{"id":27704,"date":"2021-09-18T13:00:18","date_gmt":"2021-09-18T13:00:18","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=27704"},"modified":"2024-08-10T23:57:59","modified_gmt":"2024-08-10T23:57:59","slug":"la-refraction-de-la-lumiere-et-la-loi-de-snell","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/la-refraction-de-la-lumiere-et-la-loi-de-snell\/","title":{"rendered":"La R\u00e9fraction de la Lumi\u00e8re et la Loi de Snell"},"content":{"rendered":"<p><center><\/p>\n<h1>La R\u00e9fraction de la Lumi\u00e8re et la Loi de Snell<\/h1>\n<p><em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\nDans ce cours, nous explorerons la r\u00e9fraction de la lumi\u00e8re \u00e0 travers l&#8217;analyse de la Loi de Snell. Le concept d&#8217;indice de r\u00e9fraction sera expliqu\u00e9, la Loi de Snell sera d\u00e9riv\u00e9e en utilisant le principe de Fermat, et nous \u00e9tudierons comment cette loi permet de calculer la trajectoire d&#8217;un rayon lumineux en passant d&#8217;un milieu \u00e0 un autre. De plus, nous aborderons les ph\u00e9nom\u00e8nes de r\u00e9flexion et de r\u00e9flexion totale, en appliquant ces concepts \u00e0 une s\u00e9rie d&#8217;exercices pratiques. L&#8217;objectif est de comprendre et d&#8217;appliquer la Loi de Snell dans les probl\u00e8mes d&#8217;optique.<\/em><\/p>\n<p><strong>Objectifs d&#8217;Apprentissage<\/strong><\/p>\n<ol style=\"text-align:left\">\n<li><strong>Comprendre<\/strong> le concept d&#8217;indice de r\u00e9fraction et sa relation avec la vitesse de la lumi\u00e8re dans diff\u00e9rents milieux.<\/li>\n<li><strong>Appliquer<\/strong> le principe de Fermat pour comprendre comment la lumi\u00e8re suit la trajectoire qui minimise le temps de parcours entre deux points.<\/li>\n<li><strong>D\u00e9montrer<\/strong> la Loi de Snell \u00e0 partir du principe de Fermat, pour d\u00e9terminer la trajectoire d&#8217;un rayon lumineux en traversant diff\u00e9rents milieux.<\/li>\n<li><strong>Calculer<\/strong> les angles d&#8217;incidence et de r\u00e9fraction en utilisant la Loi de Snell dans des situations avec diff\u00e9rents indices de r\u00e9fraction.<\/li>\n<li><strong>Comprendre<\/strong> le concept de r\u00e9flexion totale interne et comment elle est li\u00e9e \u00e0 l&#8217;angle critique et aux indices de r\u00e9fraction.<\/li>\n<li><strong>D\u00e9terminer<\/strong> l&#8217;angle critique pour la r\u00e9flexion totale interne \u00e0 l&#8217;interface entre deux milieux.<\/li>\n<\/ol>\n<p><strong>Sommaire<\/strong><br \/>\n<a href=\"#1\"><strong>L&#8217;indice de r\u00e9fraction<\/strong><\/a><br \/>\n<a href=\"#2\">Le principe de Fermat<\/a><br \/>\n<a href=\"#3\">La Loi de Snell de la r\u00e9fraction de la lumi\u00e8re<\/a><br \/>\n<a href=\"#4\"><strong>R\u00e9fraction, r\u00e9flexion et r\u00e9flexion totale de la lumi\u00e8re<\/strong><\/a><br \/>\n<a href=\"#5\"><strong>Exercices<\/strong><\/a><\/p>\n<p><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/LxhWbErujpo\" title=\"Lecteur vid\u00e9o YouTube\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><\/p>\n<p><a name=\"1\"><\/a><\/p>\n<h2>L&#8217;indice de r\u00e9fraction<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=LxhWbErujpo&amp;t=186s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">L&#8217;indice de r\u00e9fraction est d\u00e9fini<\/span><\/strong><\/a> comme le rapport entre la vitesse de la lumi\u00e8re dans le vide et la vitesse de la lumi\u00e8re dans ce milieu. Il s&#8217;agit d&#8217;une quantit\u00e9 sans dimension et elle est g\u00e9n\u00e9ralement repr\u00e9sent\u00e9e par la lettre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_k:<\/span><\/span><\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_k=\\displaystyle \\frac{c}{c_k}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">O\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">c<\/span><\/span> est la vitesse de la lumi\u00e8re dans le vide et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">c_k<\/span><\/span> est la vitesse de la lumi\u00e8re dans le milieu <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">k.