Seconde \u2014 Fibroscopie, r\u00e9flexion et r\u00e9fraction<\/title>\n<style>\n:root{\n --deep:#0f172a;\n --blue:#2563eb;\n --cyan:#0891b2;\n --teal:#14b8a6;\n --purple:#7c3aed;\n --pink:#db2777;\n --orange:#f97316;\n --red:#b83227;\n --green:#0f766e;\n --amber:#f59e0b;\n --ink:#1f2937;\n --paper:#ffffff;\n --line:#d9e2ec;\n}\n*{box-sizing:border-box}\nhtml{scroll-behavior:smooth}\nbody{\n margin:0;\n font-family:Arial, Helvetica, sans-serif;\n background:\n radial-gradient(circle at top left, rgba(20,184,166,.12), transparent 28%),\n linear-gradient(180deg,#f8fbff,#eef4f8);\n color:var(--ink);\n line-height:1.65;\n}\nheader{\n background:\n radial-gradient(circle at 82% 18%, rgba(245,158,11,.28), transparent 25%),\n linear-gradient(135deg,#0f172a,#1d4ed8,#14b8a6);\n color:white;\n padding:54px 22px;\n text-align:center;\n}\nheader h1{margin:0;font-size:clamp(2.1rem,4vw,3.7rem)}\nheader p{font-size:1.15rem;margin:10px 0 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.step{color:var(--deep);font-weight:800}\n.grid2{display:grid;grid-template-columns:repeat(auto-fit,minmax(320px,1fr));gap:22px;align-items:start}\n.grid3{display:grid;grid-template-columns:repeat(auto-fit,minmax(250px,1fr));gap:18px;align-items:start}\n.schema{\n background:#fbfdff;\n border:1px solid var(--line);\n border-radius:20px;\n padding:18px;\n text-align:center;\n overflow-x:auto;\n margin:14px 0;\n}\nsvg{max-width:100%;height:auto}\n.btns{text-align:center;margin:18px 0 4px}\nbutton,.linkbtn{\n border:0;\n background:var(--blue);\n color:white;\n text-decoration:none;\n padding:14px 22px;\n border-radius:14px;\n font-size:17px;\n cursor:pointer;\n margin:6px;\n box-shadow:0 7px 18px rgba(37,99,235,.20);\n display:inline-block;\n}\nbutton.stop{background:#333}.cyanbtn{background:var(--cyan)}.pinkbtn{background:var(--pink)}.greenbtn{background:#0f766e}.orangebtn{background:var(--orange)}\ntable{border-collapse:collapse;width:100%;margin:14px 0;background:white}\nth,td{border:1px solid #9aa7bd;padding:10px;text-align:center}\nth{background:#f0f4fb}\n.card{\n border:2px solid #e1e8f0;\n border-top:6px solid var(--teal);\n border-radius:18px;\n padding:18px;\n background:#fff;\n}\n.mm-center{\n grid-column:1\/-1;\n text-align:center;\n background:linear-gradient(135deg,var(--deep),var(--teal));\n color:white;\n border-radius:18px;\n padding:18px;\n font-size:1.3rem;\n font-weight:900;\n}\n.small{font-size:.92rem;color:#5b6770}\n@media print{button,.btns,.linkbtn{display:none}body{background:white}.section{box-shadow:none;break-inside:avoid}}\n<\/style>\n<\/head>\n<body>\n\n<header>\n<h1>Fibroscopie : r\u00e9flexion et r\u00e9fraction de la lumi\u00e8re<\/h1>\n<p>Seconde physique-chimie \u2022 Dioptre \u2022 R\u00e9flexion \u2022 R\u00e9fraction \u2022 Loi de Snell-Descartes \u2022 R\u00e9flexion totale \u2022 Fibre optique<\/p>\n<\/header>\n\n<div class=\"container\">\n\n<div class=\"btns\">\n<button onclick=\"playAudioSummary()\">\ud83d\udd0a \u00c9couter le r\u00e9sum\u00e9 audio clair<\/button>\n<button class=\"stop\" onclick=\"stopAudioSummary()\">\u23f9 Arr\u00eater<\/button><br>\n<a class=\"linkbtn cyanbtn\" href=\"#fiche-bilan\">\ud83d\udccc Aller \u00e0 la fiche bilan<\/a>\n<a class=\"linkbtn pinkbtn\" href=\"#exercices\">\ud83d\udca1 Exercices contextualis\u00e9s<\/a>\n<a class=\"linkbtn greenbtn\" href=\"#type-bac\">\ud83c\udf93 Partie type bac \/ \u00e9valuation<\/a>\n<a class=\"linkbtn orangebtn\" href=\"#fibre\">\ud83d\udd2c Fibre optique<\/a>\n<\/div>\n\n<div id=\"audioText\" style=\"display:none;\">\nBienvenue dans le chapitre fibroscopie, r\u00e9flexion et r\u00e9fraction de la lumi\u00e8re.\nLa fibroscopie est une technique d\u2019imagerie qui permet de guider la lumi\u00e8re \u00e0 travers une fibre optique et donc de faire une image de l\u2019int\u00e9rieur du corps.\nElle permet par exemple de d\u00e9tecter un ulc\u00e8re, de d\u00e9tecter des l\u00e9sions \u00e0 l\u2019intestin, ou de voir en chirurgie sans ouvrir.\nLorsque la lumi\u00e8re rencontre un changement de milieu transparent, une partie se r\u00e9fl\u00e9chit.\nUn dioptre est la surface qui s\u00e9pare deux milieux transparents.\nLa loi de Snell-Descartes de la r\u00e9flexion dit que l\u2019angle d\u2019incidence est \u00e9gal \u00e0 l\u2019angle de r\u00e9flexion.\nLorsqu\u2019un rayon lumineux arrive sur un dioptre, une partie de la lumi\u00e8re traverse le dioptre en \u00e9tant d\u00e9vi\u00e9e : c\u2019est la r\u00e9fraction.\nL\u2019indice de r\u00e9fraction traduit la propagation de la lumi\u00e8re dans le milieu. Il n\u2019a pas d\u2019unit\u00e9.\nLes valeurs \u00e0 conna\u00eetre sont : l\u2019air environ 1,0, l\u2019eau 1,33, le plexiglas ou le verre environ 1,5.\nLa loi de Snell-Descartes de la r\u00e9fraction s\u2019\u00e9crit n un sinus i \u00e9gale n deux sinus r.\nSi n un est plus petit que n deux, alors r est plus petit que i : le rayon se rapproche de la normale.\nSi n un est plus grand que n deux, alors r est plus grand que i : le rayon s\u2019\u00e9loigne de la normale.\nLorsque n un est plus grand que n deux, si l\u2019angle d\u2019incidence devient trop important, il n\u2019y a plus de rayon r\u00e9fract\u00e9 : toute la lumi\u00e8re se r\u00e9fl\u00e9chit sur le dioptre. On parle de r\u00e9flexion totale.\nLa fibre optique utilise la r\u00e9flexion totale pour pi\u00e9ger la lumi\u00e8re dans le c\u0153ur de la fibre.\nFin du r\u00e9sum\u00e9.\n<\/div>\n\n<section class=\"section\">\n<h2>Objectifs du chapitre<\/h2>\n<p>\nCe cours reprend tes phrases comme base : fibroscopie, r\u00e9flexion de la lumi\u00e8re, dioptre, loi de Snell-Descartes,\nr\u00e9fraction, indice de r\u00e9fraction, calculs d\u2019angle, r\u00e9flexion totale, angle limite et application \u00e0 la fibre optique.\n<\/p>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Comprendre<\/h3><p>Ce qui arrive \u00e0 un rayon lumineux lorsqu\u2019il rencontre un dioptre.<\/p><\/div>\n<div class=\"card\"><h3>Construire<\/h3><p>Sch\u00e9mas de r\u00e9flexion, r\u00e9fraction et r\u00e9flexion totale avec normale et angles.<\/p><\/div>\n<div class=\"card\"><h3>Calculer<\/h3><p>Un angle, un indice ou un angle limite avec la loi de Snell-Descartes.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>\ud83d\udd34 M\u00e9thode obligatoire pour les exercices<\/h2>\n<p>\nOn part toujours de la <span class=\"red\">formule du cours<\/span>, on fait le\n<span class=\"blue\">travail litt\u00e9ral<\/span>, puis seulement ensuite l\u2019application num\u00e9rique.\n<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Exemple : isoler l\u2019angle de r\u00e9fraction r<\/p>\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On divise par n\u2082 :<\/p>\n<div class=\"formule bluebox\">(n\u2081 sin i) \/ n\u2082 = sin r<\/div>\n<p class=\"step\">On applique la fonction arcsin pour obtenir l\u2019angle :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>I. La fibroscopie<\/h2>\n\n<p class=\"red\">\nLa fibroscopie est une technique d\u2019imagerie qui permet de guider la lumi\u00e8re \u00e0 travers une fibre, fibre optique, et donc de faire une image de l\u2019int\u00e9rieur du corps.\n<\/p>\n<ul>\n<li>d\u00e9tecter un ulc\u00e8re, un trou dans l\u2019estomac ;<\/li>\n<li>d\u00e9tecter des l\u00e9sions \u00e0 l\u2019intestin ;<\/li>\n<li>permettre de voir en chirurgie sans ouvrir.<\/li>\n<\/ul>\n\n<div class=\"grid2\">\n<div class=\"schema\">\n<svg width=\"650\" height=\"390\" viewBox=\"0 0 650 390\">\n<rect x=\"35\" y=\"35\" width=\"580\" height=\"280\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<text x=\"60\" y=\"70\" font-size=\"18\" font-weight=\"900\" fill=\"#0f172a\">Principe d\u2019une fibroscopie<\/text>\n<path d=\"M180 95 C145 120,135 170,145 215 C160 270,230 295,300 270\" fill=\"none\" stroke=\"#94a3b8\" stroke-width=\"5\"\/>\n<path d=\"M220 80 C250 125,245 170,245 215 C250 250,270 265,295 270\" fill=\"none\" stroke=\"#94a3b8\" stroke-width=\"5\"\/>\n<path d=\"M205 75 C210 120,215 160,210 210 C205 250,225 270,260 290\" fill=\"none\" stroke=\"#0F766E\" stroke-width=\"6\"\/>\n<path d=\"M225 75 C230 120,235 160,230 210 C225 250,245 270,280 290\" fill=\"none\" stroke=\"#E76F51\" stroke-width=\"6\"\/>\n<rect x=\"215\" y=\"55\" width=\"70\" height=\"32\" rx=\"6\" fill=\"#fff7ed\" stroke=\"#F97316\" stroke-width=\"3\"\/>\n<text x=\"225\" y=\"77\" font-size=\"15\" font-weight=\"900\">\u00c9cran<\/text>\n<text x=\"330\" y=\"145\" font-size=\"17\" fill=\"#B83227\" font-weight=\"900\">une fibre \u00e9claire<\/text>\n<text x=\"330\" y=\"178\" font-size=\"17\" fill=\"#0F766E\" font-weight=\"900\">une fibre capte la lumi\u00e8re<\/text>\n<text x=\"330\" y=\"215\" font-size=\"15\">\u2192 image sur un \u00e9cran<\/text>\n<\/svg>\n<\/div>\n<div>\n<p class=\"red\">\nFibroscopie : on utilise deux fibres.