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# Solution i ph o 17

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### Solution i ph o 17

1. 1. Answers Question 1 (i) Vector Diagram If the phase of the light from the first slit is zero, the phase from second slit is θ λ π φ sin 2 d= Adding the two waves with phase difference φ where       −= λ πξ x ft2 , ( ) ( ){ }βξβξφξ φξφξφξ +=++ +=++ coscos2)cos()cos( 2/)2/cos(2)cos()cos( aaa aaa This is a wave of amplitude βcos2aA = and phase β. From vector diagram, in isosceles triangle OPQ, θ λ π φβ sin 2 1 d== )2( βφ =NB and .cos2 βaA = Thus the sum of the two waves can be obtained by the addition of two vectors of amplitude a and angular directions 0 and φ. (ii) Each slit in diffraction grating produces a wave of amplitude a with phase 2β relative to previous slit wave. The vector diagram consists of a 'regular' polygon with sides of constant length a and with constant angles between adjacent sides. Let O be the centre of circumscribing circle passing through the vertices of the polygon. Then radial lines such as OS have length R and bisect the internal angles of the polygon. Figure 1.2. 1 Figure 1.2 d θ θ
2. 2. φ φ = −== SOT OTSTSO ^ ^^ and )180( 2 1 In the triangle TOS, for example βφ sin2)2/sin(2 RRa == as )2( βφ = βsin2 a R =∴ (1) As the polygon has N faces then: βφ NNZOTNZOT 2)( ^^ === Therefore in isosceles triangle TOZ, the amplitude of the resultant wave, TZ, is given by βNR sin2 . Hence form (1) this amplitude is β β sin sin Na Resultant phase is ( ) ( ) β φ φ φ )1( 1 2 1 180 2 1 2 90 ^^ ^ −= −− −−      − −= = N N N ZTOSTO STZ (iii) 2 Intensity β β 2 22 sin sin Na I = β β Na sin sin 1 β
3. 3. 37C (iv) For the principle maxima ........210where ±±== ppπβ 'and0' ' ' 222 max βπββ β β +===      = paN N aI (v) Adjacent max. estimate I1 : NN pN 2 3 i.e 2 3 2,1sin 2 π β π πββ ±===        ±= N p 2 π πβ does not give a maximum as can be observed from the graph. 23 2 3 1 22 2 2 1 Na n aI == π for N>>1 Adjacent zero intensity occurs for N π πρβ ±= i.e. N π δ ±= For phase differences much greater than sin sin aI, 22 a N =      = β β δ . (vi) θ λ θ λθθ λ πθ λ π πβ cos cos w.r.t,atingDifferenti ..........2,1,0sini.e. maximumprincipleafor d n nd nnd n ∆ =∆ ∆=∆ ±±== = Substituting m.102.1and2nnm.589.6nm,0.589 6− ×===∆+= dλλλ 2 1       − ∆ =∆ d n d n λ λ θ as 2 1cosandsin       −== d n d n λ θ λ θ 03 30.0or102.5 rads− ×=∆⇒ θ 3 I 0 π 2π 3π β
4. 4. v R tR θ θ sin2 sin2EX =∴= where v = vP for P waves and v = vS for S waves. This is valid providing X is at an angular separation less than or equal to X', the tangential ray to the liquid core. X' has an angular separation given by, from the diagram, ,cos22 1       = − R RC φ Thus ,cosfor, sin2 1       ≤= − R R v R t C θ θ where v = vP for P waves and v = vS for shear waves. (ii) 3.831.0and5447.0 == P CPC v v R R From Figure 2.2 4 θ θ R O R φ E X X’ 2.(i) r Figure2.1 Figure 2.2 α O X E BA C Rc θ i Answers Q2
5. 5. )1()90( ^^ αθθ −+−=⇒+= rAOECOA (1) 5
6. 6. (ii) Continued Snell’s Law gives: . sin sin CPv v r i P = (2) From the triangle EAO, sine rule gives . sinsin i R x RC = (3) Substituting (2) and (3) into (1)               −+      −= −− i R R ii v v C P CP sinsinsinsin90 11 θ (4) (iii) Plot of θ against i. Substituting into 4: i = 0 gives θ = 90 i = 90° gives θ = 90.8° 6 For Information Only For minimum 0, = di dθ θ . 0 sin1 cos sin1 cos 1 22 =       −       −       −       −⇒ i R R i R R i v v i v v C C P CP P CP Substituting i = 55.0o gives LHS=0, this verifying the minimum occurs at this value of i. Substituting i = 55.0o into (4) gives θ = 75.80 . 0 550 900 900 75.80
