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Unformatted text preview: Chapter 11  Problem Solutions FIND: Why are metals opaque, yet other materials may be transparent or translucent?
SOLUTION: Metals have no band gap. Incident radiation is absorbed by free electrons.
Metals do not allow visible light to pass. FIND: Identify a transparent conductor and the color of conductive polymers. SOLUTION: There are no transparent conductors. Some glasses can have reasonable
conductivity and be transparent, but not high conductivity. When polymers with highly conjugated structures are doped with an electron donor of acceptor, the polymer becomes
conductive and looks like a metal. FIND: Identify which materials are dielectrics? SOLUTION: Materials with large band gaps and high electrical resistivity are dielectrics
 BN, epoxy, and diamond. Sic is a semiconductor but probably has a relatively high impurity level with a lOQcm resistivity; CBS is a semiconductor. Ca and Sn are
conductors. FIND: Identify the types of polarization in PVdF, NaCl, Na, diamond, PE, and alumina.
SOLUTION: There are four types of polarization: 1. electronic, 2. ionic, 3. molecular,
and 4. interfacial. Ceramics can exhibit all types, metals only the ﬁrst, and polymers the
ﬁrst and third. This is the scorecard: terial Polarization Mghanisms
PVdF l and 2
NaCl 14
Na 1
C 1
PE 1, 3 weak
alumina 14 FIND: Which dipole has the large moment, CF or CaN? DATA: The electronegativities are C = 2.55, F = 3.98, and N = 3.04; the bond lengths, which can be obtained (apprordmately) by adding covalent radii, are CF = 1.49 and C2 N
= 1.16A. SOLUTION: The electronegativity difference multiplied by the charge separation is
largely responsible for the dipole moment. 1.43 x 1.49 = 2.13 for CF and 0.49 x 0.57 for
CsN. C~F has the larger moment. 277 FIND: Calculate the polarization in tin.
GIVEN: The electronic displacement is 2 x lO"°nrn
DATA: Tin is BCT; however, it can be reasonably well approximated as FCC with a
lattice parameter of6.49A3; atomic number 50; 1.6 x 10‘19 C/e '
SOLUTION: z = (4 atoms/cell)(50e/atom)/a03 = 0.73 3/a3. Thus, P = qu = (0.73 x 103° e/m3)(1.6 x 10"”C/e)(2 x 1091:.) = 2.34 x 10‘ (:Im2 FIND: The average separatio of charge when copper is placed in an electric ﬁeld
GIVEN: Polarization is 8 x 108 Gina2
DATA: Cu is FCC with a lattice parameter of3.615A; atomic number 29; 1.6 x 10'19C/e 8 x10'8C/m3 d = P/Z 2 m
q (2. 455 x 103°e/ mam: x 10” C/e) =2.04x10"A SOLUTION: z = (4 atoms/cell)(29 e/atom)/a.,’ = 2.455 e/A3 Thus,
FIND: Calculate the ionic polarization of MgO GIVEN: The applied electric ﬁeld causes a displacement of 5 x 10'9 nm between ions.
MgO has the NaCl structure. DATA: 4 ions of each type per cell; a, = 22(er + r02.) = 2(0.66 + 1.32) = 3.96A; 1.6 x
10"9 C/charge; 10 electrons in each ion. SOLUTION: Each ion carries a charge of +_2 . there are 4 ion pairs/cell. Thus, Z = (4
dipoles/cell)(20 charges/dipole)/a.,3 = o. 13 charges/A3. Hence, P = qu = (0.13 x 10”
charges/max] .6 x 10'19 C/charge)(5 x 10‘“ m) = 1.04 x 107 (:Im2 FIND: Percent increase in separation of ions in ZnS.
GIVEN: Applied ﬁeld causes ionic polarization of 7 x 10'8 cm2 DATA: 22118 has the zinc blend structure, 4 ion pairs/unit cell; 1.6 x 10':9 C/charge',
aoJE/4= rm+ + r32 = 0.74 +1.84 2:) a0 = 5.96A :s a: = 212 x10'30 m3 SOLUTION: z = (4 dipoles/celer charge 2/dipole)/a.,3 = 7.56 x 102‘ charges/m3 P = qu :> d = Pqu = (7 x 10* C/m2)/[7.56 x 102‘ charges/m3)(l.6 x 10'” orchargen =
5.8 x 10'“ m = 5.8 x 10'8A. Expressing on a fractional basis: 5.8 x 10'3/(rm+ + r52.) = 2.25 x 10'8 = 2.25 x 10‘% COMMENTS: Very small displacement! 278 10. ll. 12. 13. FIND: Voltage that develops when a hydrostatic pressure is applied to PbZrOs.
