Answers to test yourself questions

Aswers to test yourself questios Optio C C Itroductio to imagig a The focal poit of a covergig les is that poit o the pricipal axis where a ray parallel to the pricipal axis refracts through, after passage through the les. b The focal legth is the distace of the focal poit from the middle of the les. I the les equatio this is take to be a egative umber. a A real image is a image formed by actual rays of light which have refracted through a les. b A virtual image is formed ot by actual rays but by the itersectio of their mathematical extesios. 3 If a scree is placed at the positio of a real image the actual rays of light that go through that image will be reflected off the scree ad so the image will be see o the scree. I the case of a virtual image, placig a scree at the positio of the image reveals othig as there are o rays of light to reflect off the scree. 4 No oe ca explai thigs better tha Feyma ad this case is o exceptio. Search for the YouTube video Feyma: FUN TO IMAGINE 6: The Mirror where Feyma explais the apparet left-right case for the mirror. The try to see what happes with a les. 5 The distace is the focal legth so 6.0 cm. 6 See graph show. 7 a The diagrams use a vertical scale of cm per lie ad a horizotal scale of cm per lie. u = 0 cm: The formula gives: v 0 = = = v = + 0 cm. Further M = = =. So the image is real (positive v), 0 cm v f u 0 0 0 u 0 o the other side of the les, ad the image is iverted (egative M) ad has height cm ( M = ). This is what the ray diagram below also gives. physics for the IB Diploma Cambridge Uiversity Press 05 ANSWERS TO TEST YOURSELF QUESTIONS C

b u = 0 cm: The formula gives: = = = 0 v =. The image is formed at ifiity. This is what the ray diagram gives. v f u 0 0 Rays do ot meet eve whe they are exteded. Image is said to form at ifiity. c u = 5.0 cm: The formula gives: v 0 = = = v = 0 cm. Further M = = = +. So the image is virtual (egative v), v f u 0 5.0 0 u 5.0 0 cm o the same side of the les, ad the image is upright (positive M) ad has height twice as large (i.e. 4 cm) (because M = ). This is what the ray diagram below also gives. 8 The diagram uses a vertical scale of cm per lie ad a horizotal scale of cm per lie. The formula gives: v 4 = = v = + 4 cm. Further M = = = 3. So the image is real (egative v), 4 cm o v f u 6.0 8.0 u 8.0 the other side of the les, ad the image is iverted (egative M ) ad has height 3 as large (i.e. 7.5 cm) (because M = 3). This is what the ray diagram above also gives. ANSWERS TO TEST YOURSELF QUESTIONS C physics for the IB Diploma Cambridge Uiversity Press 05

9 See graph show. The diagram uses a vertical scale of cm per lie ad a horizotal scale of cm per lie. The formula gives: v 4 = = = v = 4 cm. Further M = = = + 4. So the image is virtual v f u 8.0 6.0 4 u 6.0 (egative v), 4 cm o the same side of the les, ad the image is upright (positive M) ad has height 4 as large (i.e. 6 cm) (because M = 4). This is what the ray diagram below also gives. v 0 We kow that v > 0 (real image) ad M = = (same size). Hece v = u ad so u + = = u = f = 9.0 cm. u v f u f See the diagram that follows. The rays from the top of the object have bee draw gree ad those from the bottom blue for the sake of clarity. The top of the image will be formed at a distace from the les give by: = = v =.5 cm ad the bottom at a distace of v f u 5.0 9.0 = = v = 0.0 cm. v f u 5.0 0.0 The agle the object makes with the horizotal is ta 4 76. The image makes a agle give by 4.5 ta 75 so it is slightly smaller..5 physics for the IB Diploma Cambridge Uiversity Press 05 ANSWERS TO TEST YOURSELF QUESTIONS C 3

a Sice we kow that = + we should plot versus. We expect a straight lie with gradiet equal to f u v u v ad equal vertical ad horizotal itercepts equal to f. b The graph shows the data poits ad the (small) error bars. / v 0.06 0.05 0.04 0.03 0.0 0.0 0.0 0.04 0.06 0.08 / u 0.95 The lie of best fit is = 0.096. The itercepts are 0.096 ad 0.096 givig focal legths v u 0.95 0.096 = 0.096 f = 0.4 cm ad = f = 9.896 cm. So approximately, f = 0. ± 0.3 cm. f f 0.95 3 From the diagram it should be clear that rays must hit the mirror at right agles if they are to retur to the positio of the object. This meas that the distace of the object from the les whe the object ad image coicide is the focal legth. 4 ANSWERS TO TEST YOURSELF QUESTIONS C physics for the IB Diploma Cambridge Uiversity Press 05

