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optics chapter_07_solution_manual optics chapter_07_solution_manual Document Transcript

  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.1 Chapter 7: Simple Harmonic Motion, Forced Vibrations and Origin of Refractive Index PROBLEMS 7.1 The displacement in a string is given by the following equation: ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ νπ− λ π = txatxy 2 2 cos),( where a, λ and ν represent the amplitude, wavelength and the frequency of the wave. Assume a = 0.1 cm, λ = 4 cm, ν = 1 sec –1 . Plot the time dependence of the displacement at x = 0, 0.5 cm, 1.0 cm, 1.5 cm, 2 cm, 3 cm and 4 cm. Interpret the plots physically. 7.2 The displacement associated with a standing wave on a sonometer is given by the following equation: txatxy νπ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ λ π = 2cos 2 sin2),( If the length of the string is L then the allowed values of λ are 2L, 2L/2, 2L/3, … (see Sec. 13.2). Consider the case when λ = 2L/5; study the time variation of displacement in each loop and show that alternate loops vibrate in phase (with different points in a loop having different amplitudes) and adjacent loops vibrate out of phase. 7.3 A tunnel is dug through the earth as shown in Fig. 7.15. A mass is dropped at the point A along the tunnel. Show that it will execute simple harmonic motion. What will the time period be? 7.4 A 1 g mass is suspended from a vertical spring. It executes simple harmonic motion with period 0.1 sec. By how much distance had the spring stretched when the mass was attached? 7.5 A stretched string is given simultaneous displacement in the x- and y- directions such that ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ νπ− λ π =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ νπ− λ π = tzatzytzatzx 2 2 cos),(and2 2 cos),( Study the resultant displacement (at a particular value of z) as a function of time. 7.6 In the above problem, if A O R B x Fig. 7.15 For Problem 7.3
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ νπ− λ π =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ νπ− λ π = tzatzytzatzx 2 2 sin),(and2 2 cos),( what will be the resultant displacement? 7.7 As mentioned in Sec.7.5, alkali metals are transparent to ultraviolet light. Assuming that the refractive index is primarily due to the free electrons and that there is one free electron per atom, calculate ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ω π =λ p p c2 for Li, K and Rb; you may assume that the atomic weights of Li, K and Rb are 6.94, 39.10 and 85.48 respectively; the corresponding densities are 0.534, 0.870 and 1.532 g/cm3 . Also, the values of various physical constants are: m = 9.109 × 10 – 31 kg, q = 1.602 × 10 – 19 C and ε0 = 8.854 × 10 – 12 C/N-m2 . [Ans: 1550 Å, 2884 Å and 3214 Å; the corresponding experimental values are 1551 Å, 3150 Å and 3400 Å respectively]. 7.8 (a) In a metal, the electrons can be assumed to be essentially free. Show that the drift velocity of the electron satisfies the following equation ti eqqm dt d m ω− −=−==ν+ 0EEFv v where ν represents the collision frequency. Calculate the steady state current density (J = – N q v) and show that the conductivity is given by ω−ν =ωσ im qN 1 )( 2 (b) If r represents the displacement of the electron, show that ErP )( 2 2 νω+ω −=−= im qN qN which represents the polarization. Using the above equation show that )( 1)( 2 0 2 νω+ωε −=ωκ im qN which represents the dielectric constant variation for a free electron gas. 7.9 Assuming that each atom