<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Puisque la lumi\u00e8re se d\u00e9place toujours plus lentement dans un milieu que dans le vide, l&#8217;indice de r\u00e9fraction est toujours sup\u00e9rieur ou \u00e9gal \u00e0 1.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h3>Le principe de Fermat<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=LxhWbErujpo&amp;t=397s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">La vitesse de la lumi\u00e8re d\u00e9pend<\/span><\/strong><\/a> du milieu dans lequel elle se d\u00e9place. Plus l&#8217;indice de r\u00e9fraction du milieu est \u00e9lev\u00e9, plus la vitesse de la lumi\u00e8re y est faible ; et par rapport \u00e0 cela, le principe de Fermat est \u00e9nonc\u00e9 comme suit :<\/p>\n<p style=\"text-align: center; color: #000000; background-color: #80ff80;\">Lorsque la lumi\u00e8re se d\u00e9place d&#8217;un point \u00e0 un autre, elle emprunte le chemin qui minimise le temps de parcours.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Ce principe reste valable m\u00eame lorsque la lumi\u00e8re passe \u00e0 travers diff\u00e9rents milieux.<\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h3>La Loi de Snell de la r\u00e9fraction de la lumi\u00e8re<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=LxhWbErujpo&amp;t=608s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">\u00c0 partir du principe de Fermat<\/span><\/strong><\/a>, il est possible de formuler un probl\u00e8me d&#8217;optimisation qui permettra de d\u00e9terminer la trajectoire que suivra un rayon lumineux lorsqu&#8217;il traverse diff\u00e9rents milieux. C&#8217;est ce qui conduit finalement \u00e0 la Loi de Snell, dont nous allons maintenant voir la formulation et la d\u00e9monstration.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Supposons qu&#8217;un rayon parte d&#8217;un point A et arrive \u00e0 un point B en traversant une interface qui s\u00e9pare deux milieux avec des indices de r\u00e9fraction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_2<\/span><\/span> respectivement. Notre objectif sera de trouver une relation qui nous permette de calculer la trajectoire du rayon lumineux en suivant le principe de Fermat sur le temps de parcours minimal, et pour cela nous construisons le sch\u00e9ma suivant :<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-1CdoEOp5SHc\/YUDNPnQyOxI\/AAAAAAAAFjE\/RG-kYgV4KKwAE3QwiM9nB3cA-OOXesONQCLcBGAsYHQ\/s0\/n1n2leydeSnell.PNG\" width=\"875\" height=\"518\" alt=\"Loi de Snell\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-1CdoEOp5SHc\/YUDNPnQyOxI\/AAAAAAAAFjE\/RG-kYgV4KKwAE3QwiM9nB3cA-OOXesONQCLcBGAsYHQ\/s0\/n1n2leydeSnell.PNG\" width=\"875\" height=\"518\" alt=\"Loi de Snell\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify; color: #000000;\">Le raisonnement commence par l&#8217;analyse de la forme du temps de parcours du rayon lumineux. On a :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}{Temps de parcours} &amp; =\\displaystyle \\frac{{Distance}}{{Vitesse}} \\\\ \\\\ &amp; \\displaystyle =\\frac{{Distance dans le milieu 1}}{{Vitesse dans le milieu 1}} + \\frac{{Distance dans le milieu 2}}{{Vitesse dans le milieu 2}}\\\\ \\\\&amp; =\\displaystyle \\frac{\\sqrt{a^2 + x^2}}{c_1} + \\frac{\\sqrt{b^2 + (d-x)^2}}{c_2}\\end{array}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Une fois cela fait, en maintenant fixes les points A et