meteorology-第20节

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these　matters　belongs　to　the　account　of　the　senses；　which　are　the



proper　subjects　of　such　an　inquiry；　we　need　only　state　about　them　what



is　necessary　for　us。　At　all　events；　that　is　the　reason　why　distant



objects　and　objects　seen　in　a　mirror　look　darker　and　smaller　and



smoother；　why　the　reflection　of　clouds　in　water　is　darker　than　the



clouds　themselves。　This　latter　is　clearly　the　case：　the　reflection



diminishes　the　sight　that　reaches　them。　It　makes　no　difference　whether



the　change　is　in　the　object　seen　or。　in　the　sight；　the　result　being　in



either　case　the　same。　The　following　fact　further　is　worth　noticing。



When　there　is　a　cloud　near　the　sun　and　we　look　at　it　does　not　look



coloured　at　all　but　white；　but　when　we　look　at　the　same　cloud　in　water



it　shows　a　trace　of　rainbow　colouring。　Clearly；　then；　when　sight　is



reflected　it　is　weakened　and；　as　it　makes　dark　look　darker；　so　it



makes　white　look　less　white；　changing　it　and　bringing　it　nearer　to



black。　When　the　sight　is　relatively　strong　the　change　is　to　red；　the



next　stage　is　green；　and　a　further　degree　of　weakness　gives　violet。　No



further　change　is　visible；　but　three　completes　the　series　of　colours



（as　we　find　three　does　in　most　other　things）；　and　the　change　into



the　rest　is　imperceptible　to　sense。　Hence　also　the　rainbow　appears



with　three　colours；　this　is　true　of　each　of　the　two；　but　in　a　contrary



way。　The　outer　band　of　the　primary　rainbow　is　red：　for　the　largest



band　reflects　most　sight　to　the　sun；　and　the　outer　band　is　largest。



The　middle　band　and　the　third　go　on　the　same　principle。　So　if　the



principles　we　laid　down　about　the　appearance　of　colours　are　true　the



rainbow　necessarily　has　three　colours；　and　these　three　and　no



others。　The　appearance　of　yellow　is　due　to　contrast；　for　the　red　is



whitened　by　its　juxtaposition　with　green。　We　can　see　this　from　the



fact　that　the　rainbow　is　purest　when　the　cloud　is　blackest；　and　then



the　red　shows　most　yellow。　（Yellow　in　the　rainbow　comes　between　red



and　green。）　So　the　whole　of　the　red　shows　white　by　contrast　with　the



blackness　of　the　cloud　around：　for　it　is　white　compared　to　the　cloud



and　the　green。　Again；　when　the　rainbow　is　fading　away　and　the　red　is



dissolving；　the　white　cloud　is　brought　into　contact　with　the　green　and



becomes　yellow。　But　the　moon　rainbow　affords　the　best　instance　of　this



colour　contrast。　It　looks　quite　white：　this　is　because　it　appears　on



the　dark　cloud　and　at　night。　So；　just　as　fire　is　intensified　by



added　fire；　black　beside　black　makes　that　which　is　in　some　degree



white　look　quite　white。　Bright　dyes　too　show　the　effect　of　contrast。



In　woven　and　embroidered　stuffs　the　appearance　of　colours　is



profoundly　affected　by　their　juxtaposition　with　one　another　（purple；



for　instance；　appears　different　on　white　and　on　black　wool）；　and



also　by　differences　of　illumination。　Thus　embroiderers　say　that　they



often　make　mistakes　in　their　colours　when　they　work　by　lamplight；



and　use　the　wrong　ones。



　　We　have　now　shown　why　the　rainbow　has　three　colours　and　that　these



are　its　only　colours。　The　same　cause　explains　the　double　rainbow　and



the　faintness　of　the　colours　in　the　outer　one　and　their　inverted



order。　When　sight　is　strained　to　a　great　distance　the　appearance　of



the　distant　object　is　affected　in　a　certain　way：　and　the　same　thing



holds　good　here。　So　the　reflection　from　the　outer　rainbow　is　weaker



because　it　takes　place　from　a　greater　distance　and　less　of　it



reaches　the　sun；　and　so　the　colours　seen　are　fainter。　Their　order　is



reversed　because　more　reflection　reaches　the　sun　from　the　smaller；



inner　band。　For　that　reflection　is　nearer　to　our　sight　which　is



reflected　from　the　band　which　is　nearest　to　the　primary　rainbow。　Now



the　smallest　band　in　the　outer　rainbow　is　that　which　is　nearest；　and



so　it　will　be　red；　and　the　second　and　the　third　will　follow　the　same



principle。　Let　B　be　the　outer　rainbow；　A　the　inner　one；　let　R　stand



for　the　red　colour；　G　for　green；　V　for　violet；　yellow　appears　at　the



point　Y。　Three　rainbows　or　more　are　not　found　because　even　the



second　is　fainter；　so　that　the　third　reflection　can　have　no　strength



whatever　and　cannot　reach　the　sun　at　all。　（See　diagram。）







　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　5







　　The　rainbow　can　never　be　a　circle　nor　a　segment　of　a　circle



greater　than　a　semicircle。　The　consideration　of　the　diagram　will　prove