GIVEN: PbZrOs is piezoelectric
SOLUTION: No voltage necessarily develops because the pressure, being” hydrostatic, does not cause charge displacement in the unit cell. Only the scale of the cell is reduced,
ala Hooke’s Law, in each direction. COMMENTS: If pressure were applied normal to the appropriate face of the single
crystal, voltage would develop. FIND: Describe the properties of PP ﬁber with imbedded ions. GIVEN: The ions are embedded in molten PP. ' ASSUMPTIONS: PP is a perfect insulator. SOLUTION: The imbedded ions, both negative and positive, provide the ﬁber with
locally charged areas. The ﬁber is charged for its lifetime. It will attract charged particles
ﬁ'om the atmosphere, such as dust. COMMENTS: Fiber with embedded charge is used for dust masks. FIND: Voltage required to produce a polarization of 10‘7Clm2 GIVEN: Material is polycrystalline alumina, l x 104m thick. DATA: J = 8.85 x 10%" m. is about 9 from Table 113]; the dielectric strength of
alumina is > 7 x ro‘Wcm. x=1+P/_a§__P=(rc1)_o§=(x1)_9Wd“ V = PD/(rr 1) =W= 0.126V
—° (9)(8.85 x 10‘” F/m) SOLUTION: Using equation 11.34 as modiﬁed in example problem 11.32,
This translates into a dielectric stress of 0. 126V/0.01 em = 12.6 V/cm, which is
insufﬁciently high to cause breakdown. FIND: How much thicker must a sample of mica be to achieve the same electric
polarization as mica exposed to the same electric ﬁeld? DATA: The dielectric constant of PE and mica are about 2.3 and 7.0. ,_ SOLUTION: In mica the polarization is due to both ionic and electronic mechanisms. In
PE only the electron mechanism is active, so the dielectric constant is lower. Taking a ration of the P = (r:  l)_.,§ expression: 279 I4. 15.
16. 17.  1 g 2.3  1 E / d
/ m: KPH 3:2 PE = I: :2 PE PE
PM P” [MWIJLOXEWJ [701][_0][Em/dm]
FIND: Does PAN or PVC have a higher dielectric constant? GIVEN: PAN has a higher dipole strength. DATA: Electronegativities for C, N, and C1 are 2.55, 3.04, and 3.16; bond lengths, 1,
for CCl and C s N are 1.77 and 1.16A, which can be obtained (approximately) by adding
covalent radii; ppm = 1.15 g/cm3 and ppvc = 1.4 g/cm3. SOLUTION: I am going to do this problem backwards. You want to start at the
bottom. _=_o+P/§3 P=L"“_O)E.n but “ = Cd/A :> P = (C  Cn)(de)5, but a = V/d => P = (C  COXd/AXV/d) = (CV  COW/A, but Q = cv :> P = (Q  Q.)/A.
This expression says that the change in polarization is equal to the extra charge per unit area that can be stored because of the presence of a material between the plates in a
capacitor. FIND: Should you build a capacitor ﬁ'om cordierite or PVC? How large will it be?
GIVEN: Space is limited. Capacitance need to be 0.05 uF. DATA: 1: is 5.0 and 3.2 for cordierite and PVC, according to Table 1131. Fields not
high; * = 3.35 x 10'l2F/m. SOLUTION: C = _(A/d) = 1c“.,(A/d). the material with higher It is desired. Substituting C=5x10‘3F=5.0x8.85x10'13F/m04/a) _A/d=1.13x10’m for K and Mo:
COMMENTS: All we can give for a dimension is Aid. If we knew the voltage to be used, we could put a lower bound on d by implementing the requirement to avoid
breakdown. FIND: Show the capacitance of a multiple plate assembly is C = Ic__.,(nl)A/d.
GIVEN: There are n plates separated by dielectric. SKETCH: 280 18. 19. 20. 21. SOLUTION: Count them, there are n plates and n1 cells. Each cell has characteristic separation d and area A. For one set of plates, C = 1c_.,A/d, so for nl sets of plates, C =
K_,,(nl)A/d. FIND: Calculate the relative dielectric constant of mica.