7. 7. Substituting numerical values for i = 0 → 90° one finds a minimum value at i = 55° ; the minimum values of 0, θMIN = 75•8°. Physical Consequence As θ has a minimum value of 75•8° observers at position for which 2 θ <151•6° will not observe the earthquake as seismic waves are not deviated by angles of less than 151•6°. However for 2 θ ≤ 114° the direct, non-refracted, seismic waves will reach the observer. (iv) Using the result v r t θsin2 = the time delay Δt is given by       −=∆ PS vv Rt 11 sin2 θ Substituting the given data θsin 85.10 1 31.6 1 )6370(2131       −= Therefore the angular separation of E and X is o 84.172 =θ This result is less than oC R R 114 6370 3470 cos2cos2 11 =      =      −− 7 0 cos min 1 θθ       = − R RC C 900 9001800 Refracted waves θ
8. 8. And consequently the seismic wave is not refracted through the core. 8
9. 9. (v) The observations are most likely due to reflections from the mantle-core interface. Using the symbols given in the diagram, the time delay is given by symmetrybyEXEDas 11 )ED(2' 11 )DXED(' =      −=∆       −+=∆ PS PS vv t vv t In the triangle EYD, 1cossincos2(ED) )cos()sin((ED) 22222 222 =+−+= −+= θθθ θθ CC C RRRR RRR Therefore       −−+=∆ PS CC vv RRRRt 11 cos22' 22 θ Using (ii) 37s6mor7.396 cos2 sin ' 22 s RRRR R t t CC ⇒ −+ ∆ =∆ θ θ Thus the subsequent time interval, produced by the reflection of seismic waves at the mantle core interface, is consistent with angular separation of 17.840 . 9 E XY R R D θ θ O RC
10. 10. Answer Q3 Equations of motion: )()( )()( )()( 32312 3 2 21232 2 2 13122 1 2 uukuuk dt ud m uukuuk dt ud m uukuuk dt ud m −+−= −+−= −+−= :andcos)0()(ngSubstituti 2 m k tutu onn == ωω (c)0)0()2()0()0( (b)0)0()0()2()0( (a)0)0()0()0()2( 3 22 2 2 1 2 3 2 2 22 1 2 3 2 2 2 1 22 =−+−− =−−+− =−−− uuu uuu uuu ooo ooo ooo ωωωω ωωωω ωωωω Solving for u1(0) and u2(0) in terms of u3(0) using (a) and (b) and substituting into (c) gives the equation equivalent to 0)3( 2222 =− ωωωo 22 3 oωω = , 2 3 oω and 0 oo ωωω 3,3= and 0 (ii) Equation of motion of the n’th particle: )()( )()( 1 2 12 2 112 2 nnonn n nnnn n uuuuk dt ud uukuuk dt ud m −+−= −+−= −+ −+ ω Substituting t N nsutu snn ω π cos2sin)0()(       = θθ π ωω πππ ω π ω πππ ω π ω πππ ω π ω 2cos1sin2As ),.....2,1(:2cos12 2sin2cos2sin22sin )1(2sin 2 1 2sin)1(2sin 2 1 22sin )1(2sin2sin2)1(2sin2sin 2 22 22 22 22 −= =            −=∴             −            =            −             −−      +      +=            −             −+      −      +=            − Ns N s N ns N s N ns N ns N sn N ns N sn N ns N sn N ns N sn N ns os os os os This gives ),...2,1sin2 N(s N s os =      = π ωω . 2 to1rangetoingcorrespond;when22to0fromvalueshavecan N sN m k os =∞→=ωω 10 n = 1,2……N
11. 11. (iv) For s’th mode       +       = + N sn N ns u u n n π π )1(2sin 2sin 1             +                  = + N s N ns N s N ns N ns u u n n ππππ π 2sin2cos2cos2sin 2sin 1 (a) For small 1soand,02sinand12costhus,0, 1 ≅≈      =≅      ≈      +n n u u N ns N ns N s ππ ω . (b) The highest mode, oωω 2max = , corresponds to s = N/2 as1 1 −=∴ +n n u u ( ) ( ) 1 )1(2sin 2sin −= + π π n n Case (a) Case (b) N odd N even 11
12. 12. (vi) If m' << m, one can consider the frequency associated with m' as due to vibration of m' between two adjacent, much heavier, masses which can be considered stationary relative to m'. The normal mode frequency of m', in this approximation, is given by ' 2 ' 2 ' 2' 2 m k m k kxxm = = −= ω ω  For small m', ω' will be much greater than ωmax, DIATOMIC SYSTEM More light masses, m', will increase the number of frequencies in region of ω' giving a band- gap-band spectrum. 12 m m m 0 2ωo ω ω BAND BAND GAP BAND 0 ω