v 60 4 a = = v = + 60 cm. Further M = = = 3. So the image is real (positive v), v f u 5 0 u 0 60 cm o the other side of the les, ad the image is iverted (egative M ) ad has height 3 times as large (because M = 3). This is what the ray diagram below also gives. b See graph show. 5 a We must have that u + v = 5 ad + = where distaces are i meters. The v = 50 u ad so u v 0.6 5 v + v =. This gives v 5v + 3 = 0 with solutios v = 4.30 m or v = 0.70 m. 0.6 4.30 0.70 b I the first case the magificatio is M = = 6. ad i the secod M = = 0.6 so the 0.70 4.30 magificatio is larger (i magitude) i the first case. 6 We use + = with f = 4.0 cm. Hece = u v f v 4.0 = 3.0. Hece v = 3.0 cm 3.0 ad M = =+. 4.0 The image is virtual (v < 0), upright ad smaller by a factor of 4 M = +. 4. 0 obj. F image F 7 Let u be the distace of a object from the first les. The les creates a image a distace v from the les which is give by + = hece =. This image serves as the virtual object i the secod les. Because u v f v f u the leses are thi the distace of this virtual object is also v. Hece the fial image is formed at distace of + = (the mius sig i frot of the first term is because the object is virtual) ad so f u v f = + v f f. This is what we would have obtaied if the iverse focal legth of the combiatio were u = +. Hece f = f f f f f. f + f physics for the IB Diploma Cambridge Uiversity Press 05 ANSWERS TO TEST YOURSELF QUESTIONS C 5

f 8 Usig f 0.0 4.0 f = we fid f = =.86 cm. f + f 4.0 9 a The image i the first les is foud from: + = + = v = 4.0 cm. This meas that the u v 0 40.0 v 5.0 distace of the image from the secod les is.0 cm. This image ow serves as the object for the secod les. So + = v =.0 cm. The fial image is.0 cm to the left of the secod les..0 v.0 b The overall magificatio is the product of the idividual les magificatios i.e. M = M M = 4. 0 =.. 40. 0 c Sice M < 0, the fial image is iverted relative to the origial object ad is. times larger. 0 a The image i the first les is foud from: + = + = v = 0 cm. This meas that the u v f 30.0 v 35.0 distace of the image from the secod les is 35 cm. This image ow serves as the object for the secod les. So + = v = 8.4 cm. The fial image is 8.4 cm to the left of the secod les. 35 v 0.0 b The overall magificatio is the product of the idividual les magificatios i.e. M = M M = 0 8. 4 = + 0. 548 0. 55. 30 35 c Sice M > 0, the fial image is upright relative to the origial object ad is 0.55 times smaller. 6.0 a The mirror formula gives + = = v = 6.0 cm. The magificatio is M u v f v 4.0 = 4.0 =+.5. Therefore the image is virtual, formed o the same side of the mirror as the object, upright ad.5 times taller. b Now the mirror formula gives + = = v = 3.0 cm. The magificatio is u v f v 4.0 3.0 M = 4.0 =+0.75. Therefore the image is virtual, formed o the same side of the mirror as the object, upright ad 0.75 times the object s height. c See graphs show. F obj. image obj. image F 6 ANSWERS TO TEST YOURSELF QUESTIONS C physics for the IB Diploma Cambridge Uiversity Press 05

v We are told that M = +.0 (image is upright so M > 0) hece.0 = v =.0u = 4 cm. Hece from the u mirror formula gives + = we get = f = 4 cm. Sice the focal legth is positive the mirror is u v f f 4 cocave. 3 a There are two mai les aberratios. I spherical aberratio rays that are far from the pricipal axis have a differet focal legth tha rays close to the pricipal axis. This results i images that are blurred ad curved at the edges. I chromatic aberratio, rays of differet wavelegth have slightly differet focal legths resultig i images that are blurred ad coloured. Spherical aberratio is reduced by oly allowig rays close to the pricipal axis to eter the les ad chromatic aberratio is reduced by combiig the les with a secod divergig les. b i The diagram shows (uder the simple coditios of this problem) that a differet focal legth (depedig o the distace of the paraxial rays from the pricipal axis) creates a image that is curved at the edges. ii The image draw with oe focal legth would be straight. ( f x)( f y) 4 We are told that u = f + x ad v = f + y. The f f + + x f + = + + =. Simplifyig, y f f + x + f + y f + fx + fy + xy = f ( f + x + y) = f + fx + fy xy = f 5 a The image is virtual so v = 5 cm. + = = u = 7.43 7. cm. u v 0.0 u 5 0.0 b At the focal poit of the les, 0 cm away. 5 5 c The agular magificatio i this case is M = = =.5 ad M = θ.6 0. Now θ f 0 θ 0.5 ad so θ.5 0.0064 = 0.06 rad. 6 a See graph show. 3 = 0.0064 rad h image D h F θ u F b The earest poit to the eye where the eye ca focus without straiig. c Whe the image is formed at the ear poit (5 cm away) we have that v = 5 cm. 5 + f 5 f Hece + = = + = u =. The magificatio is the u 5 f u f 5 5 f 5 + f v 5 5 f M = = + = + = + 5. u 5 f f f 5 + f physics for the IB Diploma Cambridge Uiversity Press 05 ANSWERS TO TEST YOURSELF QUESTIONS C 7