GIVEN: The sample is placed in a ﬁeld, E, of 1 lem and the measured polarization, P, i 5.4 x 108 (3/1112. ‘
DATA: The relative dielectric constant of mica, given in Table 113], is K = 5.4  8.7. The permittivity ofvacuum, so, is 8.85 x 1012 F/m.
SOLUTION: Using the expression for K developed in Example Problem 11.31: K = 1 + (P/aog) = 1 + (5.4 x 108 C/m2 )/ [(8.85 x 1012 F/rn)(103 V/m)] = 7.1 COMMENTS: This is a reasonable value.
FIND: Calculate the ﬁeld strength required in PE. will the PE survive?
GIVEN: P = 10'7C/m2 DATA: K = 2.3; _, = 8.85 x 10‘12 Pint; dielectric strength w 200 x 106 V/cm 10" C/ m2 g = PM ' 1L" = (2.3  1) (8.85 x 10” F/m) =8.7x10’V/m SOLUTION: x = I + P/__o§ :>
= 87 V/cm << dielectric strength of PE FIND: Why might the dielectric constant of a polycrystalline ceramic decrease with
frequency? SOLUTION: x = l P/_.,E,, so as P goes, so goes it. Let us examine P. The factors that
contribute to the polarization of a ceramic are electronic, ionic, and space charge.
Electronic is generally not frequency dependent; however, 10“5 Hz is very fast switching.
Ionic polarization generally shows frequency dependence in this region, since ionic masses
may be large. since the material is polycrystalline, space charge eﬁ‘ects can also be signiﬁcant. FIND: voltage required to develop charge in capacitor. 281 22. GIVEN: C = 2.5 x 10"”C; plates 20 x 20 nm; (1 = 0.01 mm; dielectric: vacuum or
PTFE. DATA: mm... = 1; mm = 2; ,0 = 8.85 x ions/m SKETCH: SOLUTION: Q = C x V and C = x“.,(A/d). Thus, Q = K_.,V(A/d) :> V = dQ/K_.,A (10"m) (2.5 x 10""C) _m = a. 71V
(8.85 x 10” F/m) (4 x 10“ m2) Vvacuum = Since the dielectric constant of PTFE is double that of vacuum and 1c appears in the
denominator, the voltage will be just half: Vm = 0.36V. FIND; What is an appropriate material and wall thickness for spark plug wire?
GIVEN: The voltage carried is 15,000 V. SKETCH: SOLUTION: A spark plug wire needs the following properties: high dielectric strength
(to avoid breakdown), low dielectric loss (to avoid losses and heat buildup), ﬂexible,
resistant to gas and oil, heat resistant, durable and long wearing. Silicone has a dielectric
strength of 220 x 10‘ V/cm, low dielectric loss, and all the other required properties. It is
the material of choice. The minimum thickness can be assessed by taking the ratio of V to
dielectric strength: 1sooow(220 x 10‘ V/cm) = 6.8 x 10'5 cm COMMENTS: The actual wall thickness is 2—5mm, which adds a degree of safety and
durability. 282 23. 24. 25. 26. 27. 28. FIND: Why might the dielectric constant of a polar polymer be temperature dependent? SOLUTION: Increasing the temperature may allow the polar groups more“ mobility,
increasing the dielectric constant with temperature. Increasing the temperature is equivalent to decreasing the frequency. Both effects give the dipoles more time to
respond. COMMENTS: We ﬁnd that most polar polymers show time and frequency dependent
dielectric properties. FIND: Why is the dielectric constant of PE frequency independent, whereas that of PVC
is not. SOLUTION: The solution to this problem is contained in that of number 21. PE shows
only electronic polarization. The electron’s response is so fast the PE shows no frequency
response at all but very high frequencies. PVC contains dipoles that do not respond as
quickly as do electrons. FIND: What is the approximate value of x for Al? 7
SOLUTION: Metals cannot store separated charge, so C a: 0 and x e 0. FIND: Is dielectric strength an intrinsic property? relative permittivity? polarization?
SOLUTION: Dielectric strength, like mechanical strength depends on the weakest link.
Hence, it is an extrinsic property? Permittivity and polarization are intrinsic properties.