\n<\/p>\n<p>\nUne fibre sert \u00e0 \u00e9clairer l\u2019int\u00e9rieur, l\u2019autre capte la lumi\u00e8re pour former une image sur un \u00e9cran.\n<\/p>\n<p>\nPour comprendre la fibre optique, il faut d\u2019abord comprendre la r\u00e9flexion, la r\u00e9fraction et la r\u00e9flexion totale.\n<\/p>\n<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>II. R\u00e9flexion de la lumi\u00e8re<\/h2>\n\n<p class=\"red\">\nLorsque la lumi\u00e8re rencontre un changement de milieu transparent, une partie se r\u00e9fl\u00e9chit.\n<\/p>\n<p class=\"red\">\nDioptre : c\u2019est la surface qui s\u00e9pare deux milieux transparents.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"760\" height=\"400\" viewBox=\"0 0 760 400\">\n<rect x=\"60\" y=\"55\" width=\"640\" height=\"275\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<line x1=\"115\" y1=\"210\" x2=\"650\" y2=\"210\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<text x=\"590\" y=\"197\" font-size=\"17\" fill=\"#B83227\" font-weight=\"900\">Dioptre<\/text>\n<line x1=\"380\" y1=\"70\" x2=\"380\" y2=\"315\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"9,8\"\/>\n<text x=\"392\" y=\"90\" font-size=\"17\" fill=\"#0f172a\" font-weight=\"900\">Normale<\/text>\n<line x1=\"205\" y1=\"90\" x2=\"380\" y2=\"210\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<polygon points=\"380,210 356,204 366,190\" fill=\"#2563EB\"\/>\n<text x=\"135\" y=\"95\" font-size=\"16\" fill=\"#2563EB\" font-weight=\"900\">rayon incident<\/text>\n<line x1=\"380\" y1=\"210\" x2=\"555\" y2=\"90\" stroke=\"#0F766E\" stroke-width=\"5\"\/>\n<polygon points=\"555,90 531,96 541,110\" fill=\"#0F766E\"\/>\n<text x=\"530\" y=\"95\" font-size=\"16\" fill=\"#0F766E\" font-weight=\"900\">rayon r\u00e9fl\u00e9chi<\/text>\n<path d=\"M340 184 A60 60 0 0 1 380 150\" fill=\"none\" stroke=\"#F97316\" stroke-width=\"4\"\/>\n<text x=\"333\" y=\"160\" fill=\"#F97316\" font-size=\"20\" font-weight=\"900\">i<\/text>\n<path d=\"M380 150 A60 60 0 0 1 420 184\" fill=\"none\" stroke=\"#7C3AED\" stroke-width=\"4\"\/>\n<text x=\"420\" y=\"165\" fill=\"#7C3AED\" font-size=\"20\" font-weight=\"900\">i’<\/text>\n<\/svg>\n<\/div>\n\n<p><span class=\"red\">i<\/span> = angle d\u2019incidence.<\/p>\n<p><span class=\"red\">i’<\/span> = angle de r\u00e9flexion, par rapport \u00e0 la normale.<\/p>\n\n<h3>Relation de Snell-Descartes de la r\u00e9flexion<\/h3>\n<div class=\"formule\">i = i’<\/div>\n<p>La lumi\u00e8re repart, se r\u00e9fl\u00e9chit, avec le m\u00eame angle.<\/p>\n<p>Si i = 70\u00b0, alors i’ = 70\u00b0.<\/p>\n<\/section>\n\n<section class=\"section\">\n<h2>III. R\u00e9fraction de la lumi\u00e8re<\/h2>\n\n<p>\nLorsqu\u2019un rayon lumineux arrive sur un dioptre, une partie de la lumi\u00e8re traverse le dioptre en \u00e9tant d\u00e9vi\u00e9e.\n<\/p>\n<p class=\"red\">\nC\u2019est la r\u00e9fraction.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"830\" height=\"500\" viewBox=\"0 0 830 500\">\n<rect x=\"55\" y=\"50\" width=\"720\" height=\"360\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<rect x=\"75\" y=\"250\" width=\"680\" height=\"140\" fill=\"#e0f2fe\" opacity=\".75\"\/>\n<line x1=\"75\" y1=\"250\" x2=\"755\" y2=\"250\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<text x=\"680\" y=\"238\" font-size=\"17\" fill=\"#B83227\" font-weight=\"900\">Dioptre<\/text>\n<text x=\"100\" y=\"125\" font-size=\"17\" fill=\"#0F766E\" font-weight=\"900\">milieu 1<\/text>\n<text x=\"100\" y=\"355\" font-size=\"17\" fill=\"#0F766E\" font-weight=\"900\">milieu 2<\/text>\n<line x1=\"405\" y1=\"75\" x2=\"405\" y2=\"390\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"9,8\"\/>\n<text x=\"420\" y=\"95\" font-size=\"17\" fill=\"#0f172a\" font-weight=\"900\">Normale<\/text>\n<line x1=\"220\" y1=\"100\" x2=\"405\" y2=\"250\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<polygon points=\"405,250 381,243 392,229\" fill=\"#2563EB\"\/>\n<text x=\"140\" y=\"100\" font-size=\"16\" fill=\"#2563EB\" font-weight=\"900\">rayon incident<\/text>\n<line x1=\"405\" y1=\"250\" x2=\"625\" y2=\"350\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<polygon points=\"625,350 600,351 608,333\" fill=\"#E76F51\"\/>\n<text x=\"610\" y=\"375\" font-size=\"16\" fill=\"#E76F51\" font-weight=\"900\">rayon r\u00e9fract\u00e9<\/text>\n<line x1=\"405\" y1=\"250\" x2=\"555\" y2=\"140\" stroke=\"#94a3b8\" stroke-width=\"3\" stroke-dasharray=\"10,8\"\/>\n<text x=\"535\" y=\"135\" font-size=\"15\" fill=\"#64748b\" font-weight=\"900\">rayon r\u00e9fl\u00e9chi<\/text>\n<path d=\"M360 214 A65 65 0 0 1 405 185\" fill=\"none\" stroke=\"#F97316\" stroke-width=\"4\"\/>\n<text x=\"355\" y=\"190\" fill=\"#F97316\" font-size=\"22\" font-weight=\"900\">i<\/text>\n<path d=\"M405 305 A60 60 0 0 0 455 276\" fill=\"none\" stroke=\"#7C3AED\" stroke-width=\"4\"\/>\n<text x=\"460\" y=\"305\" fill=\"#7C3AED\" font-size=\"22\" font-weight=\"900\">r<\/text>\n<\/svg>\n<\/div>\n\n<h3>Exp\u00e9rience : trouver la relation entre i et r<\/h3>\n<table>\n<tr><th>i<\/th><td>20\u00b0<\/td><td>30\u00b0<\/td><td>40\u00b0<\/td><td>50\u00b0<\/td><td>60\u00b0<\/td><\/tr>\n<tr><th>r<\/th><td>13\u00b0<\/td><td>20\u00b0<\/td><td>26\u00b0<\/td><td>31\u00b0<\/td><td>35\u00b0<\/td><\/tr>\n<tr><th>sin(i)<\/th><td>0,342<\/td><td>0,500<\/td><td>0,643<\/td><td>0,766<\/td><td>0,866<\/td><\/tr>\n<tr><th>sin(r)<\/th><td>0,225<\/td><td>0,342<\/td><td>0,438<\/td><td>0,515<\/td><td>0,574<\/td><\/tr>\n<tr><th>sin(i)\/sin(r)<\/th><td>1,5<\/td><td>1,5<\/td><td>1,5<\/td><td>1,5<\/td><td>1,5<\/td><\/tr>\n<\/table>\n\n<p>\nOn remarque qu\u2019il existe une relation math\u00e9matique entre sin(i) et sin(r).\n<\/p>\n<div class=\"formule bluebox\">sin(i) \/ sin(r) = constante<\/div>\n\n<h3>Loi de Snell-Descartes de la r\u00e9fraction<\/h3>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<ul>\n<li><span class=\"red\">n\u2081<\/span> : indice de r\u00e9fraction du milieu 1 ;<\/li>\n<li><span class=\"red\">n\u2082<\/span> : indice de r\u00e9fraction du milieu 2 ;<\/li>\n<li><span class=\"red\">i<\/span> : angle d\u2019incidence ;<\/li>\n<li><span class=\"red\">r<\/span> : angle de r\u00e9fraction.<\/li>\n<\/ul>\n\n<div class=\"note\">\n<p>Exemples d\u2019indices :<\/p>\n<ul>\n<li>n<sub>air<\/sub> = 1,0 ;<\/li>\n<li>n<sub>eau<\/sub> = 1,33 ;<\/li>\n<li>n<sub>plexiglas<\/sub> = 1,5 ;<\/li>\n<li>n<sub>verre<\/sub> \u2248 1,5.<\/li>\n<\/ul>\n<p class=\"red\">\nL\u2019indice de r\u00e9fraction traduit la propagation de la lumi\u00e8re dans le milieu. n n\u2019a pas d\u2019unit\u00e9.\n<\/p>\n<\/div>\n\n<p>\nOn retrouve la relation obtenue avec l\u2019exp\u00e9rience :\n<\/p>\n<div class=\"formule bluebox\">sin(i) \/ sin(r) = n\u2082 \/ n\u2081<\/div>\n<p>Avec n\u2081 = 1,0 pour l\u2019air et n\u2082 = 1,5 pour le plexiglas, on obtient environ 1,5.<\/p>\n<\/section>\n\n<section class=\"section\">\n<h2>IV. Exercices types de r\u00e9fraction<\/h2>\n\n<div class=\"exercice\">\n<h3>Exercice type 1 \u2014 Calculer r<\/h3>\n<p>\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,0<\/span>, <span class=\"red\">n\u2082 = 1,5<\/span>, <span class=\"red\">i = 30\u00b0<\/span>.\nCalculer l\u2019angle de r\u00e9fraction r.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler r. On divise par n\u2082 :<\/p>\n<div class=\"formule bluebox\">(n\u2081 sin i) \/ n\u2082 = sin r<\/div>\n<p class=\"step\">On applique la fonction arcsin :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n\n<p>Application num\u00e9rique :<\/p>\n<p>r = arcsin((1,0 \u00d7 sin 30\u00b0) \/ 1,5)<\/p>\n<p class=\"resultat\">r = 19,47\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 2 \u2014 Calculer i<\/h3>\n<p>\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,0<\/span>, <span class=\"red\">n\u2082 = 1,5<\/span>, <span class=\"red\">r = 35\u00b0<\/span>.\nCalculer l\u2019angle d\u2019incidence i.