B, le temps de parcours est d\u00e9termin\u00e9 par le point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/span> o\u00f9 le rayon touche l&#8217;interface entre les milieux. Avec cela, nous pouvons d\u00e9finir une fonction temporelle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t(x)<\/span><\/span> par<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t(x) = \\displaystyle \\frac{1}{c_1}\\sqrt{a^2 + x^2} + \\frac{1}{c_2}\\sqrt{b^2 + (d-x)^2}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Maintenant, comme le principe de Fermat stipule que la lumi\u00e8re suit la trajectoire qui minimise le temps de parcours, il est possible \u00e0 partir de l\u00e0 de trouver le <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/span> qui minimise la fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t(x).<\/span><\/span> Nous sommes en pr\u00e9sence d&#8217;un probl\u00e8me d&#8217;optimisation.<\/p>\n<p style=\"text-align: justify; color: #000000;\">En d\u00e9rivant <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t<\/span><\/span> par rapport \u00e0 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/span>, on obtient :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\begin{array}{rl}\\dfrac{dt}{dx} &amp;\\displaystyle = \\frac{1}{c_1}\\frac{d}{dx}\\sqrt{a^2 + x^2} + \\frac{1}{c_2}\\frac{d}{dx}\\sqrt{b^2+(d-x)^2}\\\\ \\\\ &amp;\\displaystyle = \\frac{1}{c_1} \\frac{2x}{2\\sqrt{a^2 + x^2}} + \\frac{1}{c_2}\\frac{2(d-x)(-1)}{2\\sqrt{b^2+(d-x)^2}} \\\\ \\\\ &amp;\\displaystyle = \\frac{1}{c_1} \\frac{x}{\\sqrt{a^2 + x^2}} - \\frac{1}{c_2}\\frac{(d-x)}{\\sqrt{b^2+(d-x)^2}} \\end{array}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Maintenant, notons que :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\\sin(\\theta_1) &amp;\\displaystyle =\\frac{x}{\\sqrt{a^2 + x^2}}\\\\ \\\\ \\sin(\\theta_2) &amp;\\displaystyle = \\frac{(d-x)}{\\sqrt{b^2+(d-x)^2}} \\\\ \\\\ c_1 &amp; \\displaystyle = \\frac{c}{n_1} \\\\ \\\\ c_2 &amp; \\displaystyle = \\frac{c}{n_2} \\end{array} <\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Ainsi, en rempla\u00e7ant ces \u00e9l\u00e9ments dans la d\u00e9riv\u00e9e du temps, on a :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\frac{dt}{dx} = \\frac{n_1}{c} \\sin(\\theta_1) - \\frac{n_2}{c}\\sin(\\theta_2)<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Enfin, si le point <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/span> minimise la fonction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">t(x),<\/span><\/span> alors la d\u00e9riv\u00e9e doit s&#8217;annuler, et on aura :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\color{blue}{n_1 \\sin(\\theta_1) = n_2 \\sin(\\theta_2)}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">C&#8217;est la Loi de Snell pour la r\u00e9fraction d&#8217;un rayon lumineux qui passe entre deux milieux et qui montre la relation entre l&#8217;angle d&#8217;incidence <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_1<\/span><\/span> et l&#8217;angle de r\u00e9fraction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_2.<\/span><\/span><\/p>\n<p><a name=\"3\"><\/a><\/p>\n<h2>R\u00e9fraction, r\u00e9flexion et r\u00e9flexion totale de la lumi\u00e8re<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=LxhWbErujpo&amp;t=1614s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Nous avons vu que lorsque la lumi\u00e8re<\/span> <\/strong><\/a>passe d&#8217;un milieu \u00e0 un autre, elle se r\u00e9fracte, mais en g\u00e9n\u00e9ral, il s&#8217;agit d&#8217;une combinaison entre r\u00e9fraction et r\u00e9flexion ; et en fonction des indices de r\u00e9fraction et de l&#8217;angle d&#8217;incidence du rayon lumineux, la r\u00e9fraction peut dispara\u00eetre, laissant place uniquement \u00e0 la r\u00e9flexion.