FIND: Bakelite® or nylon, which generates more heat in a ﬂuctuating electric ﬁeld?
GIVEN: Work, W 0: s x tan 8. W is displayed as heat. DATA: According to Table 1141, tan 6 for bakelite and nylon are 0.08 and 0.01. SOLUTION: The heat generation in Bakelite® will be 8 times higher than that in nylon.
COMMENTS: A a distributor cap in a car does not experience a high the ﬁeld, so the
loss is not large. In addition, the part is generally subjected to a temperature dictated by
its proximity to the engine. More important is the resistance to gasoline and oil, its ability
to maintain shape at temperature, its durability, and, of course, cost. FIND: Maximum voltage that can be used in it PS capacitor so that power loss does not
exceed 0.1W. GIVEN: Capacitor size is 25 x 25 x 0.01 mm; frequency is 106 Hz. DATA: For PS i: = 2.5 and tan 8 = 0.0003. SOLUTION: We use equation 11.41 to solve this problem: W a: rt__.,f‘é2tan6.
Substituting for E, = V/d and solving for V, we get V = [Wd2/1t_o.,ftan8] m. We note that
in this equation, W is per unit volume, so we need to normalize to sample volume, A X d, 283 29. 30. 31. V = [Wdz/Adzgf *,0: tan 51‘” m [Wd/Adztf _,x tan a j” 122
J = 277V COMMENTS: We should check that this is not in excess of the dielectric strength of the
PS: 2.4 x 101°V/m x 10‘5m = 240,000 v. No problem. z 0. 1(10")
(2.5 x 102 f 3. 14( 10‘ ) (8. 85 x 10”) 2.5 (00003) FIND: Is PE 3 good choice for material to be used for a container in which to reheat food
in a microwave oven? If not, suggest a better material. DATA: According to Table 11.41, tan 6 for PE is very small. Heat generation in the PE
will be minimal. SOLUTION: Oils can reach very high temperature in a microwave oven. PE melts below
140°C, and it has been reported to melt in a microwave oven when in contact with oils. The ideal material generates little heat, is stable to about 180°C, does not absorb smells,
does not degrade, does not release chemicals into the food, and so on. COMMENTS: To impart the function of browning to the top, you need a material that
absorbs heavily in the microwave region and re—radiates to cause browning. Few ceramics
absorb heavily, so a browning top may be a composite material. FIND: Calculate e’ for PC.
GIVEN: Use the data in Tables “3—1 and 11.41. DATA: 1: = 3.0 and tan a = 9 x10'4. so = 8.85 x10'12 F/m
SOLUTION: We solve the problem using equation 1 1.43: tans = e'leo :> s' = 9x 104 x 8.85 x10'12 F/m = 8.0 x 1015 F/m. COMMENTS: Recall that tan 6 is frequency dependent. The K and tan 8 values used are
at a speciﬁc frequency. FIND: Determine the refractive index of PP
GIVEN: The critical angle in it is 42.86”
SKETCH: SOLUTION: Using the concept developed in sample problem 11.51 based on equation
11.52: in, = sin" (10.) => n = (sin a)" = (sin 42.80")'1 = 1.470 284 32. 7 33.
34. 35. 36. FIND: Why is ﬁrsed silica clear and window glass green when viewed on edge? Assumptions: One common impurity in glass is iron. Iron makes oxide silica glasses
appear green, such as in Coke® bottles. FIND: Determine the angle of light passing through glass and PE in layered composite
material. GIVEN: Materials are Soda—limesilicate glass and polyethylene; angles of incidence are
(a) 10° and (b) 70°. The thickness of the PE ﬁlm is 0.2 mm and that of the glass is 1 mm.