7 The magificatio is M 5 5 = + = + = 6.0. Hece if d is the distace of the two poits we must have that f 5.0 3 0. 0 = 6.0d ad so d =.0 0 5 m. C Imagig istrumetatio 8 a The image i the objective is formed at a distace give by: + = v =.7.7 cm. The.7.50 v 0.80 magificatio of the objective is the m0 =.4.50 =. b + = u = 3.45 3.4 cm u 5 4.0 5 c The magificatio of the eyepiece is + = 7.5. The overall magificatio is the.4 7.5 = 8.3. 4.0 9 a From 5 + v = 0 we get v = 00 mm. b From = we get u 65. 65 mm u 350 80 =. D c The overall magificatio is M = m m = o e v 30 See diagram below. 00 5 350 50 65. 350 = 5. 4 5. F o F o F e F e 3 The objective forms the first image at a distace v from the objective where 30 + v = 4 ad so v = 0 mm. The magificatio of the objective is 0 mo = 4.0 30 =. The overall magificatio is D 50 M = mo me = mo + f i.e. 30 = 4. 0 + ad so + 50 = 7.5 givig fe 38.5 38 mm f =. e 3 The blue lie through the middle of the eyepiece les is a costructio ray. f e e 8 ANSWERS TO TEST YOURSELF QUESTIONS C physics for the IB Diploma Cambridge Uiversity Press 05

33 a The fial image is formed at ifiity. b M f o.0 = 4 =. Hece fe 0.4 m fe f =. e 3.5 0 34 a The agular width of the moo is θ = 3.8 0 0.009 80 θ = = 0.57 0.53.) π 6 8 = 0.009 0.009 rad. (This is b The agular magificatio is M f o 3.6 = = = 30. The diameter of the image of the moo is the fe 0. 30 0.009 = 0.76 0.8 rad. 35 a The agular magificatio is M f o 80 = = = 4.0. fe 0 65 b The agle subteded by the buildig without a telescope is θ = = 0.060 rad ad so the agle subteded 500 by the image is θ = Mθ = 4 0.060 = 0.04 rad. 36 a M f o 67 = = =.3. fe 3.0 b f + f = 70 cm o e 37 The objective focal legth must be 57 cm. If the fial image is formed at ifiity, it meas that the image i the objective is formed at a distace of 3.0 cm from the eyepiece i.e. 6.5 3.0 = 58.5 cm from the objective les. Hece + = u = 3 cm m. u 58.5 57.0 38 a A techique i radio astroomy i which radio waves emitted by distat sources are observed by a array of radio telescopes which combie the idividual sigals ito oe. λ 0. 5 b θ. =. 3 =.0 0 rad. b 5 0 5 c The smallest agular separatio that ca resolved is.0 0 rad. The smallest distace is therefore 5 7 0.0 0 = 0 m. 39 The uiverse is full of sources that emit at all parts of the electromagetic spectrum ot just optical light. 40 They do ot suffer from spherical aberratios. 4 Advatages: free of atmospheric turbulece ad light pollutio; o atmosphere to absorb specific wavelegths Disadvatages: expesive to put i orbit; expesive to service. C3 Fibre optics 8 c c 3 0 8 4 = cm = = =.07 0 m s. cm.45 43 a The pheomeo i which a ray approachig the boudary of two media reflects without ay refractio takig place. b Critical agle is that agle of icidece for which the agle of refractio is 90. c The critical agle is foud from siθc = si 90 siθc =. Sice the sie of a agle caot exceed we must have < for the critical agle to exist. So total iteral reflectio is a oe way pheomeo..46 44 siθc = si 90 siθc = = θc = 76.7..50 physics for the IB Diploma Cambridge Uiversity Press 05 ANSWERS TO TEST YOURSELF QUESTIONS C 9

