Apl 00 Numbe 7 Waes Bascs Ths Factsheet wll ntoduce wae defntons and basc popetes. Types of wae Waes may be mechancal (.e. they eque a medum such as o to popagate) o electomagnetc (whch popagate n a acuum) Waes can be classfed as: a) Tansese hee the dstubance s at ght angles to the decton of the wae. Examples nclude waes and lght (electomagnetc). wae decton Wae pulses and contnuous waes A wae pulse noles a shot o sngle dstubance of the medum t s taellng n. Fo example, doppng an object n may poduce a wae pulse. Contnuous waes nole epeated dstubances of the medum. Fo example, to poduce contnuous waes n a pple tank, the dppe would hae to be dpped nto the tank at egula nteals. The est of the Factsheet wll focus on contnuous waes. The wae fomula = f = elocty (ms - ) = waelength (m) f = fequency (Hz) Patcles n the wae bate at 90 o to wae decton b) Longtudnal hee the dstubance s n the same decton, paallel to the decton of the waes. Examples nclude sound and sesmc P waes. wae decton compesson expanson (aefacton) Patcles of the wae bate n the same decton as the wae Both longtudnal and tansese waes can be epesented gaphcally as shown below. Exam Hnt: You wll be expected to know examples of longtudnal and tansese waes and to descbe the dffeences between them. To see whee ths fomula comes fom, consde how fa the wae moes n one second. We know (fom the defnton of fequency) that thee ae f waes each second. Each wae s of length. So the total dstance moed n one second s f But elocty = dstance tme, so = f Typcal Exam Queston a) What s the elocty of a wae wth waelength 5cm and fequency Hz? [] 0.5 = = 3ms - b) A wae has a elocty of.5 ms -. Eght waes ae obseed to pass a fxed pont n seconds. Fnd ) the peod of the wae [] 8 = 0.5 ) the waelength of the wae [] f = /T = 4 = /f =.5/4 = 0.35 m. Glossay Ampltude: the maxmum dsplacement of a wae patcle fom ts undstubed poston. Waelenth: the dstance between two smla ponts on a wae. Unts: metes Fequency: the numbe of waes that pass a pont n one second. Unts: Hetz (Hz) Peod: the tme taken to complete one wae cycle. Unts: second (s) The peod (T) and the fequency (f) ae elated by T = f dsplacement ampltude waelength tough peak poston Peak(o cest): the pont of maxmum dsplacement the hghest pont Tough: the pont of mnmum dsplacement the lowest pont
Waes bascs Wae popetes All waes wll undego the followng pocesses accodng to the same laws: eflecton efacton ntefeence dffacton Reflecton and ts laws Thee ae two laws of eflecton. They apply to both plane (flat) and cued mos (o othe eflectng suface fo waes othe than lght). the and eflected ay le n a sngle plane whch s pependcula to the suface at the pont of ncdence angle of ncdence () s equal to angle of eflecton () (90 o to suface) eflected ay Refacte ndex Consde a wae passng fom mateal to mateal. In mateal ts angle to the s and n mateal t s. Fom Snell s Law, we know that sn = a constant sn Ths constant s the efacte ndex of mateal wth espect to mateal, wtten n. So Snell s Law can be wtten as: sn = n sn efacted ay mateal mateal NB: The angles ae always measued to the, NOT the eflectng suface Refacton Refacton s the change n decton of a wae as t cosses the bounday between two mateals (eg and ). Now suppose the ay s nstead taellng fom mateal to mateal. Usng the aboe equaton, we would obtan: sn = n sn Ths ges: n n bounday Wae speed, waelength and efacte ndex Refacte ndex s also elated the the wae speed and waelength n the two mateals. Ths s used to defne the efacte ndex. If the wae cossed the bounday n the opposte decton (.e. to ) then the wae decton would change n the opposte way: Refacton s goened by two laws: efacted ay bounday efacted ay the and eflected ay le n a sngle plane whch s pependcula to the suface at the pont of ncdence Snell s Law : at the bounday between any two gen mateals, the ato of the sne of the angle of ncdence to the sne of the angle of efacton s constant fo ays of any patcula waelength. n whee = elocty of wae n mateal = elocty of wae n mateal Ths defnes the efacte ndex. Also, snce fequency does not change dung efacton: n whee = waelength of wae n mateal = waelength of wae n mateal The efacte ndex s often gen elate to. So f, fo example, a queston tells you that the efacte ndex of glass s.50, t means that n glass =.50. Exam Hnt: Defnng the efacte ndex s commonly asked. Make sue you use the equaton nolng wae speeds, and defne all the tems.