DATA: For PE and this glass, 11 = 1.54 and 1.51 SKETCH: SOLUTION: All angles are deﬁned relative to the surface normal. The angle of
incidence is (a) 20° and (b) 70". we ﬁrst need to determine the angle of propagation
through the PE ﬁlm on top, then use this, the angle of reﬂection, as the angle of incidence
for the glass. The thicknesses are irrelevant information for this problem. Proceeding for
the PE layer at 20°
sin i/sin r= 112/1113 sin I = 1.54 sin 20“ :> r = 31.7” For the glass layer 31 .7“ is i: sin i/sin r = 113/112 :> sin r = (151/154) sin 31.7" :> r = 31.1"
Part (b) is done exactly the same way; however, 70" is greater than the critical angle, so
the light never enters the PE (1' = 90°) or the glass. FIND: Calculate the reﬂective index normal to the axis of a PET ﬁber I GIVEN: Birefringence is 0.90 and nun1.. = 1.660 Assumptions: We assume that the ﬁber is isotropic in its crosssection ,
SOLUTION: Using the basic deﬁnition of bireﬁingence: An = n.n,, rearranging and
substituting, nr = n.  An = 1.660 — 0.090 a 1.570 FIND: Show how a headlight reﬂects from dry and wet pavement. GIVEN: The road is rough on a scale of the wavelength of light. SOLUTION: The light incident on the rough or dry road is in large part absorbed, with
the remainder being reﬂected diﬂirsely. 0n the other hand, the ﬁght from the wet road will 285 37. 38. 39. 40. 41. be reﬂected specularly, hardly illuminating the road and blinding oncoming cars. FIND: What is required for a transparent object to disappear when immersed in water?
SOLUTION: To disappear, the object must stop reﬂecting light off its surfaces. Hence, the refractive index difference between the object and water must be zero. The refractive
index of the material in the object must be 1.33. COMMENTS: There are no limitations on shape, but a hollow object cannot disappear. FIND: Explain why optical waveguides are fabricated with a refractive index gradient.
GIVEN: Light sources are not perfectly collimated.
SKETCH: See Fig. 11.56 SOLUTION: Light that travels along the center of the ﬁber in a straight line takes a
shorter route than light that eﬁ'ectively bounces 011‘ surfaces. A refractive index gradient is
used to speed up the light that takes the longer paths, thereby insuring that all light sent at
any instant of time arrives together at the destination. FIND: How does 7t. depend on n? SOLUTION: referring to equation 11.52, we note that n = Jim/llama“ ::> ll. material = lam/n. Since n > 1, the wavelength of the light decreases as it passes from air into a
material. FIND: Critical angle of light traveling into and out of water. DATA: From Table 1152, the tellactive index of air and water are 1.0 and 1.33.
Assumption: The water is quiescen
SKETCH: ' SOLUTION: The concept of critical angle applies only when light is traveling from a
medium of high to low index. (You can try to work out the mathematics for the reverse
and you will need to take the inverse sine of a number larger than 1). Therefore, we need
calculate the solution only from a ﬁsh’s point of view. Mum = sin i/sin r. Ar the
critical angle sin r = 1‘ Sin ic = l.0/1.33 z) ic = 48.75”. COMMENTS: Next time you are swimming underwater, try the experiment. FIND: Can a ﬁsh see you on shore or in a boat?
DATA: n for water is 1.33 and for air is 1.0. 286 42. 43. SKETCH: '\ SOLUTION: We use equation 11.53 to work this problem: vwater / vair = nai, I nwater = sin i / sin r
1.0/ 1.33 = sini/sinr.
Continuing as in Example Problem 115], sin r = 1.33 sin i.
Hence, r = sin"1 (1.33 sin i).
This expression requires that 1 2 1.33 sin i. Hence, 7 sin ism = 1/133 :> i = 48.8". So long as the angle from which the ﬁsh looks at the point on the surface is less than 48.8"
, the ﬁsh can theoretically see out of the water. When i = 0°, r = 0°; when i = 48.8°, r : 90°. Hence, the limitation is on the ﬁsh's view of the surface, not your position on land or
in a boat. COMMENTS: I don't know enough about ﬁsh vision to know just what ﬁsh do see.
FIN D: Determine the reﬂectivity from ice and water at normal incidence DATA: From Table 1155, the refractive index of ice and water are 1.31 and 1.33 2
SOLUTION: Fresnel Formula is 12:93?) For ice,R=[0.31/2.31]2= 1.8%, For
n water, R = [033/233]2 = 2.0% FIND: Show the principle of a refractometer
GIVEN: It uses the principle of critical angle
SKETCH: 287 45. SOLUTION: Light is collimated and directed to propagate through a sample ﬁlm. The angle of propagation is increased until the detector, which sits at 90°, senses a strong
intensity. COMMENTS: Monochromatic light at different wave lengths can be used to determine
dispersion as well. FIND: Design a waveguide assembly that can bear high tensile loads.
