherent and have the same quantum state. Since photons satisfy Bose-Einstein statisticsinstead of Fermi-Dirac statistics, it is not forbidden for many photons to have the same state.These two photons are now twice as likely to stimulate another photon in their state as anyof the singlet photons in the device. Once one of them does stimulate another emission, thethree photons will be three times more likely to stimulate another like themselves, etc. Verysoon a very large fraction of the photons in the laser have the same wavelength, direction, andphase. The light emitted by the semiconductor diode laser is coherent and monochromatic.

From this discussion one should not deduce that all semiconductor diode lasers are madefrom silicon. In fact, only recently have we figured out how to make lasers from silicon at

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all. Additionally, the color of the light is determined by the semiconductor’s energy gap;since there are lasers having many different colors, they must be made from many differentsemiconductors. Since the light wavelength is determined by the semiconductor’s band gap,it is possible to tune the wavelength over a few dozen nanometers range by adjusting thediode’s temperature. At higher temperature the crystal expands so that the atoms arefurther apart and the attraction between them (the energy gap) is reduced slightly.

8.5.2 The PIN Diode Photodetector

When energetic particles irradiate a semiconductor, the resistivity of the semiconductordecreases. The energetic particles break molecular bonds by colliding with the valenceelectrons. Each broken bond yields an electron free to conduct electricity in the conductionband; each broken bond also yields a free hole in the valence band and these increases infree charge carriers increase the electric conductivity (σ = 1

ρ) and reduce the resistivity.

Eventually, these carriers reach the edge of the semiconductor or recombine with anotherpartner to recover the status quo when the irradiance is removed. This makes semiconductorsgood photoconductors.

In order to generate an electron-hole pair, the incident particle must deposit enough en-ergy upon the bound electron to break the chemical bond. This energy is the semiconductor’sband gap energy, Eg. Very energetic particles might break several bonds each before runningout of energy; each collision costs the particle Eg so that counting the broken bonds can giveinformation about the energy of the incident particle. This property makes semiconductorsvery valuable to high energy physicists.

Alternatively, if the energies of incident particles are known to be near the gap energy,then counting the broken bonds yields information about the brightness or intensity of thebeam. In our discussion of the pn-junction above, we noted that electrons from the n-typematerial diffuse across the junction as do holes from the p-type material to recombine andto make several microns thickness of intrinsically pure semiconductor at the junction with ahigh electric field.

When light with sufficiently high frequency encounters this junction, it breaks thesebonds. This high electric field then pushes the alleviated holes toward the p-type material andthe electrons toward the n-type material to generate a small potential difference between thetwo sides of the pn-junction. Solar cells can generate electric power utilizing this photoelectriceffect, but pn-junctions also make sensitive and linear photodetectors. Since each photoncreates one electron-hole pair, the generated current is directly proportional to the light’sintensity.

The biggest problem with pn-junction photodetectors is that such a thin junction absorbsvery few of the incident photons; most pass through without interacting with the junction.To increase the junction’s thickness, we deposit a layer of intrinsically pure (i) materialbetween the p-type and the n-type materials. These pin-junctions (PIN diodes) are moresensitive and have faster response than regular pn-junctions.

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