this　and　the　other　properties　of　the　rainbow。　（See　diagram。）



　　Let　A　be　a　hemisphere　resting　on　the　circle　of　the　horizon；　let



its　centre　be　K　and　let　H　be　another　point　appearing　on　the　horizon。



Then；　if　the　lines　that　fall　in　a　cone　from　K　have　HK　as　their　axis；



and；　K　and　M　being　joined；　the　lines　KM　are　reflected　from　the



hemisphere　to　H　over　the　greater　angle；　the　lines　from　K　will　fall



on　the　circumference　of　a　circle。　If　the　reflection　takes　place　when



the　luminous　body　is　rising　or　setting　the　segment　of　the　circle　above



the　earth　which　is　cut　off　by　the　horizon　will　be　a　semi…circle；　if



the　luminous　body　is　above　the　horizon　it　will　always　be　less　than　a



semicircle；　and　it　will　be　smallest　when　the　luminous　body　culminates。



First　let　the　luminous　body　be　appearing　on　the　horizon　at　the　point



H；　and　let　KM　be　reflected　to　H；　and　let　the　plane　in　which　A　is；



determined　by　the　triangle　HKM；　be　produced。　Then　the　section　of　the



sphere　will　be　a　great　circle。　Let　it　be　A　（for　it　makes　no　difference



which　of　the　planes　passing　through　the　line　HK　and　determined　by



the　triangle　KMH　is　produced）。　Now　the　lines　drawn　from　H　and　K　to　a



point　on　the　semicircle　A　are　in　a　certain　ratio　to　one　another；　and



no　lines　drawn　from　the　same　points　to　another　point　on　that



semicircle　can　have　the　same　ratio。　For　since　both　the　points　H　and



K　and　the　line　KH　are　given；　the　line　MH　will　be　given　too；



consequently　the　ratio　of　the　line　MH　to　the　line　MK　will　be　given



too。　So　M　will　touch　a　given　circumference。　Let　this　be　NM。　Then　the



intersection　of　the　circumferences　is　given；　and　the　same　ratio　cannot



hold　between　lines　in　the　same　plane　drawn　from　the　same　points　to　any



other　circumference　but　MN。



　　Draw　a　line　DB　outside　of　the　figure　and　divide　it　so　that



D：B=MH：MK。　But　MH　is　greater　than　MK　since　the　reflection　of　the



cone　is　over　the　greater　angle　（for　it　subtends　the　greater　angle　of



the　triangle　KMH）。　Therefore　D　is　greater　than　B。　Then　add　to　B　a　line



Z　such　that　B＋Z：D=D：B。　Then　make　another　line　having　the　same　ratio　to



B　as　KH　has　to　Z；　and　join　MI。



　　Then　I　is　the　pole　of　the　circle　on　which　the　lines　from　K　fall。　For



the　ratio　of　D　to　IM　is　the　same　as　that　of　Z　to　KH　and　of　B　to　KI。　If



not；　let　D　be　in　the　same　ratio　to　a　line　indifferently　lesser　or



greater　than　IM；　and　let　this　line　be　IP。　Then　HK　and　KI　and　IP　will



have　the　same　ratios　to　one　another　as　Z；　B；　and　D。　But　the　ratios



between　Z；　B；　and　D　were　such　that　Z＋B：D=D：　B。　Therefore



IH：IP=IP：IK。　Now；　if　the　points　K；　H　be　joined　with　the　point　P　by　the



lines　HP；　KP；　these　lines　will　be　to　one　another　as　IH　is　to　IP；　for



the　sides　of　the　triangles　HIP；　KPI　about　the　angle　I　are



homologous。　Therefore；　HP　too　will　be　to　KP　as　HI　is　to　IP。　But　this



is　also　the　ratio　of　MH　to　MK；　for　the　ratio　both　of　HI　to　IP　and　of



MH　to　MK　is　the　same　as　that　of　D　to　B。　Therefore；　from　the　points



H；　K　there　will　have　been　drawn　lines　with　the　same　ratio　to　one



another；　not　only　to　the　circumference　MN　but　to　another　point　as



well；　which　is　impossible。　Since　then　D　cannot　bear　that　ratio　to



any　line　either　lesser　or　greater　than　IM　（the　proof　being　in　either



case　the　same）；　it　follows　that　it　must　stand　in　that　ratio　to　MI



itself。　Therefore　as　MI　is　to　IK　so　IH　will　be　to　MI　and　finally　MH　to



MK。



　　If；　then；　a　circle　be　described　with　I　as　pole　at　the　distance　MI　it



will　touch　all　the　angles　which　the　lines　from　H　and　K　make　by　their



reflection。　If　not；　it　can　be　shown；　as　before；　that　lines　drawn　to



different　points　in　the　semicircle　will　have　the　same　ratio　to　one



another；　which　was　impossible。　If；　then；　the　semicircle　A　be



revolved　about　the　diameter　HKI；　the　lines　reflected　from　the　points



H；　K　at　the　point　M　will　have　the　same　ratio；　and　will　make　the



angle　KMH　equal；　in　every　plane。　Further；　the　angle　which　HM　and　MI



make　with　HI　will　always　be　the　same。　So　there　are　a　number　of



triangles　on　HI　and　KI　equal　to　the　triangles　HMI　and　KMI。　Their



perpendiculars　will　fall　on　HI　at　the　same　point　and　will　be　equal。



Let　O　be　the　point　on　which　they　fall。　Then　O　is　the　centre　of　the



circle；　half　of　which；　MN；　is　cut　off　by　the　horizon。　（See　diagram。）



　　Next　let　the　horizon　be　ABG　but　let　H　have　risen　above　the



horizon。　Let　the　axis　now　be　HI。　The　proof　will　be　the　same　for　the



rest　as　before；　but　the　pole　I　of　the　circ