of copper contributes one free electron and that the low frequency conductivity σ is about 6 ×10 7 mhos/meter, show that ν ≈ 4 × 10 13 s –1 . Using this value of ν, show that the conductivity is almost real for ω < 10 11 s –1 . For ω = 10 8 s–1 calculate the complex dielectric constant and compare its value with the one obtained for infra-red frequencies. It may be noted that for small frequencies, only one of the electrons of a copper atom can be considered to be free. On the other hand, for X-ray frequencies all the electrons may be
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.3 assumed to be free (see Problems 7.10, 7.11 and 7.12). Discuss the validity of the above argument. 7.10 Show that for high frequencies (ω >> ν) the dielectric constant (as derived in Problem 7.8) is essentially real with frequency dependence of the form 2 2 1 ω ω −=κ p where 2/1 0 2 ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ε =ω m qN p is known as the plasma frequency. The above dielectric constant variation is indeed valid for X-ray wavelengths in many metals. Assuming that at such frequencies all the electrons an be assumed to be free, calculate ωp for copper for which the atomic number is 29, mass number is 63 and density is 9 g/cm3 . [Ans: ~ 9 × 10 16 sec – 1 ] 7.11 Obtain an approximate value for the refractive index of metallic sodium corresponding to λ= 1 Å. Assume all the electrons of sodium to be free. 7.12 In an ionic crystal (like NaCl, CaF2, etc.) one has to take into account infra-red resonance oscillations of the ions. Show that Eq. (68) modifies to )()( 1 22 20 2 22 10 2 2 ω−ωε + ω−ωε += M qNp m qN n where M represents the reduced mass of the two ions and p represents the valency of the ion (p = 1 for Na+ , Cl – ; p = 2 for Ca++ , F2 – – ). Show that the above equation can be written in the form* 2 2 2 2 2 1 2 122 λ−λ + λ−λ += ∞ AA nn where 2 2 1 1 2 , 2 ω π =λ ω π =λ cc 4 2 0 22 2 2 4 1 0 22 2 1 4 , 4 λ επ =λ επ = Mc qNp A mc qN A 7.13 The refractive index variation for CaF2 (in the visible region of the spectrum) can be written in the form * Quoted from Ref. 9; measurements are of Paschen.
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.4 92 9 152 15 2 1026.1 1010.5 1088.8 1012.6 09.6 − − − − ×−λ × + ×−λ × +=n where λ is in meters. (a) Plot the variation of n2 with λ in the visible region. (b) From the values of A1 and A2 show that m/M ≈ 2.07 × 10 – 5 and compare this with the exact value. (c) Show that the value of n∞ obtained by using the constants A1, A2, λ1 and λ2 agrees reasonably well with the experimental value. 7.14 (a) The refractive index of a plasma (neglecting collisions) is approximately given by (see Sec. 7.6) 2 2 2 1 ω ω −= p n where 2/1 2/1 0 2 414.56 N m qN p ≈⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ε =ω s –1 is known as the plasma frequency. In the ionosphere the maximum value of N0 is ≈ 10 10 – 10 12 electrons/m3 . Calculate the plasma frequency. Notice that at high frequencies n2 ≈ 1; thus high frequency waves (like the one used in TV) are not reflected by the ionosphere. On the other hand, for low frequencies, the refractive index is imaginary (like in a conductor – see Sec. 24.3) and the beam gets reflected. This fact is used in long distance radio communications (see also Fig. 3.27). (b) Assume that for x ≈ 200 km, N = 10 12 electrons/m3 and that the electron density increases to 2 × 10 12 electrons/m3 at x ≈ 300 km. For x < 300 km, the electron density decreases. Assuming a parabolic variation of N, plot the corresponding refractive index variation.