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler i. On divise par n\u2081 :<\/p>\n<div class=\"formule bluebox\">sin i = (n\u2082 sin r) \/ n\u2081<\/div>\n<p class=\"step\">On applique la fonction arcsin :<\/p>\n<div class=\"formule bluebox\">i = arcsin((n\u2082 sin r) \/ n\u2081)<\/div>\n<\/div>\n\n<p>Application num\u00e9rique :<\/p>\n<p>i = arcsin((1,5 \u00d7 sin 35\u00b0) \/ 1,0)<\/p>\n<p class=\"resultat\">i = 59,36\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 3 \u2014 Calculer n\u2082<\/h3>\n<p>\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,5<\/span>, <span class=\"red\">i = 65\u00b0<\/span>, <span class=\"red\">r = 70\u00b0<\/span>.\nCalculer n\u2082.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler n\u2082. On divise par sin r :<\/p>\n<div class=\"formule bluebox\">n\u2082 = (n\u2081 sin i) \/ sin r<\/div>\n<\/div>\n\n<p>Application num\u00e9rique :<\/p>\n<p>n\u2082 = (1,5 \u00d7 sin 65\u00b0) \/ sin 70\u00b0<\/p>\n<p class=\"resultat\">n\u2082 \u2248 1,4<\/p>\n<\/div>\n\n<div class=\"note\">\n<h3>Remarque importante<\/h3>\n<p class=\"red\">Si n\u2081 < n\u2082 alors r < i.<\/p>\n<p>Le rayon se rapproche de la normale.<\/p>\n<div class=\"schema\">\n<svg width=\"620\" height=\"250\" viewBox=\"0 0 620 250\">\n<line x1=\"80\" y1=\"125\" x2=\"540\" y2=\"125\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<line x1=\"310\" y1=\"35\" x2=\"310\" y2=\"220\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"8,7\"\/>\n<line x1=\"190\" y1=\"45\" x2=\"310\" y2=\"125\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<line x1=\"310\" y1=\"125\" x2=\"420\" y2=\"190\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<text x=\"100\" y=\"55\" fill=\"#2563EB\" font-weight=\"900\">n\u2081 < n\u2082<\/text>\n<text x=\"430\" y=\"190\" fill=\"#B83227\" font-weight=\"900\">r < i<\/text>\n<\/svg>\n<\/div>\n\n<p class=\"red\">Si n\u2081 > n\u2082 alors r > i.<\/p>\n<p>Le rayon s\u2019\u00e9loigne de la normale.<\/p>\n<div class=\"schema\">\n<svg width=\"620\" height=\"250\" viewBox=\"0 0 620 250\">\n<line x1=\"80\" y1=\"125\" x2=\"540\" y2=\"125\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<line x1=\"310\" y1=\"35\" x2=\"310\" y2=\"220\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"8,7\"\/>\n<line x1=\"235\" y1=\"45\" x2=\"310\" y2=\"125\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<line x1=\"310\" y1=\"125\" x2=\"500\" y2=\"190\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<text x=\"100\" y=\"55\" fill=\"#2563EB\" font-weight=\"900\">n\u2081 > n\u2082<\/text>\n<text x=\"430\" y=\"190\" fill=\"#B83227\" font-weight=\"900\">r > i<\/text>\n<\/svg>\n<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>V. R\u00e9flexion totale<\/h2>\n\n<p class=\"red\">\nLorsque n\u2081 > n\u2082, si i devient trop important, alors d\u00e8s que r = 90\u00b0, il n\u2019y a plus de rayon r\u00e9fract\u00e9 :\ntoute la lumi\u00e8re se r\u00e9fl\u00e9chit sur le dioptre. On dit qu\u2019il y a r\u00e9flexion totale.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"760\" height=\"390\" viewBox=\"0 0 760 390\">\n<rect x=\"70\" y=\"60\" width=\"610\" height=\"255\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<rect x=\"90\" y=\"190\" width=\"570\" height=\"100\" fill=\"#e0f2fe\" opacity=\".75\"\/>\n<line x1=\"90\" y1=\"190\" x2=\"660\" y2=\"190\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<line x1=\"375\" y1=\"75\" x2=\"375\" y2=\"290\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"9,8\"\/>\n<line x1=\"210\" y1=\"90\" x2=\"375\" y2=\"190\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<line x1=\"375\" y1=\"190\" x2=\"530\" y2=\"100\" stroke=\"#0F766E\" stroke-width=\"5\"\/>\n<line x1=\"375\" y1=\"190\" x2=\"600\" y2=\"190\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<polygon points=\"600,190 578,178 578,202\" fill=\"#E76F51\"\/>\n<text x=\"510\" y=\"82\" fill=\"#0F766E\" font-size=\"16\" font-weight=\"900\">r\u00e9flexion<\/text>\n<text x=\"545\" y=\"215\" fill=\"#E76F51\" font-size=\"16\" font-weight=\"900\">r = 90\u00b0<\/text>\n<text x=\"420\" y=\"250\" fill=\"#B83227\" font-size=\"18\" font-weight=\"900\">pas de rayon r\u00e9fract\u00e9 au-del\u00e0 de i limite<\/text>\n<\/svg>\n<\/div>\n\n<p>\nLorsque <span class=\"red\">i > i<sub>limite<\/sub><\/span>, alors on dit qu\u2019il y a r\u00e9flexion totale.\n<\/p>\n\n<h3>Calcul de i<sub>limite<\/sub><\/h3>\n<p>Il y a r\u00e9flexion totale lorsque <span class=\"red\">r = 90\u00b0<\/span>.<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">Pour l\u2019angle limite, r = 90\u00b0 :<\/p>\n<div class=\"formule bluebox\">n\u2081 sin i<sub>limite<\/sub> = n\u2082 sin 90\u00b0<\/div>\n<p class=\"step\">Or sin 90\u00b0 = 1 :<\/p>\n<div class=\"formule bluebox\">n\u2081 sin i<sub>limite<\/sub> = n\u2082<\/div>\n<p class=\"step\">On divise par n\u2081 :<\/p>\n<div class=\"formule bluebox\">sin i<sub>limite<\/sub> = n\u2082 \/ n\u2081<\/div>\n<p class=\"step\">On applique arcsin :<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Calculer l\u2019angle limite<\/h3>\n<p>Donn\u00e9es : <span class=\"red\">n\u2081 = 1,5<\/span> et <span class=\"red\">n\u2082 = 1,0<\/span>.<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<p>i<sub>limite<\/sub> = arcsin(1,0 \/ 1,5)<\/p>\n<p class=\"resultat\">i<sub>limite<\/sub> = 41,8\u00b0<\/p>\n<p class=\"red\">D\u00e8s que i > i<sub>limite<\/sub>, il y a r\u00e9flexion totale. Il n\u2019y a pas de r\u00e9fraction.<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"fibre\">\n<h2>VI. Application : la fibre optique<\/h2>\n\n<p class=\"red\">\nC\u2019est une fibre transparente qui permet de guider la lumi\u00e8re.\n<\/p>\n<p class=\"red\">\nL\u2019indice de la gaine et du c\u0153ur sont choisis pour qu\u2019il y ait r\u00e9flexion totale sur la gaine\net que toute la lumi\u00e8re reste pi\u00e9g\u00e9e dans la fibre optique.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"850\" height=\"360\" viewBox=\"0 0 850 360\">\n<rect x=\"70\" y=\"115\" width=\"700\" height=\"120\" rx=\"25\" fill=\"#dbeafe\" stroke=\"#2563EB\" stroke-width=\"3\"\/>\n<rect x=\"70\" y=\"145\" width=\"700\" height=\"60\" rx=\"20\" fill=\"#fefce8\" stroke=\"#F59E0B\" stroke-width=\"3\"\/>\n<text x=\"110\" y=\"135\" fill=\"#2563EB\" font-size=\"17\" font-weight=\"900\">Gaine : n\u2082<\/text>\n<text x=\"110\" y=\"180\" fill=\"#B83227\" font-size=\"17\" font-weight=\"900\">C\u0153ur : n\u2081<\/text>\n<text x=\"420\" y=\"105\" fill=\"#B83227\" font-size=\"18\" font-weight=\"900\">n\u2081 > n\u2082<\/text>\n<polyline points=\"40,250 115,175 235,175 330,205 445,145 560,205 670,145 810,145\" fill=\"none\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<g fill=\"#E76F51\">\n<polygon points=\"115,175 92,180 105,160\"\/>\n<polygon points=\"235,175 212,170 220,190\"\/>\n<polygon points=\"330,205 305,198 322,185\"\/>\n<polygon points=\"445,145 420,152 430,130\"\/>\n<polygon points=\"560,205 535,198 552,185\"\/>\n<polygon points=\"670,145 645,152 655,130\"\/>\n<\/g>\n<text x=\"255\" y=\"132\" fill=\"#0F766E\" font-size=\"16\" font-weight=\"900\">r\u00e9flexion totale<\/text>\n<text x=\"505\" y=\"255\" fill=\"#0F766E\" font-size=\"16\" font-weight=\"900\">la lumi\u00e8re reste pi\u00e9g\u00e9e<\/text>\n<\/svg>\n<\/div>\n\n<div class=\"note\">\n<p class=\"red\">Fibroscopie :<\/p>\n<p>On utilise deux fibres : une fibre \u00e9claire l\u2019int\u00e9rieur du corps ; l\u2019autre capte la lumi\u00e8re pour former une image sur un \u00e9cran.<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"exercices\">\n<h2>VII. Exercices contextualis\u00e9s<\/h2>\n\n<div class=\"exercice\">\n<h3>1. Air \u2192 eau : rayon entrant dans le lagon<\/h3>\n<p>\nOn fait une r\u00e9fraction de l\u2019air vers l\u2019eau.\nDonn\u00e9es : <span class=\"red\">n<sub>air<\/sub> = 1,0<\/span>, <span class=\"red\">n<sub>eau<\/sub> = 1,33<\/span>, <span class=\"red\">i = 25\u00b0<\/span>.\nCalculer r.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On isole r :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n<p>r = arcsin((1,0 \u00d7 sin 25\u00b0) \/ 1,33)<\/p>\n<p class=\"resultat\">r = 18,5\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>2. Eau \u2192 air : rayon qui sort de l\u2019eau<\/h3>\n<p>\nOn fait une r\u00e9fraction de l\u2019eau vers l\u2019air.\nDonn\u00e9es : <span class=\"red\">n<sub>eau<\/sub> = 1,33<\/span>, <span class=\"red\">n<sub>air<\/sub> = 1,0<\/span>, <span class=\"red\">r = 25\u00b0<\/span>.\nCalculer i.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On isole i :<\/p>\n<div class=\"formule bluebox\">i = arcsin((n\u2082 sin r) \/ n\u2081)<\/div>\n<\/div>\n<p>i = arcsin((1,0 \u00d7 sin 25\u00b0) \/ 1,33)<\/p>\n<p class=\"resultat\">i = 18,5\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>3. Angle limite eau \u2192 air<\/h3>\n<p>Donn\u00e9es : n\u2081 = 1,33 pour l\u2019eau et n\u2082 = 1,0 pour l\u2019air.