Waes bascs Typcal Exam Queston The efacte ndex of s.33. The speed of lght n s 3 0 8 ms -. a) Calculate the speed of lght n. [].33 = 30 8 8 3 0 = =.60 8 ms -.33 b) State the effect of the efacton on the fequency [] none t s unchanged Tp: The speed of lght n any othe medum should always be lowe than n acuum (o ). Use ths to check you answe f you hae got too hgh a speed, you hae pobably got the and mateal the wong way up n the equaton Intefeence When two sets of waes combne, we use: Pncple of Supeposton The esultant dsplacement at a pont s equal to the ecto sum of the nddual dsplacements at that pont. To see how ths woks, we wll look at some examples:. Constucte Intefeence Wae + Wae Ctcal angle and total ntenal eflecton When lght taels fom a mateal wth a hghe efacte ndex to one wth a lowe efacte ndex (eg fom glass to ), t s possble fo the angle of efacton to be 90 o. Poduces In ths case, the peaks and toughs n the two waes concde, and hence enfoce each othe. glass efacted ay. Destucte Intefeence Wae + Wae The angle at whch ths occus s called the ctcal angle, c. sn sn c Usng the equaton = n, we fnd: glassn sn sn90 But snce we ae usually gen n glass, not glass n, t s moe useful to wte ths as: sn c sn90 n glass Snce sn90 o =, we hae: sn c n glass Gen that the efacte ndex of glass s.50, we can calculate: snc = 0. 667 c = 4.8 o.50 (3 SF) Note: ths cannot occu fo lght taellng fom to glass, snce the angle to the deceases when taellng n ths decton. What happens fo angles lage than the ctcal angle? If the angle of ncdence s geate than the ctcal angle, then the ay cannot be efacted nstead t s totally ntenally eflected eflected ay glass In fact, a cetan amount of eflecton wll always occu at the nteface, but fo angles geate than the ctcal angle, only eflecton can occu. The ncdent and eflected ay obey the laws of eflecton. 3 Poduces In ths case, the peaks n one wae concde wth the toughs n the othe, to poduce no esultant dsplacement the two cancel. 3. Wae + Wae Poduces In ths case, peaks and toughs do not exactly concde. Phase dffeence Phase dffeence s a way of measung how fa ahead one wae s of anothe (eg half a waelength, quate of a waelength etc). It s gen as an angle, wth a whole waelength coespondng to 360 o. So f one wae leads anothe by a quate of a waelength, ths s a phase dffeence of ¼ 360 o = 90 o. If the phase dffeence s zeo (example aboe), they ae n phase. If the phase dffeence s 80 o (= half a waelength), they ae completely out of phase (example ).
Waes bascs Souces of waes ae coheent f they mantan a constant phase dffeence and hae the same fequency (eg lases). A waefont s a lne o suface n the path of a wae moton on whch all the dstubances ae n phase. It s pependcula to the decton tael of the wae. Dffacton of lght and the sngle slt Dffacton of lght though a slt onto a sceen leads to the poducton of lght and dak fnges. The bghtness and wdth of the fnges can be epesented gaphcally: Typcal Exam Queston a) Explan what s meant by supeposton of waes. [] Waes concdng at a pont n space Dstubances add togethe,.e. supepose b) Dstngush between constucte and destucte ntefeence. [4] Constucte: waes n phase as they supepose Dstubances add to ge a lage ampltude Destucte: waes 80 o out of phase Dstubances cancel to ge zeo ampltude c) State the condtons necessay fo souces of waes to be coheent.[] Same fequency Constant phase elatonshp Dffacton When waes pass an edge of an obstacle, o though a gap, they spead out and change shape. The waelength, fequency and elocty emans constant. The extent of the speadng depends on the sze of the gap, as shown below: bghtness cental fnge The fnge patten has the followng popetes: It s symmetcal The cental bght fnge s much bghte than the othe fnges It s twce as wde as the othe fnges The bghtness (ntensty of lght) deceases wth dstance fom the cental fnge so the oute fnges ae the fantest. The poston of the dak fnges can be calculated usng: sn = w a whee = angle subtended n the cente (see dagam) = waelength w = slt wdth a =,, 3. (fnge numbe) Cental lght fnge st dak fnge nd lght fnge nd dak fnge Double slt dffacton Ths poduces a dffeence n ntensty wthn each bght fnge seen n the sngle slt patten. Ths esults fom ntefeence between lght fom one slt the othe ognal one-slt fnge Exam Hnt: When dawng dagams of dffacton, make sue you keep the spacng between the waefonts the same ths shows the fequency of the waes s unchanged. Appecable dffacton only occus f the gap s no bgge than the waelength of the wae. Vaaton n ntensty due to two slts The naowe the slt, the geate the dffacton fo a patcula waelength The longe the waelength fo a constant slt wdth, the geate the dffacton Ths ntefeence esults fom the fact that lght fom the dffeent slts taels a dffeent dstance to each a gen pont on the sceen; ths s efeed to as the path dffeence. 4
Waes bascs At the cente of the sceen (pont A), waes fom slts X and Y hae taelled the same dstance and theefoe ae n phase, and hence ntefee constuctely leadng to a bght fnge. Pogesse and statonay waes The waes dscussed so fa hae been pogesse.e. they moe n a patcula decton, tansfeng enegy along the decton the wae s taellng. In a statonay (o standng) wae, the wae does not moe n a patcula decton and enegy s stoed by the wae. slts Y X A sceen Statonay waes ae the esult of two pogesse waes of the same fequency taellng n opposte dectons along the same lne. The dagam below shows an example of a statonay wae. B Each pont on the we can oscllate between the two postons shown. As the dstance fom the cente changes, so does the path dffeence between the waes fom each slt. At pont B, fo example, the waes fom slt Y hae taelled substantally futhe than those fom slt X. N Oscllates between these ponts N N When the path dffeence becomes half the waelength, then destucte ntefeence occus, poducng a dak fnge. Futhe nceases n dstance fom the sceen nceases the path dffeence untl t becomes a whole waelength ths makes the waes n phase agan, so constucte ntefeence occus. Futhe nceases poduce a path dffeence of ½ waelengths gng destucte ntefeence agan. So the bght fnges ae poduced fom path dffeence m, and the dak fnges fom path dffeence (m + ½), whee s the waelength and m s any whole numbe. The fnge spacng between two adjacent bght fnges s gen by D y = d D = dstance to sceen; = waelength; d = slt spacng In between the lght and dak fnges, the ntefeence s not pefectly constucte o destucte, so the ntensty of the lght changes gadually. Polasaton Nomally, the oscllatons n a tansese wae may be n many dffeent dectons. Fo example, fo a wae taellng out of ths page towads you, the oscllatons wll be n the plane of the page, and could be left to ght, up and down, dagonally etc. Tansese waes may undego polasaton. A polased wae oscllates n one decton only. Longtudnal waes cannot be polased because the oscllatons ae aleady n one decton only. Exam Hnt: Ths s a key dffeence between longtudnal and tansese waes, and s often asked. Polased lght s most easly poduced usng a pece of Polaod (as used n sunglasses). Polaod woks by only allowng though lght whch oscllates n a patcula decton A The nodes (maked N) nee moe. These occu whee the two ognal waes ntefee destuctely The antnodes (maked A) ae ponts of maxmum dsplacement. They occu whee the two ognal waes ntefee constuctely. Table below compaes pogesse and statonay waes. Table. Pogesse and statonay waes Statonay wae Pogesse wae Stoes batonal enegy Tansmts batonal enegy Ampltude aes Ampltude s constant All ponts between any two adjacent nodes ae n phase Phase aes smoothly wth dstance along the path of the Nodes ae half a waelength apat; antnodes ae mdway between nodes Typcal Exam Queston a) Explan the tems node and antnode [] Node: pont of no baton Antnode: pont of maxmum baton wae No nodes o antnodes b) Two dentcal pogesse waes ae taellng along the same staght lne n opposte dectons. () Explan how a statonay wae patten s fomed [3] Statonay wae s fomed by the supeposton of the two waes Nodes ae ceated by destucte ntefeence and antnodes ae ceated by constucte ntefeence. () Compae the ampltude and phase of patcles along a statonay wae wth those of a pogesse wae. [] All ponts on a statonay wae ae n phase, ponts on a pogesse wae ae out of phase wth each othe All ponts on a pogesse wae hae the same ampltude, dffeent ponts on statonay wae hae dffeent ampltudes A If the lght s passed though a etcal pece of Polaod, then the emegng ay wll be polased etcally. It wll hae half the ntensty of the ognal beam ths s why sunglasses wok. If ths ay then meets a hozontal pece of Polaod, no lght wll pass though. 5