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.5 SOLUTIONS 7.1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ λν− λ π = txatxy 2 2 cos),( ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π− π = tx 2 2 cos1.0 where x and y are measured in centimeters and t in seconds. Thus )2cos(1.0),0( ttxy π== ; ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π− π == ttxy 2 4 cos1.0),5.0( )2sin(1.0),0.1( ttxy π== ; ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π− π == ttxy 2 4 3 cos1.0),5.1( ),0()2(cos1.0),0.2( txyttxy =−=π−== ; ),0.1(2 2 cos1.0),0.3( txyttxy =−=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π− π −== because the two points are 2 λ apart. ),0(2cos1.0),0.4( txyttxy ==π== because the two points are one wavelength apart. 7.2 For 5 2L =λ , t L x atxy πν⎟ ⎠ ⎞ ⎜ ⎝ ⎛ π = 2cos 5 sin2),( Thus the loops in the region 5 3 5 2 , 5 0 L x LL x <<<< and 5 4 5 3 L x L << will be vibrating in phase. The loops in the region 5 2 5 L x L << and 5 4 5 3 L x L << will also vibrate in phase but will be out of phase with adjacent loops. 7.3 If the acceleration due to gravity at the points B and Q are g and g΄ then g R OQ g OQ R g g =′⇒= ′ where we have assumed the earth to have uniform density. We assume the origin at the point P; the acceleration will be directed towards P and will have magnitude A O P B x y Q
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.6 y R g OQ y g R OQ g =⋅=θ′cos Thus the equation of the motion will be yy R g dt yd 2 02 2 ω−=−= and the time period will be g R T π= ω π = 2 2 0 . 7.4 m k tx dt xd kx dt xd m =ω=ω+⇒−= 0 2 02 2 2 2 ;0)( Thus k m T π= ω π = 2 2 0 . If T = 0.1 s, m = 1 g, 32 1095.3400 ×≈π=k dynes/cm cm25.0 dynes/cm1095.3 g.cm/s980 3 2 ≈ × ==∆⇒=∆ k mg xmgxk 7.5 Since ),(),( tzxtzy = , the string will vibrate along a direction making an angle 4 π with the x and y axes. 7.6 Since 222 ),(),( atzytzx =+ , each point on the string will rotate on the circumference of a circle (of radius a). Such a wave is known as a circularly polarized wave. 7.7 For Li 22 23 10617.4 94.6 534.0106 ×≈ ×× =N cm-3 ≈ 4.617 × 1028 m-3 Thus ( ) 1-16 2 1 1231 219282 1 0 2 s1021.1 10854.810109.9 10602.110617.4 ×≈ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ××× ××× =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ε =ω −− − m Nq p m1055.1 2 7− ×≈ ω π =λ p p c For K, 328 23 m10335.1 10.39 870.0106 − ×≈ ×× =N Thus m1089.2s10518.6 7115 −− ×≈λ⇒×≈ω pp For Rb, 128 23 m10075.1 48.85 532.1106 − ×≈ ×× =N Thus m1022.3s10849.5 7115 −− ×≈λ⇒×≈ω pp 7.8 If the steady state solution of the equation
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.7 ti eqqm dt d m ω− −=−=ν+ 0EEv v is written in the form ti e ω− = 0vv then 0)( Ev0 qmi −=ν+ω− Thus )()( 0 ω−ν −= ω−ν −= ω− im q e im q ti EE v and EvJ )( 2 ω−ν +=−= im Nq Nq Since EJ σ= , where σ is the conductivity, we get )( 2 ω−ν =σ im Nq (b) ti e im q dt d ω− ω−ν −== )( 0E v r we get )()( 0 ω−νω = ω−νω += ω− imi q e imi q ti EE r Thus EErP χ= ων+ω −=−== )( 2 2 im qN qN where )( 2 2 ων+ω −=χ im qN is the susceptibility. Thus )( 1 2 0 2 0 0 ων+ωε −= ε ε =κ⇒χ+ε=ε im qN 7.9 At low frequencies ν ≈σ m qN 2 23 23 1086.0 63 9106 ×≈ ×× =N atoms/cm3 = 0.86 ×1029 m–3 Thus ( ) 113 731 219292 s104 106101.9 106.11086.0 − − − ×≈ ××× ××× ≈ σ ≈ν m qN
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.8 Since )( 2 ω−ν =σ im qN , the complex part of σ can be neglected for ω << ν or ω < 1011 s−1 . Now )( 1 0 2 0 ν+ωωε −= ε ε =κ im qN or, )()( 1 22 0 2 22 0 2 ν+ωωε ν + ν+ωε −=κ m qN i m qN Thus the real part of κ is given by (for ω <<ν) 5 2 0 2 107.11 ×−≈ νε −≈κ m qN r If we write ir innn += then 22 irr nn −=κ Thus ni >>> nr and it is the imaginary part of the refractive index which dominates over the real part. 7.10 From Problem 7.8(b) 2 2 2 0 2 1 )( 1)( ω ω −≈ ων+ωε −=ωκ p im qN the last expression is valid for high frequencies when ω>>ν; here 0 2 ε =ω m qN p [see Eq.(76)]. If all the electrons are assumed to be free then (using Z = 29) 303 23 105.2cm 63 910629 ×≈ ××× ≈ − N m-3 Thus ( ) 16 1231 30 19 109 10854.8101.9 105.2 106.1 ×≈ ××× × ×=ω −− − p s-1 7.11 For metallic sodium the density is 0.97 g/cm3 , A ≈ 23 and Z = 11 [ see Sec. 7.5]; thus 329323 23 m1078.2cm1078.2 23 97.010611 −− ×=×≈ ××× =N Thus 116 s103 − ×≈ωp For λ = 1 Å = 10-10 m 119 10 8 s109.1 10 10322 − − ×≈ ××π = λ π =ω c Thus 12 ≈n and the metal will be completely transparent.