45 a We kow that: siθ C = si 90 hece siθ C =. The cosθc = si θc = = b From the diagram, si A = si a. But a = 90 θc ad so si a = si(90 θc) = cosθc. Hece. si A = cosθ = C =. c A = arcsi = arcsi.50.40 = 3.6. 46 A = arcsi = arcsi.5.44 = 9.. 47 We must have.4 = =.7367.74. 48 It has to be exceptioally pure. 49 a Dispersio is the pheomeo i which the speed of a wave depeds o wavelegth. This meas that the differet wavelegth compoets of a beam of light will take differet times to travel the same distace. b Material dispersio is the dispersio discussed i (a). Waveguide dispersio has to do with rays of light followig differet paths i a optic fibre ad hece takig differet times to arrive at their destiatio. 8 c c 3 0 8 8 50 a = cm = = =.9737 0.97 0 m s. cm.5 b The shortest time will be for a ray that travels dow the legth of the fibre o a straight lie of legth 3 8.0 0 5 8.0 km, i.e. the time of travel will be 8 = 4.05 0 s. The logest time of travel will be for.9737 0 that ray that udergoes as may iteral reflectios as possible. The legth of the path travelled is the 3 8.0 8.0786 0 5 = 8.0786 8.08 km (see diagram) ad so the time is 8 = 4.09 0 s. si 8.9737 0 8 x = y/si 8 y 5 The height of the pulses will be less ad the width of the pulses greater. 5 a A moomode optic fibre is a fibre with a very thi core so that effectively all rays eterig the fibre follow the same path. I a multimode fibre (which is thicker tha a moomode fibre) rays follow very may paths of differet legth i gettig to their destiatio. b The trasitio from multimode to moomode fibres offers a very large icrease i badwidth. As discussed also i questio 3, dispersio limits the maximum frequecy that ca be trasmitted ad hece the badwidth. A very small diameter moomode fibre will suffer the least from modal dispersio (ad hece the distortio ad wideig of the pulse) ad material dispersio is also miimised by usig lasers rather tha LED s. Hece the badwidth is icreased as the moomode fibre diameter is decreased ad laser light is used. 0 ANSWERS TO TEST YOURSELF QUESTIONS C physics for the IB Diploma Cambridge Uiversity Press 05

53 Advatages iclude: (i) the low atteuatio per uit legth which meas that a sigal ca travel large distaces before amplificatio (ii) icreased security because the sigal ca be ecrypted ad the trasmissio lie itself caot easily be tampered with (iii) large badwidth ad so a large iformatio carryig capacity (iv) ot susceptible to oise (v) they are thi ad light ad (vi) do ot radiate so there is o crosstalk betwee lies that are close to each other. 54 The mai cause of atteuatio i a optic fibre is scatterig of light off impurities i the glass makig up the core of the fibre. 55 Let P i be the power i to the first amplifier. The the power out of the first amplifier is P = P 0 0. This is iput to the secod amplifier so its output is Pout = Pi Pi = + 0 0 0 that the gai overall is G + G. Pout 56 The power loss is 0 log = 0 log 3.0 =.58 db. P 4.60 i G G G G G + G 0 0 0 0 0 i G = 0 Pi showig Pout 57 The power loss is 0 log = 0 log 5.0 =.67 db. So the loss per km is.67 = 0.087 dbkm. Pi 8.40 5 58 The power loss whe the power falls to 70% of the origial iput power is Pout P 0 log = 0 log 0.70 =.55 db. So, L =.55 L = 0.3 km. Pi P 59 There is o overall gai i power sice + 5 = 3.0 db. Let the iput power be P. The the output power is G P P 0 0 0.3 = = P 0.0P. 60 There is o overall gai or loss i power sice + 7 0 + 3 = 0 db. So the output power is the same as the iput power, the ratio is. P P 6 The overall gai is out 0 log = 0 log = 0 log 3.0 db. Hece + G 6.0 = 3.0 db givig G = db. 6 a See graph. atteuatio per uit legth/db km.5.0.5.0 0.5 0 900 000 00 00 300 400 500 600 700 wavelegth/m P i P b The atteuatio per uit legth is least for log wavelegths, i particular 30 m ad 550 m, ad these are ifrared wavelegths. physics for the IB Diploma Cambridge Uiversity Press 05 ANSWERS TO TEST YOURSELF QUESTIONS C