<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<p>i<sub>limite<\/sub> = arcsin(1,0 \/ 1,33)<\/p>\n<p class=\"resultat\">i<sub>limite<\/sub> \u2248 48,7\u00b0<\/p>\n<p>Il y aura r\u00e9flexion totale lorsque i > i<sub>limite<\/sub>.<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>4. Milieu inconnu<\/h3>\n<p>\nOn fait une r\u00e9fraction de l\u2019air dans un milieu inconnu.\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,0<\/span>, <span class=\"red\">i = 20\u00b0<\/span>, <span class=\"red\">r = 24\u00b0<\/span>.\nCalculer l\u2019indice n\u2082 du milieu inconnu.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler n\u2082. On divise par sin r :<\/p>\n<div class=\"formule bluebox\">n\u2082 = (n\u2081 sin i) \/ sin r<\/div>\n<\/div>\n<p>n\u2082 = (1,0 \u00d7 sin 20\u00b0) \/ sin 24\u00b0<\/p>\n<p class=\"resultat\">n\u2082 \u2248 0,84<\/p>\n<p class=\"red\">Impossible pour un milieu transparent usuel car un indice de r\u00e9fraction doit \u00eatre sup\u00e9rieur ou \u00e9gal \u00e0 1.<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>5. Fr\u00e9quence d\u2019un signal<\/h3>\n<p>On mesure une p\u00e9riode <span class=\"red\">T = 20,3 ms<\/span>. Calculer la fr\u00e9quence f.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">f = 1 \/ T<\/div>\n<p class=\"step\">On cherche f : la formule est d\u00e9j\u00e0 sous la bonne forme.<\/p>\n<div class=\"formule bluebox\">f = 1 \/ T<\/div>\n<\/div>\n<p class=\"red\">Conversion :<\/p>\n<p>T = 20,3 ms = 20,3 \u00d7 10\u207b\u00b3 s<\/p>\n<p>f = 1 \/ (20,3 \u00d7 10\u207b\u00b3)<\/p>\n<p class=\"resultat\">f = 49,2 Hz<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>6. R\u00e9seau \u00e9lectrique EEWF<\/h3>\n<p>EEWF fournit une \u00e9lectricit\u00e9 \u00e0 <span class=\"red\">f = 50 Hz<\/span>. Calculer la p\u00e9riode du signal.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">f = 1 \/ T<\/div>\n<p class=\"step\">On veut isoler T. On multiplie par T :<\/p>\n<div class=\"formule bluebox\">f \u00d7 T = 1<\/div>\n<p class=\"step\">On divise par f :<\/p>\n<div class=\"formule bluebox\">T = 1 \/ f<\/div>\n<\/div>\n<p>T = 1 \/ 50<\/p>\n<p class=\"resultat\">T = 0,020 s<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>7. \u00c9chographie<\/h3>\n<p>\nOn mesure avec un \u00e9chographe un temps aller-retour <span class=\"red\">t<sub>A\/R<\/sub> = 2,0 ms<\/span>.\nOn donne <span class=\"red\">v<sub>son<\/sub> = 1500 m\u00b7s\u207b\u00b9<\/span>.\nCalculer la profondeur de l\u2019obstacle.\n<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">v = d \/ t<\/div>\n<p class=\"step\">On isole la distance parcourue :<\/p>\n<div class=\"formule bluebox\">d = v \u00d7 t<\/div>\n<\/div>\n<p>t = 2,0 \u00d7 10\u207b\u00b3 s<\/p>\n<p>d<sub>aller-retour<\/sub> = 1500 \u00d7 2,0 \u00d7 10\u207b\u00b3 = 3,0 m<\/p>\n<p class=\"red\">Le son fait un aller-retour, donc la profondeur vaut la moiti\u00e9.<\/p>\n<p class=\"resultat\">profondeur = 1,5 m<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"type-bac\">\n<h2>VIII. Partie type bac \/ \u00e9valuation \u2014 niveau seconde<\/h2>\n\n<div class=\"exercice\">\n<h3>Situation 1 \u2014 Fibroscopie et r\u00e9flexion totale<\/h3>\n<p>\nUne fibre optique poss\u00e8de un c\u0153ur d\u2019indice n\u2081 = 1,50 et une gaine d\u2019indice n\u2082 = 1,40.\n<\/p>\n<ol>\n<li>Expliquer le r\u00f4le de la fibre optique en fibroscopie.<\/li>\n<li>Dire pourquoi il faut que n\u2081 > n\u2082.<\/li>\n<li>Calculer l\u2019angle limite.<\/li>\n<li>Conclure : que se passe-t-il si i est sup\u00e9rieur \u00e0 cet angle ?<\/li>\n<\/ol>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<p class=\"step\">On remplace par les indices de la gaine et du c\u0153ur :<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(1,40 \/ 1,50)<\/div>\n<\/div>\n<p>i<sub>limite<\/sub> \u2248 69,0\u00b0<\/p>\n<p class=\"resultat\">Si i > 69,0\u00b0, il y a r\u00e9flexion totale et la lumi\u00e8re reste guid\u00e9e dans la fibre.<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Situation 2 \u2014 Rayon lumineux dans l\u2019eau<\/h3>\n<p>\nUn rayon lumineux passe de l\u2019air dans l\u2019eau avec i = 40\u00b0. On donne n<sub>air<\/sub> = 1,0 et n<sub>eau<\/sub> = 1,33.\n<\/p>\n<ol>\n<li>Faire le sch\u00e9ma de la r\u00e9fraction avec normale, rayon incident et rayon r\u00e9fract\u00e9.<\/li>\n<li>Calculer r.<\/li>\n<li>Dire si le rayon se rapproche ou s\u2019\u00e9loigne de la normale.<\/li>\n<\/ol>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On isole r :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n<p>r = arcsin((1,0 \u00d7 sin 40\u00b0) \/ 1,33)<\/p>\n<p class=\"resultat\">r \u2248 28,9\u00b0<\/p>\n<p>Comme n\u2081 < n\u2082, alors r < i : le rayon se rapproche de la normale.<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"fiche-bilan\">\n<h2>\ud83d\udccc Fiche bilan \u2014 Fibroscopie, r\u00e9flexion et r\u00e9fraction<\/h2>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Fibroscopie<\/h3><p>Technique d\u2019imagerie qui guide la lumi\u00e8re dans une fibre optique.<\/p><\/div>\n<div class=\"card\"><h3>Dioptre<\/h3><p class=\"red\">Surface qui s\u00e9pare deux milieux transparents.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9flexion<\/h3><p>Une partie de la lumi\u00e8re se r\u00e9fl\u00e9chit.<\/p><div class=\"formule\">i = i’<\/div><\/div>\n<div class=\"card\"><h3>R\u00e9fraction<\/h3><p>Une partie traverse le dioptre en \u00e9tant d\u00e9vi\u00e9e.<\/p><\/div>\n<div class=\"card\"><h3>Snell-Descartes<\/h3><div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div><\/div>\n<div class=\"card\"><h3>Indice<\/h3><p>n traduit la propagation de la lumi\u00e8re dans un milieu.<\/p><p class=\"red\">n n\u2019a pas d\u2019unit\u00e9.<\/p><\/div>\n<div class=\"card\"><h3>Valeurs<\/h3><p>air 1,0 ; eau 1,33 ; verre \/ plexiglas \u2248 1,5.<\/p><\/div>\n<div class=\"card\"><h3>Cas n\u2081 < n\u2082<\/h3><p class=\"red\">r < i : le rayon se rapproche de la normale.<\/p><\/div>\n<div class=\"card\"><h3>Cas n\u2081 > n\u2082<\/h3><p class=\"red\">r > i : le rayon s\u2019\u00e9loigne de la normale.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9flexion totale<\/h3><p>Si n\u2081 > n\u2082 et i > i<sub>limite<\/sub>, il n\u2019y a plus de rayon r\u00e9fract\u00e9.<\/p><\/div>\n<div class=\"card\"><h3>Angle limite<\/h3><div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082\/n\u2081)<\/div><\/div>\n<div class=\"card\"><h3>Fibre optique<\/h3><p>La lumi\u00e8re reste pi\u00e9g\u00e9e par r\u00e9flexion totale.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>Carte mentale<\/h2>\n<div class=\"grid3\">\n<div class=\"mm-center\">FIBROSCOPIE \u2014 R\u00c9FLEXION \u2014 R\u00c9FRACTION<\/div>\n<div class=\"card\"><h3>Fibroscopie<\/h3><p>Image de l\u2019int\u00e9rieur du corps gr\u00e2ce aux fibres optiques.<\/p><\/div>\n<div class=\"card\"><h3>Dioptre<\/h3><p>S\u00e9paration entre deux milieux transparents.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9flexion<\/h3><p>i = i’.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9fraction<\/h3><p>n\u2081 sin i = n\u2082 sin r.<\/p><\/div>\n<div class=\"card\"><h3>Angle limite<\/h3><p>r = 90\u00b0 ; i<sub>limite<\/sub> = arcsin(n\u2082\/n\u2081).<\/p><\/div>\n<div class=\"card\"><h3>Fibre optique<\/h3><p>R\u00e9flexion totale dans le c\u0153ur.<\/p><\/div>\n<\/div>\n<\/section>\n\n<\/div>\n\n<script>\nlet utterance;\nfunction playAudioSummary(){\n speechSynthesis.cancel();\n const text=document.getElementById(\"audioText\").innerText;\n utterance=new SpeechSynthesisUtterance(text);\n utterance.lang=\"fr-FR\";\n utterance.rate=0.90;\n speechSynthesis.speak(utterance);\n}\nfunction stopAudioSummary(){speechSynthesis.cancel();}\n<\/script>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Seconde \u2014 Fibroscopie, r\u00e9flexion et r\u00e9fraction Fibroscopie : r\u00e9flexion et r\u00e9fraction de la lumi\u00e8re Seconde physique-chimie \u2022 Dioptre \u2022 R\u00e9flexion \u2022 R\u00e9fraction \u2022 Loi de Snell-Descartes...