<\/p>\n<p style=\"text-align: justify; color: #000000;\">Supposons qu&#8217;un rayon lumineux se propage d&#8217;un mat\u00e9riau <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/span> \u00e0 un autre <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">b<\/span><\/span> avec des indices de r\u00e9fraction <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_a<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_b<\/span><\/span> respectivement. Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_a \\gt n_b,<\/span><\/span> selon la Loi de Snell, on a :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\sin(\\theta_b) = \\frac{n_a}{n_b}\\sin(\\theta_a)<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">\u00c9tant donn\u00e9 que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_a\/n_b \\gt 1,<\/span><\/span> il en r\u00e9sulte que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin(\\theta_b) \\gt \\sin(\\theta_a),<\/span><\/span> ce qui signifie que le rayon r\u00e9fract\u00e9 s&#8217;\u00e9loigne de la normale. Cela implique qu&#8217;il doit exister un <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_a\\lt 90^o<\/span><\/span> pour lequel <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\sin(\\theta_b)=1<\/span><\/span> et, par cons\u00e9quent, <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_b=90^o,<\/span><\/span> comme illustr\u00e9 dans la figure suivante.<\/p>\n<p style=\"text-align: justify; color: #000000;\">L&#8217;angle d&#8217;incidence qui fait que le rayon se r\u00e9fracte sur l&#8217;interface est connu sous le nom d&#8217;angle critique et satisfait la relation suivante :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\sin(\\theta_{critique}) = \\frac{n_b}{n_a}<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Ce qui \u00e9quivaut \u00e0 dire :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\theta_{critique} = \\arcsin\\left( \\frac{n_b}{n_a} \\right)<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_a \\gt \\theta_{critique},<\/span><\/span> alors il y a r\u00e9flexion totale.<\/p>\n<p><a name=\"4\"><\/a><\/p>\n<h2>Exercices :<\/h2>\n<ol style=\"text-align: justify; color: #000000;\">\n<li>Consid\u00e9rons un rayon lumineux qui passe de l&#8217;eau au verre comme illustr\u00e9 dans la figure suivante :<br \/>\n<center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-98FbTK-_FZo\/YT-61nYxRnI\/AAAAAAAAFiM\/JbBsuAnS6IA8aB-4hvroeZ1qDF2ebxQUwCLcBGAsYHQ\/s0\/n1n2snell.PNG\" width=\"442\" height=\"321\" alt=\"rayon lumineux passant de l'eau au verre\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-98FbTK-_FZo\/YT-61nYxRnI\/AAAAAAAAFiM\/JbBsuAnS6IA8aB-4hvroeZ1qDF2ebxQUwCLcBGAsYHQ\/s0\/n1n2snell.PNG\" width=\"442\" height=\"321\" alt=\"rayon lumineux passant de l'eau au verre\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><br \/>\nL&#8217;indice de r\u00e9fraction de l&#8217;eau est de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1 = 1,33,<\/span><\/span> et celui du verre est de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_2=1,52.<\/span><\/span> Si un rayon lumineux qui passe de l&#8217;eau au verre frappe l&#8217;interface qui s\u00e9pare les deux milieux avec un angle d&#8217;inclinaison de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_1 = 60^o<\/span><\/span> par rapport \u00e0 la normale, avec quel angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_2<\/span><\/span> le rayon r\u00e9fract\u00e9 sortira-t-il ? <span class=\"collapseomatic \" id=\"id69e03a2493ba2\"  tabindex=\"0\" title=\"SOLUTION\"    >SOLUTION<\/span><div id=\"target-id69e03a2493ba2\" class=\"collapseomatic_content \">\nEn utilisant la Loi de Snell, on obtient :<\/p>\n<table>\n<tbody>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(1)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1 \\sin(\\theta_1) = n_2 \\sin(\\theta_2)<\/span><\/span><\/td>\n<td>; Loi de Snell<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv <\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\sin(\\theta_2) = \\frac{n_1}{n_2}\\sin(\\theta_1)<\/span><\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv <\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\theta_2 = \\arcsin\\left(\\frac{n_1}{n_2}\\sin(\\theta_1)\\right)<\/span><\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(2)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1=1,33<\/span><\/span><\/td>\n<td>; Indice de r\u00e9fraction de l&#8217;eau<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(3)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_2=1,52<\/span><\/span><\/td>\n<td>; Indice de r\u00e9fraction du verre<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(4)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_1=60^o<\/span><\/span><\/td>\n<td>; Angle d&#8217;incidence \u00e0 l&#8217;interface du rayon lumineux<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(5)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\theta_2 = \\arcsin\\left(\\frac{1,33}{1,52}\\sin(60^o)\\right) \\approx 49,268^o<\/span><\/span><\/td>\n<td>; De (1,2,3,4), Angle de r\u00e9fraction<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div><a href=\"https:\/\/www.youtube.com\/watch?v=LxhWbErujpo&amp;t=1363s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">[vid\u00e9o]<\/span><\/strong><\/a><\/li>\n<li>Trois liquides s\u00e9par\u00e9s par deux interfaces ont les indices de r\u00e9fraction suivants : <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1=1,33,<\/span><\/span> <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_2=1,41<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_3=1,68,<\/span><\/span>, et ils sont dispos\u00e9s comme illustr\u00e9 dans la figure suivante :<center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-AAZlxjqC4s4\/YUAdiGNMSQI\/AAAAAAAAFiU\/mOE-xMfybOoxenNH2O8sufjpTuzH6-WIwCLcBGAsYHQ\/s0\/n1n2n3snell.PNG\" width=\"443\" height=\"430\" alt=\"Loi de Snell appliqu\u00e9e \u00e0 trois milieux\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-AAZlxjqC4s4\/YUAdiGNMSQI\/AAAAAAAAFiU\/mOE-xMfybOoxenNH2O8sufjpTuzH6-WIwCLcBGAsYHQ\/s0\/n1n2n3snell.PNG\" width=\"443\" height=\"430\" alt=\"Loi de Snell appliqu\u00e9e \u00e0 trois milieux\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center>Si le rayon qui passe du milieu avec indice <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1<\/span><\/span> \u00e0 celui avec indice <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_2<\/span><\/span> le fait en frappant l&#8217;interface avec un angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta_1=70^o<\/span><\/span>, avec quel angle se r\u00e9fractera-t-il lorsqu&#8217;il passera dans le milieu avec indice <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_3<\/span><\/span> ? <span class=\"collapseomatic \" id=\"id69e03a2493ed8\"  tabindex=\"0\" title=\"SOLUTION\"    >SOLUTION<\/span><div id=\"target-id69e03a2493ed8\" class=\"collapseomatic_content \">\nDe mani\u00e8re analogue \u00e0 l&#8217;exercice pr\u00e9c\u00e9dent, nous avons le raisonnement suivant :<\/p>\n<table>\n<tbody>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(1)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1 \\sin(\\theta_1) = n_2 \\sin(\\theta_2) <\/span><\/span><\/td>\n<td>; Loi de Snell pour le passage du milieu n1 au n2<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(2)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_2 \\sin(\\theta_2) = n_3 \\sin(\\theta_3) <\/span><\/span><\/td>\n<td>; Loi de Snell pour le passage du milieu n2 au n3<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(3)<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">n_1 \\sin(\\theta_1) = n_3 \\sin(\\theta_3) <\/span><\/span><\/td>\n<td>; De (1,2)<\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\sin(\\theta_3) = \\frac{n_1}{n_3}\\sin(\\theta_1) <\/span><\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv<\/span><\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\theta_3 = \\arcsin\\left(\\frac{n_1}{n_3}\\sin(\\theta_1)\\right) <\/span><\/span><\/td>\n<td><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Enfin, en rempla\u00e7ant les donn\u00e9es, on obtient :<\/p>\n<p><center><br \/>\n<span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\theta_3= \\arcsin\\left(\\frac{1,33}{1,68}\\sin(70^o)\\right) \\approx 48,0667^o <\/span><\/span><\/center><br \/>\nNotez que ce raisonnement montre que nous pouvons effectuer les calculs en prenant uniquement en compte les milieux d&#8217;entr\u00e9e et de sortie du rayon, en ignorant compl\u00e8tement celui situ\u00e9 au milieu.<br \/>\n<\/div> <a href=\"https:\/\/www.youtube.com\/watch?v=LxhWbErujpo&amp;t=1417s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">[vid\u00e9o]<\/span><\/strong><\/a><\/li>\n<li>Depuis le fond d&#8217;une piscine, un rayon lumineux est dirig\u00e9 vers l&#8217;interface entre l&#8217;air et l&#8217;eau. D\u00e9terminez l&#8217;angle d&#8217;incidence pour qu&#8217;il se produise une r\u00e9flexion totale.<center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-GiDk_G3uybI\/YUDhONr0MaI\/AAAAAAAAFjM\/aSGUOspZeCsm7Cz7DG4r-JrCr03QhYyBgCLcBGAsYHQ\/s0\/%25C3%25A1ngulocr%25C3%25ADtico.PNG\" width=\"588\" height=\"358\" alt=\"rayon lumineux frappant \u00e0 l'angle critique\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-GiDk_G3uybI\/YUDhONr0MaI\/AAAAAAAAFjM\/aSGUOspZeCsm7Cz7DG4r-JrCr03QhYyBgCLcBGAsYHQ\/s0\/%25C3%25A1ngulocr%25C3%25ADtico.PNG\" width=\"588\" height=\"358\" alt=\"rayon lumineux frappant \u00e0 l'angle critique\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><br \/>\n<span class=\"collapseomatic \" id=\"id69e03a249405c\"  tabindex=\"0\" title=\"SOLUTION\"    >SOLUTION<\/span><div id=\"target-id69e03a249405c\" class=\"collapseomatic_content \">\nL&#8217;angle critique est donn\u00e9 par :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\theta_{critique}= \\arcsin\\left(\\frac{1,00}{1,33}\\right) \\approx 48,7535^o<\/span><\/span><\/p>\n<\/div><a href=\"https:\/\/www.youtube.com\/watch?v=LxhWbErujpo&amp;t=1869s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">[vid\u00e9o]<\/span><\/strong><\/a><\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>La R\u00e9fraction de la Lumi\u00e8re et la Loi de Snell R\u00e9sum\u00e9 : Dans ce cours, nous explorerons la r\u00e9fraction de la lumi\u00e8re \u00e0 travers l&#8217;analyse de la Loi de Snell. Le concept d&#8217;indice de r\u00e9fraction sera expliqu\u00e9, la Loi de Snell sera d\u00e9riv\u00e9e en utilisant le principe de Fermat, et nous \u00e9tudierons comment cette loi [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":27682,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":4,"footnotes":""},"categories":[847,647],"tags":[],"class_list":["post-27704","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-optique-geometrique","category-physique"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>La R\u00e9fraction de la Lumi\u00e8re et la Loi de Snell - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"La R\u00e9fraction et la Loi de Snell expliqu\u00e9es : apprenez comment la lumi\u00e8re change de direction en passant d&#039;un milieu 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