GIVEN: Seven optical ﬁber are required.
DATA: Pound for pound, Kevlar® and Spectra® ﬁbers are stronger than steel. SKETCH: SOLUTION: this is just one of many possible solutions. The large ﬁbers are the optical
ﬁbers, which are coated with a protective polymer. The small ﬁbers are the polymeric
reinforcement ﬁbers. The entire assembly is protected with another polymer coating. FIND: Why are immersion lenses used in optical microscopy.
DATA: The refractive index of oil is greater than that of air.
SKETCH:  ,. . SOLUTION: Light travels from the sample on the stage upwards through the cover slip
through air and into the glass compound objective lens. Reﬂection will occur at each
surface. The oil ﬁlls the space between the cover slip and the objective lens. If the indices
match  that of the oil equals that of the glass  then no retraction occurs and more light
enters the objective lens. (Normally as you increase magniﬁcation, your light intensity
decreases with the area of viewing.) COMMENTS: With better dry objectives  better coatings, higher aperture  oil
immersion techniques, which can be messy, are being used less and less. 288 46. 47. 48. FIND: Estimate the amount of light that would reach the ﬁlm in a camera that uses a 5
lens assembly. GIVEN: Lenses are not treated with anti reﬂective coating.
DATA: From Table 1155, Miami,“ = 1.51
Assumptions: The glass is a sodalimesilicate of n — 1.51. Normal incidence. SKETCH: R=[”' ]=(o.51/2.51f =95.87%
n+ 1 SOLUTION: Light is reﬂected from each lens. Assuming normal incidence,
is transmitted from each of 5 lenses (0.9587)’ = 0.81 = 81% is transmitted to the ﬁlm. COMMENTS: This effect is signiﬁcant and cameras often have more lenses than this. In
addition, we have ignored the effects of the reﬂected light. Consider, for example, light
inﬂected off an interior lens and then reﬂected back by a prior lens onto the ﬁlm. The
image will not be sharp. FIND: What glass makes the best beveled glass window? GIVEN: The purpose of the bevel is to produce mini rainbows inside the house. SKETCH: See Fig. “55 SOLUTION: Unfortunately refractive index does not determine which glass provides the
best rainbow. The cut of the glass and the optical dispersion are the critical issues. for
any given geometry of cut, select the glass with the highest dispersion. COMMENTS: Lead glass is used for crystal. Because of its high refractive index, it has
high reﬂectivity. It sparkles. FIND: Why can people who normally wear glasses see better under water? SOLUTION: Light traveling through your eye is traveling though an aqueous solution.
When it emerges from your eye into the air, it is refracted. Ifit goes into water from your
eye, refraction is greatly reduced. Errors in the eye's lens are magniﬁed when the interface
facilitates signiﬁcant light refraction, since the light rays passing through water subtend a 289 49. 50. 51. 52. smaller angle than do those passing through air. COMMENTS: This is the principle of oil immersion microscopy. FIND: What determines the color separation of the sunlight passing through beveled
glass? SKETCH: Refer to Fig. 11.55. SOLUTION: The color separation is determined ﬁrst a few factors:
1. The refractive index of the glass 2. The perfection of the surfaces of the glass 3. The distance the light travels through the glass 4. The various incident angles 5. The distance between the lens and the point of analysis. COMMENTS: It is beautiful when the light is resolved by a beveled glass window. The
best windows are "crystal", or high lead silica glasses. FIND: Why does light effect only electronic polarization? SOLUTION: The response time of a process is largely effected by the masses that must
move. The larger the mass, the longer the response time. Only electrons with their small
mass can respond to frequencies associated with light in the visible region. FIND: What is the maximum wavelength to stimulate photoconduction in tin?