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-775","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=775"}],"version-history":[{"count":1,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/775\/revisions"}],"predecessor-version":[{"id":777,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/775\/revisions\/777"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":775,"date":"2026-05-28T03:47:12","date_gmt":"2026-05-28T01:47:12","guid":{"rendered":"https:\/\/pcwallis.malo.wf\/?page_id=775"},"modified":"2026-05-28T03:47:13","modified_gmt":"2026-05-28T01:47:13","slug":"fibre-optique","status":"publish","type":"page","link":"https:\/\/pcwallis.malo.wf\/index.php\/fibre-optique\/","title":{"rendered":"fibre optique"},"content":{"rendered":"\n\n\n\n\n\nSeconde \u2014 Fibroscopie, r\u00e9flexion et r\u00e9fraction<\/title>\n<style>\n:root{\n --deep:#0f172a;\n --blue:#2563eb;\n --cyan:#0891b2;\n --teal:#14b8a6;\n --purple:#7c3aed;\n --pink:#db2777;\n --orange:#f97316;\n --red:#b83227;\n --green:#0f766e;\n --amber:#f59e0b;\n --ink:#1f2937;\n --paper:#ffffff;\n --line:#d9e2ec;\n}\n*{box-sizing:border-box}\nhtml{scroll-behavior:smooth}\nbody{\n margin:0;\n font-family:Arial, Helvetica, sans-serif;\n background:\n radial-gradient(circle at top left, rgba(20,184,166,.12), transparent 28%),\n linear-gradient(180deg,#f8fbff,#eef4f8);\n 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display:inline-block;\n}\nbutton.stop{background:#333}.cyanbtn{background:var(--cyan)}.pinkbtn{background:var(--pink)}.greenbtn{background:#0f766e}.orangebtn{background:var(--orange)}\ntable{border-collapse:collapse;width:100%;margin:14px 0;background:white}\nth,td{border:1px solid #9aa7bd;padding:10px;text-align:center}\nth{background:#f0f4fb}\n.card{\n border:2px solid #e1e8f0;\n border-top:6px solid var(--teal);\n border-radius:18px;\n padding:18px;\n background:#fff;\n}\n.mm-center{\n grid-column:1\/-1;\n text-align:center;\n background:linear-gradient(135deg,var(--deep),var(--teal));\n color:white;\n border-radius:18px;\n padding:18px;\n font-size:1.3rem;\n font-weight:900;\n}\n.small{font-size:.92rem;color:#5b6770}\n@media print{button,.btns,.linkbtn{display:none}body{background:white}.section{box-shadow:none;break-inside:avoid}}\n<\/style>\n<\/head>\n<body>\n\n<header>\n<h1>Fibroscopie : r\u00e9flexion et r\u00e9fraction de la lumi\u00e8re<\/h1>\n<p>Seconde physique-chimie \u2022 Dioptre \u2022 R\u00e9flexion \u2022 R\u00e9fraction \u2022 Loi de Snell-Descartes \u2022 R\u00e9flexion totale \u2022 Fibre optique<\/p>\n<\/header>\n\n<div class=\"container\">\n\n<div class=\"btns\">\n<button onclick=\"playAudioSummary()\">\ud83d\udd0a \u00c9couter le r\u00e9sum\u00e9 audio clair<\/button>\n<button class=\"stop\" onclick=\"stopAudioSummary()\">\u23f9 Arr\u00eater<\/button><br>\n<a class=\"linkbtn cyanbtn\" href=\"#fiche-bilan\">\ud83d\udccc Aller \u00e0 la fiche bilan<\/a>\n<a class=\"linkbtn pinkbtn\" href=\"#exercices\">\ud83d\udca1 Exercices contextualis\u00e9s<\/a>\n<a class=\"linkbtn greenbtn\" href=\"#type-bac\">\ud83c\udf93 Partie type bac \/ \u00e9valuation<\/a>\n<a class=\"linkbtn orangebtn\" href=\"#fibre\">\ud83d\udd2c Fibre optique<\/a>\n<\/div>\n\n<div id=\"audioText\" style=\"display:none;\">\nBienvenue dans le chapitre fibroscopie, r\u00e9flexion et r\u00e9fraction de la lumi\u00e8re.\nLa fibroscopie est une technique d\u2019imagerie qui permet de guider la lumi\u00e8re \u00e0 travers une fibre optique et donc de faire une image de l\u2019int\u00e9rieur du corps.\nElle permet par exemple de d\u00e9tecter un ulc\u00e8re, de d\u00e9tecter des l\u00e9sions \u00e0 l\u2019intestin, ou de voir en chirurgie sans ouvrir.\nLorsque la lumi\u00e8re rencontre un changement de milieu transparent, une partie se r\u00e9fl\u00e9chit.\nUn dioptre est la surface qui s\u00e9pare deux milieux transparents.\nLa loi de Snell-Descartes de la r\u00e9flexion dit que l\u2019angle d\u2019incidence est \u00e9gal \u00e0 l\u2019angle de r\u00e9flexion.\nLorsqu\u2019un rayon lumineux arrive sur un dioptre, une partie de la lumi\u00e8re traverse le dioptre en \u00e9tant d\u00e9vi\u00e9e : c\u2019est la r\u00e9fraction.\nL\u2019indice de r\u00e9fraction traduit la propagation de la lumi\u00e8re dans le milieu. Il n\u2019a pas d\u2019unit\u00e9.\nLes valeurs \u00e0 conna\u00eetre sont : l\u2019air environ 1,0, l\u2019eau 1,33, le plexiglas ou le verre environ 1,5.\nLa loi de Snell-Descartes de la r\u00e9fraction s\u2019\u00e9crit n un sinus i \u00e9gale n deux sinus r.\nSi n un est plus petit que n deux, alors r est plus petit que i : le rayon se rapproche de la normale.\nSi n un est plus grand que n deux, alors r est plus grand que i : le rayon s\u2019\u00e9loigne de la normale.\nLorsque n un est plus grand que n deux, si l\u2019angle d\u2019incidence devient trop important, il n\u2019y a plus de rayon r\u00e9fract\u00e9 : toute la lumi\u00e8re se r\u00e9fl\u00e9chit sur le dioptre. On parle de r\u00e9flexion totale.\nLa fibre optique utilise la r\u00e9flexion totale pour pi\u00e9ger la lumi\u00e8re dans le c\u0153ur de la fibre.\nFin du r\u00e9sum\u00e9.\n<\/div>\n\n<section class=\"section\">\n<h2>Objectifs du chapitre<\/h2>\n<p>\nCe cours reprend tes phrases comme base : fibroscopie, r\u00e9flexion de la lumi\u00e8re, dioptre, loi de Snell-Descartes,\nr\u00e9fraction, indice de r\u00e9fraction, calculs d\u2019angle, r\u00e9flexion totale, angle limite et application \u00e0 la fibre optique.\n<\/p>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Comprendre<\/h3><p>Ce qui arrive \u00e0 un rayon lumineux lorsqu\u2019il rencontre un dioptre.<\/p><\/div>\n<div class=\"card\"><h3>Construire<\/h3><p>Sch\u00e9mas de r\u00e9flexion, r\u00e9fraction et r\u00e9flexion totale avec normale et angles.<\/p><\/div>\n<div class=\"card\"><h3>Calculer<\/h3><p>Un angle, un indice ou un angle limite avec la loi de Snell-Descartes.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>\ud83d\udd34 M\u00e9thode obligatoire pour les exercices<\/h2>\n<p>\nOn part toujours de la <span class=\"red\">formule du cours<\/span>, on fait le\n<span class=\"blue\">travail litt\u00e9ral<\/span>, puis seulement ensuite l\u2019application num\u00e9rique.\n<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Exemple : isoler l\u2019angle de r\u00e9fraction r<\/p>\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On divise par n\u2082 :<\/p>\n<div class=\"formule bluebox\">(n\u2081 sin i) \/ n\u2082 = sin r<\/div>\n<p class=\"step\">On applique la fonction arcsin pour obtenir l\u2019angle :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>I. La fibroscopie<\/h2>\n\n<p class=\"red\">\nLa fibroscopie est une technique d\u2019imagerie qui permet de guider la lumi\u00e8re \u00e0 travers une fibre, fibre optique, et donc de faire une image de l\u2019int\u00e9rieur du corps.\n<\/p>\n<ul>\n<li>d\u00e9tecter un ulc\u00e8re, un trou dans l\u2019estomac ;<\/li>\n<li>d\u00e9tecter des l\u00e9sions \u00e0 l\u2019intestin ;<\/li>\n<li>permettre de voir en chirurgie sans ouvrir.<\/li>\n<\/ul>\n\n<div class=\"grid2\">\n<div class=\"schema\">\n<svg width=\"650\" height=\"390\" viewBox=\"0 0 650 390\">\n<rect x=\"35\" y=\"35\" width=\"580\" height=\"280\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<text x=\"60\" y=\"70\" font-size=\"18\" font-weight=\"900\" fill=\"#0f172a\">Principe d\u2019une fibroscopie<\/text>\n<path d=\"M180 95 C145 120,135 170,145 215 C160 270,230 295,300 270\" fill=\"none\" stroke=\"#94a3b8\" stroke-width=\"5\"\/>\n<path d=\"M220 80 C250 125,245 170,245 215 C250 250,270 265,295 270\" fill=\"none\" stroke=\"#94a3b8\" stroke-width=\"5\"\/>\n<path d=\"M205 75 C210 120,215 160,210 210 C205 250,225 270,260 290\" fill=\"none\" stroke=\"#0F766E\" stroke-width=\"6\"\/>\n<path d=\"M225 75 C230 120,235 160,230 210 C225 250,245 270,280 290\" fill=\"none\" stroke=\"#E76F51\" stroke-width=\"6\"\/>\n<rect x=\"215\" y=\"55\" width=\"70\" height=\"32\" rx=\"6\" fill=\"#fff7ed\" stroke=\"#F97316\" stroke-width=\"3\"\/>\n<text x=\"225\" y=\"77\" font-size=\"15\" font-weight=\"900\">\u00c9cran<\/text>\n<text x=\"330\" y=\"145\" font-size=\"17\" fill=\"#B83227\" font-weight=\"900\">une fibre \u00e9claire<\/text>\n<text x=\"330\" y=\"178\" font-size=\"17\" fill=\"#0F766E\" font-weight=\"900\">une fibre capte la lumi\u00e8re<\/text>\n<text x=\"330\" y=\"215\" font-size=\"15\">\u2192 image sur un \u00e9cran<\/text>\n<\/svg>\n<\/div>\n<div>\n<p class=\"red\">\nFibroscopie : on utilise deux fibres.\n<\/p>\n<p>\nUne fibre sert \u00e0 \u00e9clairer l\u2019int\u00e9rieur, l\u2019autre capte la lumi\u00e8re pour former une image sur un \u00e9cran.