C4 Medical Imagig 63 a Atteuatio is the loss of eergy i a beam as it travels through a medium. b For X-rays the mai mechaism for atteuatio is the photoelectric effect i which X-ray photos kock electros off the medium s atoms, losig eergy i the process. This effect is depedet o the medium atoms atomic umber Z. This meas that media with differet Z have differet atteuatio which allows a image of the boudary of the two media to be made. 64 The image ca be formed faster by usig itesifyig scree. This is doe by havig X-rays that have goe through the patiet strike a scree cotaiig fluorescet crystals which the emit visible light. The visible light helps form the image o photographic film faster. 65 If eighbourig orgas have the same atomic umber the boudary of the orgas will ot appear clearly. By havig the patiet swallow a barium meal, the atomic umber of orgas such as the stomach or the itestie is ow greater ad ca be distiguished from its surroudig tissue. 66 The blurry images are caused by X-rays that have scattered i the patiet s body ad thus ow deviate from their origial directios. This may be miimised by placig lead strips betwee the patiet ad the film, alog the directio of the icidet X-ray beam. I this way cattred X-rays will be blocked by the strip ad ot fall o the film. 67 a For the top curve the HVT is 6.0 mm ad for the other it is about 4.0 mm. b The larger eergy correspods to the curve with the loger HVT. 68 a The HVT is about 5.0 mm ad so the lier atteuatio coefficiet is about l = 0.39 0.4 mm. 5.0 b The trasmitted itesity must be 0% of the icidet ad from the graph this correspods to a legth of about.5 mm. 69 0.60 = e µ 4 ad so µ= l 0.6 = 0.8 mm. The, 0.0 = e µ x ad so 4 x = l 0. = l 0. =.6 3 mm. µ 0.8 70 µ = = 3 l 0.3 mm 0.3 ad so I = I 0e = 0.794I 0 0.8I 0. 7 It meas that as the beam moves through the metal the proportio of the total eergy of the X-rays carried by high eergy photos icreases. This is because the low eergy photos get absorbed leavig oly the high eergy photos move through. For the 0 kev photos the trasmitted itesity is I 0 = I 0e. 5 = 0.07I 0. For the 5 l kev photos it is I 5 I 0e.8 = 5 = 0.90I 0. Hece I 5 0.90I 0 = =.4. I 0 0.07I 0 7 Ultrasoud is soud of frequecy higher tha about 0 khz. It is produced by applyig a alteratig voltage to certai crystals which vibrate as a result emittig ultrasoud. v 540 4 73 The wavelegth of this ultrasoud is λ = = 6 = 3. 0 m = 0.3 mm. The order of magitude of the f 5 0 size that ca be resolved is of the order of the wavelegth ad so about 0.3 mm. l 74 a Impedace is the product of the desity of a medium ad the speed of soud i that medium. 6 Z.4 0 3 b Z = ρc c = = =.5 0 m s. ρ 940 4ZZ 75 a The fractio of the trasmitted itesity is give by ad i this case equals 6 4 40.6 0 ( Z + Z ) 3 6.0 0. (40 +.6 0 ) b This is a very small fractio of the icidet itesity ad ot eough to be useful for diagostic purposes. More itesity has to be trasmitted which is why the gel like substace is put i betwee the ski ad the trasducer; the gel has a itesity closer to the tissue s so more itesity gets trasmitted. ANSWERS TO TEST YOURSELF QUESTIONS C physics for the IB Diploma Cambridge Uiversity Press 05

76 I the A the sigal stregth may be coverted to a dot whose brightess is proportioal to the sigal stregth. We ow imagie a series of trasducers alog a area of the body. Puttig together the images (as dots) of each trasducer forms a two-dimesioal image of the surfaces that cause reflectio of the ultrasoud pulses. This creates a B sca. 77 The differece i eergy for spi up ad dow states depeds o the magetic field stregth; oly those protos i regios where the magetic field has the right value will be absorbed ad so it is possible to determie where these photos have bee emitted from. This is achieved by exposig the patiet to a additioal o-uiform field, the gradiet field. 78 I this imagig techique, the patiet is ot exposed to ay harmful radiatio. The method is based o the fact that protos have a property called spi. The proto s spi will alig parallel or ati-parallel to a magetic field ad the eergy of the proto will deped o whether its spi is up (i.e. parallel to the magetic field) or dow (opposite to the field). The state with spi up has a lower eergy tha that of spi dow. The differece i eergy depeds o the magetic field stregth. The patiet is placed i a eclosure that creates a very uiform magetic field throughout the body. A source of radio frequecy forces protos with spi up to make a trasitio to a state with spi dow. As soo as this happes the protos will make a trasitio back dow to the spi up state emittig a photo i the process. The idea is to detect these photos ad correlate them with the poit from which they were emitted. This is doe with the help of a gradiet field as discussed i the previous questio. physics for the IB Diploma Cambridge Uiversity Press 05 ANSWERS TO TEST YOURSELF QUESTIONS C 3