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.9 7.12 Consider the term ( )22 10 2 1 ω−ωε = m qN T At large frequencies it takes the value 2 10 2 1 ωε =∞ m qN T Thus ( ) ( ) 4 12 1 222 0 2 2 1 22 10 22 2 1 22 10 2 11 4 11 λ λ−λπε = ωω−ωε ω = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ω − ω−ωε =− ∞ cm qN m qN m qN TT Similarly we can evaluate the third term to get 2 2 2 0 22 4 2 2 2 1 2 0 22 4 1 2 21 2 1 4 1 4 1 λ−λ ⋅ επ λ + λ−λ ⋅ επ λ +++= ∞∞ Mc qpN mc qN TTn or 2 2 2 2 2 1 2 122 λ−λ + λ−λ += ∞ AA nn where ∞∞∞ ++= 21 2 1 TTn 2 2 2 2 1 1 2 22 0 2 2 2 12 0 2 2 2 20 2 2 10 2 1 44 1 1 λ + λ += λ επ +λ επ += ωε + ωε += AA cM qpN cm qN M qpN m qN and A1 and A2 are defined in the problem. 7.13 (b) m1042.9m1088.8 8 1 2152 1 −− ×≈λ⇒×=λ m1055.3m1026.1 5 2 292 2 −− ×≈λ⇒×=λ Now m M m M p M mA A 11 4 5 8 4 2 4 1 2 1 1048.2 1055.3 1042.9 2 − − − ×≈⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × × ⋅= λ ⋅ λ = From the data given 215 1 m1012.6 − ×=A and 29 2 m1010.5 − ×=A Thus 5 1 211 1007.21048.2 −− ×≈×= A A M m Now
  • OPTICS Ch07 on Simple Harmonic Motion Problems & Solutions 20Mar09 7.10 .108.2 78 1067.13840 101.9 11 5 27 31 Ca Ca 2 2 − − − ×≈⇒ ××× ⋅ × ≈ + = M m mm mm mm M F F (c) 2 0 2 4 2 2 22 0 2 4 1 2 1 4 and 4 cM qpN A cm qN A επ λ = επ λ = Thus ∞∞∞ ++= 21 2 1 TTn 73.5 1026.1 1010.5 1088.8 1012.6 1 1 9 9 15 15 2 2 2 2 1 1 ≈ × × + × × += λ + λ += − − − − AA which compares reasonably well with the experimental value of 6.09. 7.14 (a) 12 1 s4.56 − ≈ω Np For N = 1010 electrons/m3 16 s106.5 − ×≈ωp and for N = 1012 electrons/m3 17 s106.5 − ×≈ωp For TV waves, ν ≈ 500 MHz; thus for N = 1012 m−3 02.0 105002 106.5 6 7 ≈ ××π × ≈ ω ωp and 9996.01 2 2 2 ≈ ω ω −≈ p n For ω < ωp, the refractive index gets imaginary and the wave gets reflected [see Sec. 2.4.3]. (b) We write ( )2512 103102)( ×−α−×≈ xxN for 2 ×105 m δ x δ 4 ×105 m where N is measured in m−3 and x in meters. At x =200 km = 2 ×105 m, N = 1012 m−3 giving 51012 m1001010 − =α⇒×α= Thus ( )[ ]2512 2 0 2 2 1031021)( ×−α−×⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ωε −≈ x m q xn ; 2 ×105 m < x < 4 ×105 m ( )[ ]2511 2 15 1031051 104.6 1 ×−×− ω × −≈ − x where ω is measured in s-1 and x in m.