GIVEN: The bandgap in tin is 0.09 eV DATA: 1.6 x 10'191/eV; c = 3 x 10“ n/s; h 2 6.62 x 10'3‘Js SOLUTION: Equation 11.72 states that E = helix. = E; :>
a = hc/EG = (6.62 x 106‘ Js)(3 x 103 m/s)/(l .6 x 10'19 J/eV)(0.08 eV)
= 15.5 urn =15512.5 nm COMMENTS: This is longer and hence less energetic than the visible region. Hence,
energy from the visible region is absorbed, leading to the generalization that electrical
conductors are opaque. FIND: What materials are transparent? SOLUTION: High band gap materials are transparent and we can calculate the bandgap
necessary to avoid absorption of visible light. From Fig. 11.1—1, the most energetic visible
light has a wavelength of 0.4 pm. We substitute this into E = hell and solve for E: EG = (6.62 x 10‘“ Js)(3 x 10" m/s)/(1.6 x 10"” J/eV)(4 x 10'1 m) = 3.16v
All materials with bandgaps greater than 3.1 eV have the chance of being transparent to
visible light. ' 290 53. 54. 55. 56. 57. COMMENTS: As you know, other eﬁ'ects can prevent a materials from being
transparent. FIND: Might the materials in problem 441ase? SOLUTION: To lase materials needs a AE that corresponds to an energy of 0.40.7 um,
which assessed using E = hem. We calculated in problem 44 what bandgap is required to absorb radiation at 0.4 pm = 3.1 eV. For it = 0.7 urn, the bandgap is 1.8eV. We need to
have 3 energy levels, with an energy rise of 1.8  3.1 EV from ground to a metastable excited state. From the table, only tin cannot lase in the visible and it may not be possible
with CDS. FIND: Determine the concentration of Fe ions in a Coke® bottle. GIVEN: The extinction coefﬁcient is 573 mm'1 Assumptions: Assume a Coke® bottle is about 3 mm thick and about 20% of incident
light is transmitted. SOLUTION: The BeerLambert equation, 11.68 is 111., = exp(_cx). Substituting
values, 0.20 = exp(573 x c x 3). Taking the natural log of each side and solving for c:
c = ln(0.20)/(573 x 3) = 0.94% FIND: Why is beer normally sold in brown or green bottles or cans?
GIVEN: Beer degrades in UV light SOLUTION: Clearly, the brown or green glass absorbs UV radiation, thereby preventing
it from reaching the beer. COMMENTS: When you buy beer in clear bottles, make sure that it has not been on the
shelf long. FIND: Might polymers be photosensitive glasses?
SOLUTION: The silver halide crystals in photosensitive glasses disassociate. The fragments need to remain relatively close to one another so that they can recombine in a reasonably short period of time. Since the diffusion coefﬁcient of ions in polymers may be high, the fragments may move far apart. Glasses with low diffusion coefﬁcients need to be
selected. COMMENTS: Some photochromic polymer glasses have been made, but the colors
were not pleasing. FIND: Sunburn through a car window?
SOLUTION: Sunburn is chieﬂy caused by UV radiation that gets through the Earth’s
atmosphere. Fig. 11.61 shows that dielectric materials, such as oxide glasse and ceramics, are generally opaque to UV light. Hence, a sunburn will take many long hours
behind ordinary glass. COMMENTS: The bare arm in the open window will tan much more rapidly. 291 58. 59. 60. 61. FIND: What are the required characteristics of a high power laser window?
SOLUTION: No absorption at the lasing frequency, since high power will give rise to high heating. No scattering to attenuate or degrade the beam’s collimation. Stable to the
environmean conditions. COMMENTS: Materials that satisfy these requirements are hard to come by. FIND: Plot optical transmission as a function of grain size for silica crystals. SKETCH: 1
Transmitttvity Grain Slze SOLUTION: The sample will pass the light at small grain size and large grain size.
COMMENTS: Many gemstones, which are often single crystals, and Lucalox®, which is
small grained alumina, are transparent. FIND: Explain how solgel based glassy silica can be both transparent and porous.
SOLUTION: Typically in these zerogels, as they are called, the pores are very small, on
the order of 100A. The pores are so small that they do not scatter visible light. COMMENTS: These glasses have potential for use as window glass, since the thermal
conductivity is very low. FIND: How does doubling the concentration of a chromophore aﬁ‘ect the absorption of
light at the absorption ﬁequency? GIVEN: At a concentration of 0.001 g/cm3, the intensity decreases 10 %.
ASSUMPTIONS: All the intensity decrease is due to the chromophore. SOLUTION: Equation 11.68 shows the relationship between intensity and
concentration: I / Io = exp (ecx). Hence, when c = 0.00! g/cm3, I/ I0 = 0.90. Hence, ex = 105.4 cm3/g, a constant in this
problem. Thus, substituting the new values, m0 = exp (—ecx) = exp (405.4 x 0.002) = 31 %. The intensity decreases 19 % when the concentration is doubled. COMMENTS: The problem could have been completed faster by using the ratio
technique used throughout the text. 292 62. 63. 65. 66. 67. 68. FIND: Why is snow white?