\n<\/p>\n<p>\nPour comprendre la fibre optique, il faut d\u2019abord comprendre la r\u00e9flexion, la r\u00e9fraction et la r\u00e9flexion totale.\n<\/p>\n<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>II. R\u00e9flexion de la lumi\u00e8re<\/h2>\n\n<p class=\"red\">\nLorsque la lumi\u00e8re rencontre un changement de milieu transparent, une partie se r\u00e9fl\u00e9chit.\n<\/p>\n<p class=\"red\">\nDioptre : c\u2019est la surface qui s\u00e9pare deux milieux transparents.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"760\" height=\"400\" viewBox=\"0 0 760 400\">\n<rect x=\"60\" y=\"55\" width=\"640\" height=\"275\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<line x1=\"115\" y1=\"210\" x2=\"650\" y2=\"210\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<text x=\"590\" y=\"197\" font-size=\"17\" fill=\"#B83227\" font-weight=\"900\">Dioptre<\/text>\n<line x1=\"380\" y1=\"70\" x2=\"380\" y2=\"315\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"9,8\"\/>\n<text x=\"392\" y=\"90\" font-size=\"17\" fill=\"#0f172a\" font-weight=\"900\">Normale<\/text>\n<line x1=\"205\" y1=\"90\" x2=\"380\" y2=\"210\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<polygon points=\"380,210 356,204 366,190\" fill=\"#2563EB\"\/>\n<text x=\"135\" y=\"95\" font-size=\"16\" fill=\"#2563EB\" font-weight=\"900\">rayon incident<\/text>\n<line x1=\"380\" y1=\"210\" x2=\"555\" y2=\"90\" stroke=\"#0F766E\" stroke-width=\"5\"\/>\n<polygon points=\"555,90 531,96 541,110\" fill=\"#0F766E\"\/>\n<text x=\"530\" y=\"95\" font-size=\"16\" fill=\"#0F766E\" font-weight=\"900\">rayon r\u00e9fl\u00e9chi<\/text>\n<path d=\"M340 184 A60 60 0 0 1 380 150\" fill=\"none\" stroke=\"#F97316\" stroke-width=\"4\"\/>\n<text x=\"333\" y=\"160\" fill=\"#F97316\" font-size=\"20\" font-weight=\"900\">i<\/text>\n<path d=\"M380 150 A60 60 0 0 1 420 184\" fill=\"none\" stroke=\"#7C3AED\" stroke-width=\"4\"\/>\n<text x=\"420\" y=\"165\" fill=\"#7C3AED\" font-size=\"20\" font-weight=\"900\">i’<\/text>\n<\/svg>\n<\/div>\n\n<p><span class=\"red\">i<\/span> = angle d\u2019incidence.<\/p>\n<p><span class=\"red\">i’<\/span> = angle de r\u00e9flexion, par rapport \u00e0 la normale.<\/p>\n\n<h3>Relation de Snell-Descartes de la r\u00e9flexion<\/h3>\n<div class=\"formule\">i = i’<\/div>\n<p>La lumi\u00e8re repart, se r\u00e9fl\u00e9chit, avec le m\u00eame angle.<\/p>\n<p>Si i = 70\u00b0, alors i’ = 70\u00b0.<\/p>\n<\/section>\n\n<section class=\"section\">\n<h2>III. R\u00e9fraction de la lumi\u00e8re<\/h2>\n\n<p>\nLorsqu\u2019un rayon lumineux arrive sur un dioptre, une partie de la lumi\u00e8re traverse le dioptre en \u00e9tant d\u00e9vi\u00e9e.\n<\/p>\n<p class=\"red\">\nC\u2019est la r\u00e9fraction.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"830\" height=\"500\" viewBox=\"0 0 830 500\">\n<rect x=\"55\" y=\"50\" width=\"720\" height=\"360\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<rect x=\"75\" y=\"250\" width=\"680\" height=\"140\" fill=\"#e0f2fe\" opacity=\".75\"\/>\n<line x1=\"75\" y1=\"250\" x2=\"755\" y2=\"250\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<text x=\"680\" y=\"238\" font-size=\"17\" fill=\"#B83227\" font-weight=\"900\">Dioptre<\/text>\n<text x=\"100\" y=\"125\" font-size=\"17\" fill=\"#0F766E\" font-weight=\"900\">milieu 1<\/text>\n<text x=\"100\" y=\"355\" font-size=\"17\" fill=\"#0F766E\" font-weight=\"900\">milieu 2<\/text>\n<line x1=\"405\" y1=\"75\" x2=\"405\" y2=\"390\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"9,8\"\/>\n<text x=\"420\" y=\"95\" font-size=\"17\" fill=\"#0f172a\" font-weight=\"900\">Normale<\/text>\n<line x1=\"220\" y1=\"100\" x2=\"405\" y2=\"250\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<polygon points=\"405,250 381,243 392,229\" fill=\"#2563EB\"\/>\n<text x=\"140\" y=\"100\" font-size=\"16\" fill=\"#2563EB\" font-weight=\"900\">rayon incident<\/text>\n<line x1=\"405\" y1=\"250\" x2=\"625\" y2=\"350\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<polygon points=\"625,350 600,351 608,333\" fill=\"#E76F51\"\/>\n<text x=\"610\" y=\"375\" font-size=\"16\" fill=\"#E76F51\" font-weight=\"900\">rayon r\u00e9fract\u00e9<\/text>\n<line x1=\"405\" y1=\"250\" x2=\"555\" y2=\"140\" stroke=\"#94a3b8\" stroke-width=\"3\" stroke-dasharray=\"10,8\"\/>\n<text x=\"535\" y=\"135\" font-size=\"15\" fill=\"#64748b\" font-weight=\"900\">rayon r\u00e9fl\u00e9chi<\/text>\n<path d=\"M360 214 A65 65 0 0 1 405 185\" fill=\"none\" stroke=\"#F97316\" stroke-width=\"4\"\/>\n<text x=\"355\" y=\"190\" fill=\"#F97316\" font-size=\"22\" font-weight=\"900\">i<\/text>\n<path d=\"M405 305 A60 60 0 0 0 455 276\" fill=\"none\" stroke=\"#7C3AED\" stroke-width=\"4\"\/>\n<text x=\"460\" y=\"305\" fill=\"#7C3AED\" font-size=\"22\" font-weight=\"900\">r<\/text>\n<\/svg>\n<\/div>\n\n<h3>Exp\u00e9rience : trouver la relation entre i et r<\/h3>\n<table>\n<tr><th>i<\/th><td>20\u00b0<\/td><td>30\u00b0<\/td><td>40\u00b0<\/td><td>50\u00b0<\/td><td>60\u00b0<\/td><\/tr>\n<tr><th>r<\/th><td>13\u00b0<\/td><td>20\u00b0<\/td><td>26\u00b0<\/td><td>31\u00b0<\/td><td>35\u00b0<\/td><\/tr>\n<tr><th>sin(i)<\/th><td>0,342<\/td><td>0,500<\/td><td>0,643<\/td><td>0,766<\/td><td>0,866<\/td><\/tr>\n<tr><th>sin(r)<\/th><td>0,225<\/td><td>0,342<\/td><td>0,438<\/td><td>0,515<\/td><td>0,574<\/td><\/tr>\n<tr><th>sin(i)\/sin(r)<\/th><td>1,5<\/td><td>1,5<\/td><td>1,5<\/td><td>1,5<\/td><td>1,5<\/td><\/tr>\n<\/table>\n\n<p>\nOn remarque qu\u2019il existe une relation math\u00e9matique entre sin(i) et sin(r).\n<\/p>\n<div class=\"formule bluebox\">sin(i) \/ sin(r) = constante<\/div>\n\n<h3>Loi de Snell-Descartes de la r\u00e9fraction<\/h3>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<ul>\n<li><span class=\"red\">n\u2081<\/span> : indice de r\u00e9fraction du milieu 1 ;<\/li>\n<li><span class=\"red\">n\u2082<\/span> : indice de r\u00e9fraction du milieu 2 ;<\/li>\n<li><span class=\"red\">i<\/span> : angle d\u2019incidence ;<\/li>\n<li><span class=\"red\">r<\/span> : angle de r\u00e9fraction.<\/li>\n<\/ul>\n\n<div class=\"note\">\n<p>Exemples d\u2019indices :<\/p>\n<ul>\n<li>n<sub>air<\/sub> = 1,0 ;<\/li>\n<li>n<sub>eau<\/sub> = 1,33 ;<\/li>\n<li>n<sub>plexiglas<\/sub> = 1,5 ;<\/li>\n<li>n<sub>verre<\/sub> \u2248 1,5.<\/li>\n<\/ul>\n<p class=\"red\">\nL\u2019indice de r\u00e9fraction traduit la propagation de la lumi\u00e8re dans le milieu. n n\u2019a pas d\u2019unit\u00e9.\n<\/p>\n<\/div>\n\n<p>\nOn retrouve la relation obtenue avec l\u2019exp\u00e9rience :\n<\/p>\n<div class=\"formule bluebox\">sin(i) \/ sin(r) = n\u2082 \/ n\u2081<\/div>\n<p>Avec n\u2081 = 1,0 pour l\u2019air et n\u2082 = 1,5 pour le plexiglas, on obtient environ 1,5.<\/p>\n<\/section>\n\n<section class=\"section\">\n<h2>IV. Exercices types de r\u00e9fraction<\/h2>\n\n<div class=\"exercice\">\n<h3>Exercice type 1 \u2014 Calculer r<\/h3>\n<p>\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,0<\/span>, <span class=\"red\">n\u2082 = 1,5<\/span>, <span class=\"red\">i = 30\u00b0<\/span>.\nCalculer l\u2019angle de r\u00e9fraction r.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler r. On divise par n\u2082 :<\/p>\n<div class=\"formule bluebox\">(n\u2081 sin i) \/ n\u2082 = sin r<\/div>\n<p class=\"step\">On applique la fonction arcsin :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n\n<p>Application num\u00e9rique :<\/p>\n<p>r = arcsin((1,0 \u00d7 sin 30\u00b0) \/ 1,5)<\/p>\n<p class=\"resultat\">r = 19,47\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 2 \u2014 Calculer i<\/h3>\n<p>\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,0<\/span>, <span class=\"red\">n\u2082 = 1,5<\/span>, <span class=\"red\">r = 35\u00b0<\/span>.\nCalculer l\u2019angle d\u2019incidence i.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler i. On divise par n\u2081 :<\/p>\n<div class=\"formule bluebox\">sin i = (n\u2082 sin r) \/ n\u2081<\/div>\n<p class=\"step\">On applique la fonction arcsin :<\/p>\n<div class=\"formule bluebox\">i = arcsin((n\u2082 sin r) \/ n\u2081)<\/div>\n<\/div>\n\n<p>Application num\u00e9rique :<\/p>\n<p>i = arcsin((1,5 \u00d7 sin 35\u00b0) \/ 1,0)<\/p>\n<p class=\"resultat\">i = 59,36\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type 3 \u2014 Calculer n\u2082<\/h3>\n<p>\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,5<\/span>, <span class=\"red\">i = 65\u00b0<\/span>, <span class=\"red\">r = 70\u00b0<\/span>.\nCalculer n\u2082.