Waes bascs Exam Wokshop Ths s a typcal poo student s answe to an exam queston. The comments explan what s wong wth the answes and how they can be mpoed. The examne s answe s gen below. a) () Explan the tem 'wae font' [] t s at ght-angles to the wae decton 0/ () Ths statement s tue, but does not explan waefont. When the student ead the next pat of the queston, s/he should hae ealsed that ths was not an adequate answe. State the elatonshp between the oentaton of a wae font and the decton n whch the wae s taellng [] at ght angles / b) A longtudnal wae of fequency 30 khz has a speed of 340ms - when taellng n. Its waelength when taellng n s 0.05m. () Calculate the mnmum dstance between two ponts on the wae that dffe n phase by 60 o when t s taellng though.[3] 60 o = one sxth of waelength 0.05 6 = 0.008m /3 The fst pat of the method s coect, but the student has used the waelength fo waes. Read the queston! Ths shows the adantage of showng wokng wthout t, no maks would hae been awaded. () Calculate the speed of the wae n [].5ms - 0/ The student has pobably faled to conet khz to Hz. If s/he had shown wokng, one mak mght hae been awaded, snce s/he could hae demonstated the knowledge that fequency s unchanged. The answe should hae woed hm/he! () A pulse of the wae lasts fo 0ms. Calculate the numbe of complete waes that t contans. [] 30 000 0.0 = 300 / The student has edently now ealsed that the fequency s 30 khz, not 30 Hz, but has neglected to change the eale answe! Always make tme to check. Examne s Answes a) () A suface n whch all oscllatons ae n phase () They ae pependcula b) () = /f =340/30000 =0.03m Phase dffeence of coesponds to /6 =.89mm () speed n = f waelength n = 30 000 0.05 =500ms - () Numbe of waes = tme of pulse fequency = 0.0 30000 =300 Questons. a) Explan the dffeence between tansese and longtudnal waes. b) Ge two examples of tansese waes and two examples of longtudnal waes.. Explan the dffeence between a wae pulse and a contnuous wae. 3. Explan what s meant by the followng tems: Ampltude Waelength Peak Tough Fequency 4. Defne the efacte ndex fo a wae taellng fom mateal to mateal. 5. Explan what s meant by the Pncple of Supeposton. 6. Explan what s meant by dffacton. 7. Explan why sound waes cannot be polased. 8. Ge two dffeences between a statonay wae and a pogesse wae. 9. A wae has speed 50ms - and waelength m. Calculate ts peod. 0. The efacte ndex of glass s.50. a) A ay of lght passes fom to glass. It makes an angle of 0 o to the just befoe enteng the glass. Calculate the angle the efacted ay makes wth the. b) The speed of lght n s 3 0 8 ms -. Calculate the speed of lght n glass. c) Calculate the ctcal angle fo glass. Answes. See page. See page 3. See page 4. See page 5. See page 6. See page 4 7. See page 5 8. See page 5 9. f = 50/ = 5Hz T = /f = 0.04 s 0. a) sn0/sn =.5 sn = sn0/.5 = 0.8 = 3 o b) c / c glass =.50 c glass = 3 0 8 /.50 = 0 8 ms - c) snc = /.5 = 0.667 c = 4 o ( SF) Acknowledgements: Ths Factsheet was eseached and wtten by Nnde Hunjan Cuculum Pess, Unt 305B The Bg Peg, 0 Vyse Steet, Bmngham B8 6NF. Physcs Factsheets may be coped fee of chage by teachng staff o students, poded that the school s a egsteed subscbe. They may be netwoked fo use wthn the school. No pat of these Factsheets may be epoduced, stoed n a eteal system o tansmtted n any othe fom o by any othe means wthout the po pemsson of the publshe. ISSN 35-536 6