SOLUTION: Snow is a collection of small ice crystals. The total surface area of ice
crystals in snow is enormous. Light is scattered from the surfaces of the crystals, giving snow a white appearance. Thus, even though ice is transparent, new fallen snow is white.
COMMENTS: If you shine a blue light on snow, it will appear bluish. FlN D: Why are there no K; for elements with low atomic number? SOLUTION: A K; implies an outer as well as an inner electron shell. The elements of . interest have so few electrons that only one shell is occupied. FIND: Why do some materials glow white in UV light? SOLUTION: Many materials contain “impurities” that ﬂuoresce in the UV, that is, they
absorb UV radiation and reernit in the visible, causing them to glow. COMMENTS: To paper and other materials often are added optical brighteners that give this effect and make the material look “whiter”. Fluorescence microscopy uses this
effect too. FIND: Why are household mirrors silvered on the back side and most laser mirrors
silvered on the front surface? SOLUTION: Lasers are usually used for their special properties  monochromatic,
coherent, wellcollimated. If the laser passes through a material, these properties can only
degrade. In addition, any losses in the glass will cause it to heat up. Even low losses in
high power lasers can be important. FIND: How does laser or LED light interact with a prism?
SOLUTION: Both lasers and LED’s are monochromatic sources. Thus, the light cannot
be ﬁrrther resolved into wavelengths. It simply retracts as it passes through the prism. FIND: How long will it take for the intensity of a phosphor to degrade to Lie?
GIVEN: The intensity after 10 sis 1J2. SOLUTION: Using the equation: 1/1., = exp(t/1), where ‘t is a relaxation or decay time.
We use the given information to calculate 1:: 0.5 = exp(lO/1:) :> t = 14.4 3
Hence, when t = 14.4 sec, I = lJe. FIND: What is required to make a colored LED?
DATA: h = 6.62 x 10‘34 1—3; c = 3 x ro‘m/s; 1.6 x 10'” MN
SOLUTION: We need to make a p—n junction with an E5 that corresponds to an emission
at wavelengths of 5.3 (green), 6.8 (red), and 4.6 (blue) all x 10'7m. The conversion from
AEG is done using: HG = rte/7t = (6.62 x 10'3‘Jsx3 x lO’m/s)/(l .6 x 10“”J/evxr in 10'7m) 293 69. 70. 71. 72. a 2.3 (green), 1.8 (red), and 2.7 (blue) eV FIND: What is a suitable detector for laser light? SOLUTION: An LED, since electrons will be excited, producing a voltage that can be
ampliﬁed. FIND: What property of a laser are utilized in bar code readers?
SOLUTION: Presumably a HeNe laser beam is split. Half the beam reﬂects OR the bar
code on the package and it is somehow recombined with the reference half. The coherence of the laser allows a pattern to establish between the two beams that can be
sensed. FIND: Show how doping with different ions produces different color emission.
GIVEN: ZnS doped with Cl' is violet and doped with Cu” is red. :1 SKETCH: II] gapf Cl ' levels SOLUTION: The dopant introduces new levels. The new gap in the Cl' doped ZnS is
larger than that of the Cu2+ doped ZnS, since the latter emits at a longer wavelength. Note
that Cl' is donertype and Cu2+ is acceptortype. FIND: What is the operating principle of a UV sensor for estimating temperature?
SOLUTION: You have probably seen pictures of houses and other objects that show, via
a color map, the sources of heat loss. Those sources are generally shown as the red areas,
the ones that are "hot". When an object is above OK, it emits radiation characteristic of its
temperature. An instrument that can measure the wavelength of the emitted radiation can
be used to determine an object's temperature. This is the principle of black body radiation,
which you can study in heat transfer classes. COMMENTS: The same principle is used to image soldiers using night vision glasses, to
ﬁnd people in smokeﬁlled rooms, or to nonintrusiver measure the temperature of a
moving hot ﬁber. 294 ...
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This note was uploaded on 09/22/2011 for the course MSE 2001 taught by Professor Tannebaum during the Fall '08 term at Georgia Tech.
 Fall '08
 Tannebaum

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