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler n\u2082. On divise par sin r :<\/p>\n<div class=\"formule bluebox\">n\u2082 = (n\u2081 sin i) \/ sin r<\/div>\n<\/div>\n\n<p>Application num\u00e9rique :<\/p>\n<p>n\u2082 = (1,5 \u00d7 sin 65\u00b0) \/ sin 70\u00b0<\/p>\n<p class=\"resultat\">n\u2082 \u2248 1,4<\/p>\n<\/div>\n\n<div class=\"note\">\n<h3>Remarque importante<\/h3>\n<p class=\"red\">Si n\u2081 < n\u2082 alors r < i.<\/p>\n<p>Le rayon se rapproche de la normale.<\/p>\n<div class=\"schema\">\n<svg width=\"620\" height=\"250\" viewBox=\"0 0 620 250\">\n<line x1=\"80\" y1=\"125\" x2=\"540\" y2=\"125\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<line x1=\"310\" y1=\"35\" x2=\"310\" y2=\"220\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"8,7\"\/>\n<line x1=\"190\" y1=\"45\" x2=\"310\" y2=\"125\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<line x1=\"310\" y1=\"125\" x2=\"420\" y2=\"190\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<text x=\"100\" y=\"55\" fill=\"#2563EB\" font-weight=\"900\">n\u2081 < n\u2082<\/text>\n<text x=\"430\" y=\"190\" fill=\"#B83227\" font-weight=\"900\">r < i<\/text>\n<\/svg>\n<\/div>\n\n<p class=\"red\">Si n\u2081 > n\u2082 alors r > i.<\/p>\n<p>Le rayon s\u2019\u00e9loigne de la normale.<\/p>\n<div class=\"schema\">\n<svg width=\"620\" height=\"250\" viewBox=\"0 0 620 250\">\n<line x1=\"80\" y1=\"125\" x2=\"540\" y2=\"125\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<line x1=\"310\" y1=\"35\" x2=\"310\" y2=\"220\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"8,7\"\/>\n<line x1=\"235\" y1=\"45\" x2=\"310\" y2=\"125\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<line x1=\"310\" y1=\"125\" x2=\"500\" y2=\"190\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<text x=\"100\" y=\"55\" fill=\"#2563EB\" font-weight=\"900\">n\u2081 > n\u2082<\/text>\n<text x=\"430\" y=\"190\" fill=\"#B83227\" font-weight=\"900\">r > i<\/text>\n<\/svg>\n<\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>V. R\u00e9flexion totale<\/h2>\n\n<p class=\"red\">\nLorsque n\u2081 > n\u2082, si i devient trop important, alors d\u00e8s que r = 90\u00b0, il n\u2019y a plus de rayon r\u00e9fract\u00e9 :\ntoute la lumi\u00e8re se r\u00e9fl\u00e9chit sur le dioptre. On dit qu\u2019il y a r\u00e9flexion totale.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"760\" height=\"390\" viewBox=\"0 0 760 390\">\n<rect x=\"70\" y=\"60\" width=\"610\" height=\"255\" rx=\"20\" fill=\"#fbfdff\" stroke=\"#D9E2EC\"\/>\n<rect x=\"90\" y=\"190\" width=\"570\" height=\"100\" fill=\"#e0f2fe\" opacity=\".75\"\/>\n<line x1=\"90\" y1=\"190\" x2=\"660\" y2=\"190\" stroke=\"#B83227\" stroke-width=\"4\"\/>\n<line x1=\"375\" y1=\"75\" x2=\"375\" y2=\"290\" stroke=\"#0f172a\" stroke-width=\"3\" stroke-dasharray=\"9,8\"\/>\n<line x1=\"210\" y1=\"90\" x2=\"375\" y2=\"190\" stroke=\"#2563EB\" stroke-width=\"5\"\/>\n<line x1=\"375\" y1=\"190\" x2=\"530\" y2=\"100\" stroke=\"#0F766E\" stroke-width=\"5\"\/>\n<line x1=\"375\" y1=\"190\" x2=\"600\" y2=\"190\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<polygon points=\"600,190 578,178 578,202\" fill=\"#E76F51\"\/>\n<text x=\"510\" y=\"82\" fill=\"#0F766E\" font-size=\"16\" font-weight=\"900\">r\u00e9flexion<\/text>\n<text x=\"545\" y=\"215\" fill=\"#E76F51\" font-size=\"16\" font-weight=\"900\">r = 90\u00b0<\/text>\n<text x=\"420\" y=\"250\" fill=\"#B83227\" font-size=\"18\" font-weight=\"900\">pas de rayon r\u00e9fract\u00e9 au-del\u00e0 de i limite<\/text>\n<\/svg>\n<\/div>\n\n<p>\nLorsque <span class=\"red\">i > i<sub>limite<\/sub><\/span>, alors on dit qu\u2019il y a r\u00e9flexion totale.\n<\/p>\n\n<h3>Calcul de i<sub>limite<\/sub><\/h3>\n<p>Il y a r\u00e9flexion totale lorsque <span class=\"red\">r = 90\u00b0<\/span>.<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">Pour l\u2019angle limite, r = 90\u00b0 :<\/p>\n<div class=\"formule bluebox\">n\u2081 sin i<sub>limite<\/sub> = n\u2082 sin 90\u00b0<\/div>\n<p class=\"step\">Or sin 90\u00b0 = 1 :<\/p>\n<div class=\"formule bluebox\">n\u2081 sin i<sub>limite<\/sub> = n\u2082<\/div>\n<p class=\"step\">On divise par n\u2081 :<\/p>\n<div class=\"formule bluebox\">sin i<sub>limite<\/sub> = n\u2082 \/ n\u2081<\/div>\n<p class=\"step\">On applique arcsin :<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Exercice type \u2014 Calculer l\u2019angle limite<\/h3>\n<p>Donn\u00e9es : <span class=\"red\">n\u2081 = 1,5<\/span> et <span class=\"red\">n\u2082 = 1,0<\/span>.<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<p>i<sub>limite<\/sub> = arcsin(1,0 \/ 1,5)<\/p>\n<p class=\"resultat\">i<sub>limite<\/sub> = 41,8\u00b0<\/p>\n<p class=\"red\">D\u00e8s que i > i<sub>limite<\/sub>, il y a r\u00e9flexion totale. Il n\u2019y a pas de r\u00e9fraction.<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"fibre\">\n<h2>VI. Application : la fibre optique<\/h2>\n\n<p class=\"red\">\nC\u2019est une fibre transparente qui permet de guider la lumi\u00e8re.\n<\/p>\n<p class=\"red\">\nL\u2019indice de la gaine et du c\u0153ur sont choisis pour qu\u2019il y ait r\u00e9flexion totale sur la gaine\net que toute la lumi\u00e8re reste pi\u00e9g\u00e9e dans la fibre optique.\n<\/p>\n\n<div class=\"schema\">\n<svg width=\"850\" height=\"360\" viewBox=\"0 0 850 360\">\n<rect x=\"70\" y=\"115\" width=\"700\" height=\"120\" rx=\"25\" fill=\"#dbeafe\" stroke=\"#2563EB\" stroke-width=\"3\"\/>\n<rect x=\"70\" y=\"145\" width=\"700\" height=\"60\" rx=\"20\" fill=\"#fefce8\" stroke=\"#F59E0B\" stroke-width=\"3\"\/>\n<text x=\"110\" y=\"135\" fill=\"#2563EB\" font-size=\"17\" font-weight=\"900\">Gaine : n\u2082<\/text>\n<text x=\"110\" y=\"180\" fill=\"#B83227\" font-size=\"17\" font-weight=\"900\">C\u0153ur : n\u2081<\/text>\n<text x=\"420\" y=\"105\" fill=\"#B83227\" font-size=\"18\" font-weight=\"900\">n\u2081 > n\u2082<\/text>\n<polyline points=\"40,250 115,175 235,175 330,205 445,145 560,205 670,145 810,145\" fill=\"none\" stroke=\"#E76F51\" stroke-width=\"5\"\/>\n<g fill=\"#E76F51\">\n<polygon points=\"115,175 92,180 105,160\"\/>\n<polygon points=\"235,175 212,170 220,190\"\/>\n<polygon points=\"330,205 305,198 322,185\"\/>\n<polygon points=\"445,145 420,152 430,130\"\/>\n<polygon points=\"560,205 535,198 552,185\"\/>\n<polygon points=\"670,145 645,152 655,130\"\/>\n<\/g>\n<text x=\"255\" y=\"132\" fill=\"#0F766E\" font-size=\"16\" font-weight=\"900\">r\u00e9flexion totale<\/text>\n<text x=\"505\" y=\"255\" fill=\"#0F766E\" font-size=\"16\" font-weight=\"900\">la lumi\u00e8re reste pi\u00e9g\u00e9e<\/text>\n<\/svg>\n<\/div>\n\n<div class=\"note\">\n<p class=\"red\">Fibroscopie :<\/p>\n<p>On utilise deux fibres : une fibre \u00e9claire l\u2019int\u00e9rieur du corps ; l\u2019autre capte la lumi\u00e8re pour former une image sur un \u00e9cran.<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"exercices\">\n<h2>VII. Exercices contextualis\u00e9s<\/h2>\n\n<div class=\"exercice\">\n<h3>1. Air \u2192 eau : rayon entrant dans le lagon<\/h3>\n<p>\nOn fait une r\u00e9fraction de l\u2019air vers l\u2019eau.\nDonn\u00e9es : <span class=\"red\">n<sub>air<\/sub> = 1,0<\/span>, <span class=\"red\">n<sub>eau<\/sub> = 1,33<\/span>, <span class=\"red\">i = 25\u00b0<\/span>.\nCalculer r.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On isole r :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n<p>r = arcsin((1,0 \u00d7 sin 25\u00b0) \/ 1,33)<\/p>\n<p class=\"resultat\">r = 18,5\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>2. Eau \u2192 air : rayon qui sort de l\u2019eau<\/h3>\n<p>\nOn fait une r\u00e9fraction de l\u2019eau vers l\u2019air.\nDonn\u00e9es : <span class=\"red\">n<sub>eau<\/sub> = 1,33<\/span>, <span class=\"red\">n<sub>air<\/sub> = 1,0<\/span>, <span class=\"red\">r = 25\u00b0<\/span>.\nCalculer i.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On isole i :<\/p>\n<div class=\"formule bluebox\">i = arcsin((n\u2082 sin r) \/ n\u2081)<\/div>\n<\/div>\n<p>i = arcsin((1,0 \u00d7 sin 25\u00b0) \/ 1,33)<\/p>\n<p class=\"resultat\">i = 18,5\u00b0<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>3. Angle limite eau \u2192 air<\/h3>\n<p>Donn\u00e9es : n\u2081 = 1,33 pour l\u2019eau et n\u2082 = 1,0 pour l\u2019air.<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<p>i<sub>limite<\/sub> = arcsin(1,0 \/ 1,33)<\/p>\n<p class=\"resultat\">i<sub>limite<\/sub> \u2248 48,7\u00b0<\/p>\n<p>Il y aura r\u00e9flexion totale lorsque i > i<sub>limite<\/sub>.<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>4. Milieu inconnu<\/h3>\n<p>\nOn fait une r\u00e9fraction de l\u2019air dans un milieu inconnu.\nDonn\u00e9es : <span class=\"red\">n\u2081 = 1,0<\/span>, <span class=\"red\">i = 20\u00b0<\/span>, <span class=\"red\">r = 24\u00b0<\/span>.\nCalculer l\u2019indice n\u2082 du milieu inconnu.\n<\/p>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On veut isoler n\u2082. On divise par sin r :<\/p>\n<div class=\"formule bluebox\">n\u2082 = (n\u2081 sin i) \/ sin r<\/div>\n<\/div>\n<p>n\u2082 = (1,0 \u00d7 sin 20\u00b0) \/ sin 24\u00b0<\/p>\n<p class=\"resultat\">n\u2082 \u2248 0,84<\/p>\n<p class=\"red\">Impossible pour un milieu transparent usuel car un indice de r\u00e9fraction doit \u00eatre sup\u00e9rieur ou \u00e9gal \u00e0 1.<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>5. Fr\u00e9quence d\u2019un signal<\/h3>\n<p>On mesure une p\u00e9riode <span class=\"red\">T = 20,3 ms<\/span>. Calculer la fr\u00e9quence f.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">f = 1 \/ T<\/div>\n<p class=\"step\">On cherche f : la formule est d\u00e9j\u00e0 sous la bonne forme.<\/p>\n<div class=\"formule bluebox\">f = 1 \/ T<\/div>\n<\/div>\n<p class=\"red\">Conversion :<\/p>\n<p>T = 20,3 ms = 20,3 \u00d7 10\u207b\u00b3 s<\/p>\n<p>f = 1 \/ (20,3 \u00d7 10\u207b\u00b3)<\/p>\n<p class=\"resultat\">f = 49,2 Hz<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>6. R\u00e9seau \u00e9lectrique EEWF<\/h3>\n<p>EEWF fournit une \u00e9lectricit\u00e9 \u00e0 <span class=\"red\">f = 50 Hz<\/span>. Calculer la p\u00e9riode du signal.<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">f = 1 \/ T<\/div>\n<p class=\"step\">On veut isoler T. On multiplie par T :<\/p>\n<div class=\"formule bluebox\">f \u00d7 T = 1<\/div>\n<p class=\"step\">On divise par f :<\/p>\n<div class=\"formule bluebox\">T = 1 \/ f<\/div>\n<\/div>\n<p>T = 1 \/ 50<\/p>\n<p class=\"resultat\">T = 0,020 s<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>7. \u00c9chographie<\/h3>\n<p>\nOn mesure avec un \u00e9chographe un temps aller-retour <span class=\"red\">t<sub>A\/R<\/sub> = 2,0 ms<\/span>.\nOn donne <span class=\"red\">v<sub>son<\/sub> = 1500 m\u00b7s\u207b\u00b9<\/span>.\nCalculer la profondeur de l\u2019obstacle.\n<\/p>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">v = d \/ t<\/div>\n<p class=\"step\">On isole la distance parcourue :<\/p>\n<div class=\"formule bluebox\">d = v \u00d7 t<\/div>\n<\/div>\n<p>t = 2,0 \u00d7 10\u207b\u00b3 s<\/p>\n<p>d<sub>aller-retour<\/sub> = 1500 \u00d7 2,0 \u00d7 10\u207b\u00b3 = 3,0 m<\/p>\n<p class=\"red\">Le son fait un aller-retour, donc la profondeur vaut la moiti\u00e9.<\/p>\n<p class=\"resultat\">profondeur = 1,5 m<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"type-bac\">\n<h2>VIII. Partie type bac \/ \u00e9valuation \u2014 niveau seconde<\/h2>\n\n<div class=\"exercice\">\n<h3>Situation 1 \u2014 Fibroscopie et r\u00e9flexion totale<\/h3>\n<p>\nUne fibre optique poss\u00e8de un c\u0153ur d\u2019indice n\u2081 = 1,50 et une gaine d\u2019indice n\u2082 = 1,40.\n<\/p>\n<ol>\n<li>Expliquer le r\u00f4le de la fibre optique en fibroscopie.<\/li>\n<li>Dire pourquoi il faut que n\u2081 > n\u2082.<\/li>\n<li>Calculer l\u2019angle limite.<\/li>\n<li>Conclure : que se passe-t-il si i est sup\u00e9rieur \u00e0 cet angle ?<\/li>\n<\/ol>\n\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">i<sub>limite<\/sub> = arcsin(n\u2082 \/ n\u2081)<\/div>\n<p class=\"step\">On remplace par les indices de la gaine et du c\u0153ur :<\/p>\n<div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(1,40 \/ 1,50)<\/div>\n<\/div>\n<p>i<sub>limite<\/sub> \u2248 69,0\u00b0<\/p>\n<p class=\"resultat\">Si i > 69,0\u00b0, il y a r\u00e9flexion totale et la lumi\u00e8re reste guid\u00e9e dans la fibre.<\/p>\n<\/div>\n\n<div class=\"exercice\">\n<h3>Situation 2 \u2014 Rayon lumineux dans l\u2019eau<\/h3>\n<p>\nUn rayon lumineux passe de l\u2019air dans l\u2019eau avec i = 40\u00b0. On donne n<sub>air<\/sub> = 1,0 et n<sub>eau<\/sub> = 1,33.\n<\/p>\n<ol>\n<li>Faire le sch\u00e9ma de la r\u00e9fraction avec normale, rayon incident et rayon r\u00e9fract\u00e9.<\/li>\n<li>Calculer r.<\/li>\n<li>Dire si le rayon se rapproche ou s\u2019\u00e9loigne de la normale.<\/li>\n<\/ol>\n<div class=\"litteral\">\n<p class=\"start\">Formule du cours :<\/p>\n<div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div>\n<p class=\"step\">On isole r :<\/p>\n<div class=\"formule bluebox\">r = arcsin((n\u2081 sin i) \/ n\u2082)<\/div>\n<\/div>\n<p>r = arcsin((1,0 \u00d7 sin 40\u00b0) \/ 1,33)<\/p>\n<p class=\"resultat\">r \u2248 28,9\u00b0<\/p>\n<p>Comme n\u2081 < n\u2082, alors r < i : le rayon se rapproche de la normale.<\/p>\n<\/div>\n<\/section>\n\n<section class=\"section\" id=\"fiche-bilan\">\n<h2>\ud83d\udccc Fiche bilan \u2014 Fibroscopie, r\u00e9flexion et r\u00e9fraction<\/h2>\n<div class=\"grid3\">\n<div class=\"card\"><h3>Fibroscopie<\/h3><p>Technique d\u2019imagerie qui guide la lumi\u00e8re dans une fibre optique.<\/p><\/div>\n<div class=\"card\"><h3>Dioptre<\/h3><p class=\"red\">Surface qui s\u00e9pare deux milieux transparents.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9flexion<\/h3><p>Une partie de la lumi\u00e8re se r\u00e9fl\u00e9chit.<\/p><div class=\"formule\">i = i’<\/div><\/div>\n<div class=\"card\"><h3>R\u00e9fraction<\/h3><p>Une partie traverse le dioptre en \u00e9tant d\u00e9vi\u00e9e.<\/p><\/div>\n<div class=\"card\"><h3>Snell-Descartes<\/h3><div class=\"formule\">n\u2081 sin i = n\u2082 sin r<\/div><\/div>\n<div class=\"card\"><h3>Indice<\/h3><p>n traduit la propagation de la lumi\u00e8re dans un milieu.<\/p><p class=\"red\">n n\u2019a pas d\u2019unit\u00e9.<\/p><\/div>\n<div class=\"card\"><h3>Valeurs<\/h3><p>air 1,0 ; eau 1,33 ; verre \/ plexiglas \u2248 1,5.<\/p><\/div>\n<div class=\"card\"><h3>Cas n\u2081 < n\u2082<\/h3><p class=\"red\">r < i : le rayon se rapproche de la normale.<\/p><\/div>\n<div class=\"card\"><h3>Cas n\u2081 > n\u2082<\/h3><p class=\"red\">r > i : le rayon s\u2019\u00e9loigne de la normale.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9flexion totale<\/h3><p>Si n\u2081 > n\u2082 et i > i<sub>limite<\/sub>, il n\u2019y a plus de rayon r\u00e9fract\u00e9.<\/p><\/div>\n<div class=\"card\"><h3>Angle limite<\/h3><div class=\"formule bluebox\">i<sub>limite<\/sub> = arcsin(n\u2082\/n\u2081)<\/div><\/div>\n<div class=\"card\"><h3>Fibre optique<\/h3><p>La lumi\u00e8re reste pi\u00e9g\u00e9e par r\u00e9flexion totale.<\/p><\/div>\n<\/div>\n<\/section>\n\n<section class=\"section\">\n<h2>Carte mentale<\/h2>\n<div class=\"grid3\">\n<div class=\"mm-center\">FIBROSCOPIE \u2014 R\u00c9FLEXION \u2014 R\u00c9FRACTION<\/div>\n<div class=\"card\"><h3>Fibroscopie<\/h3><p>Image de l\u2019int\u00e9rieur du corps gr\u00e2ce aux fibres optiques.<\/p><\/div>\n<div class=\"card\"><h3>Dioptre<\/h3><p>S\u00e9paration entre deux milieux transparents.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9flexion<\/h3><p>i = i’.<\/p><\/div>\n<div class=\"card\"><h3>R\u00e9fraction<\/h3><p>n\u2081 sin i = n\u2082 sin r.<\/p><\/div>\n<div class=\"card\"><h3>Angle limite<\/h3><p>r = 90\u00b0 ; i<sub>limite<\/sub> = arcsin(n\u2082\/n\u2081).<\/p><\/div>\n<div class=\"card\"><h3>Fibre optique<\/h3><p>R\u00e9flexion totale dans le c\u0153ur.<\/p><\/div>\n<\/div>\n<\/section>\n\n<\/div>\n\n<script>\nlet utterance;\nfunction playAudioSummary(){\n speechSynthesis.cancel();\n const text=document.getElementById(\"audioText\").innerText;\n utterance=new SpeechSynthesisUtterance(text);\n utterance.lang=\"fr-FR\";\n utterance.rate=0.90;\n speechSynthesis.speak(utterance);\n}\nfunction stopAudioSummary(){speechSynthesis.cancel();}\n<\/script>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Seconde \u2014 Fibroscopie, r\u00e9flexion et r\u00e9fraction Fibroscopie : r\u00e9flexion et r\u00e9fraction de la lumi\u00e8re Seconde physique-chimie \u2022 Dioptre \u2022 R\u00e9flexion \u2022 R\u00e9fraction \u2022 Loi de Snell-Descartes...<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-775","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/775","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/comments?post=775"}],"version-history":[{"count":1,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/775\/revisions"}],"predecessor-version":[{"id":777,"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/pages\/775\/revisions\/777"}],"wp:attachment":[{"href":"https:\/\/pcwallis.malo.wf\/index.php\/wp-json\/wp\/v2\/media?parent=775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}