{"id":27620,"date":"2021-08-26T13:00:48","date_gmt":"2021-08-26T13:00:48","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=27620"},"modified":"2024-08-07T09:39:20","modified_gmt":"2024-08-07T09:39:20","slug":"problemes-resolus-des-miroirs-plans","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/","title":{"rendered":"Probl\u00e8mes R\u00e9solus des Miroirs Plans"},"content":{"rendered":"<p><center><\/p>\n<h1>Miroirs Plans, Probl\u00e8mes R\u00e9solus<\/h1>\n<p><em><strong>R\u00e9sum\u00e9 :<\/strong><br \/>\nDans cette le\u00e7on, nous examinerons quelques probl\u00e8mes r\u00e9solus des miroirs plans. L&#8217;angle de r\u00e9flexion <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma<\/span><\/span> est d\u00e9termin\u00e9 en fonction de l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span><\/span> entre deux miroirs plans reli\u00e9s par une charni\u00e8re, et des exemples sp\u00e9cifiques sont calcul\u00e9s. Les valeurs critiques de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span><\/span> sont examin\u00e9es pour que le rayon rebondisse une fois sur chaque miroir, et la formule pour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma<\/span><\/span> est valid\u00e9e. De plus, les angles d&#8217;incidence qui font revenir le rayon sur lui-m\u00eame sont identifi\u00e9s, et une s\u00e9quence d&#8217;angles de retour <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_n = n\\theta<\/span><\/span> est calcul\u00e9e.<br \/>\n<\/em><br \/>\n<u><strong>Objectifs d&#8217;apprentissage<\/strong><\/u><br \/>\n\u00c0 la fin de cette le\u00e7on, l&#8217;\u00e9tudiant sera capable de :<\/p>\n<ol style=\"text-align:left;\">\n<li><strong>Comprendre<\/strong> les formules fondamentales de l&#8217;optique des miroirs plans.<\/li>\n<li><strong>Appliquer<\/strong> la loi de r\u00e9flexion aux probl\u00e8mes avec des miroirs plans.<\/li>\n<li><strong>D\u00e9terminer<\/strong> l&#8217;angle de r\u00e9flexion <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma<\/span><\/span> en fonction de l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span><\/span> entre deux miroirs plans.<\/li>\n<li><strong>Analyser<\/strong> les limites des formules pour les miroirs et leurs conditions de validit\u00e9.<\/li>\n<\/ol>\n<p><strong>TABLE DES MATI\u00c8RES<\/strong><br \/>\n<a href=\"#1\"><strong>Introduction<\/strong><\/a><br \/>\n<a href=\"#2\"><strong>Miroirs reli\u00e9s par une charni\u00e8re<\/strong><\/a><br \/>\n<a href=\"#3\">Examen des limites du raisonnement<\/a><br \/>\n<a href=\"#4\">Angles de retour<\/a><\/p>\n<p><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/sX--0tertWI\" 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>Introduction<\/h2>\n<p style=\"text-align: justify; color: #000000;\">Dans la <a href=\"http:\/\/toposuranos.com\/material\/fr\/reflexion-en-miroirs-plans-et-spheriques\/\" rel=\"noopener\" target=\"_blank\">le\u00e7on pr\u00e9c\u00e9dente<\/a>, nous avons examin\u00e9 la plupart des formules li\u00e9es \u00e0 l&#8217;optique des miroirs plans et sph\u00e9riques ; cependant, pour une meilleure compr\u00e9hension de ces sujets, il est n\u00e9cessaire de revoir leur application dans la r\u00e9solution de probl\u00e8mes associ\u00e9s \u00e0 ces sujets. C&#8217;est pourquoi nous consacrerons cette partie exclusivement \u00e0 la r\u00e9vision de la solution de certains probl\u00e8mes. Cette fois, nous nous concentrerons exclusivement sur les miroirs plans.<\/p>\n<p><a name=\"2\"><\/a><\/p>\n<h2>Miroirs reli\u00e9s par une charni\u00e8re<\/h2>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=sX--0tertWI&amp;t=64s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">Deux miroirs plans reli\u00e9s par<\/span><\/strong><\/a> une extr\u00e9mit\u00e9 forment un angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta.<\/span><\/span> Si un rayon de lumi\u00e8re tombe sur un des miroirs avec un angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span><\/span> par rapport \u00e0 la normale de sorte que la lumi\u00e8re rebondisse une fois sur chaque miroir et se croise en formant un angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma:<\/span><\/span><\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-CA7kLZjaxyo\/YSMSe1ea97I\/AAAAAAAAFc8\/i5snILhLzT8XlVM88Hs8JDZiJvum5zmlgCLcBGAsYHQ\/s0\/ESPEJOPLANO1.PNG\" width=\"400\" height=\"100\" alt=\"angles dans les miroirs plans\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-CA7kLZjaxyo\/YSMSe1ea97I\/AAAAAAAAFc8\/i5snILhLzT8XlVM88Hs8JDZiJvum5zmlgCLcBGAsYHQ\/s0\/ESPEJOPLANO1.PNG\" width=\"400\" height=\"100\" alt=\"angles dans les miroirs plans\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<ol style=\"text-align: justify; color: #000000;\">\n<li type=\"a\">Trouvez une formule pour d\u00e9terminer l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma<\/span><\/span> en fonction des autres donn\u00e9es.<\/li>\n<li type=\"a\">Si le rayon de lumi\u00e8re tombe sur le premier miroir avec un angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha=30^o<\/span><\/span> et que l&#8217;angle entre les miroirs est de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta=50^o<\/span><\/span>, quel sera l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma<\/span><\/span>?<\/li>\n<\/ol>\n<span class=\"collapseomatic \" id=\"id69e37b67be849\"  tabindex=\"0\" title=\"SOLUTION\"    >SOLUTION<\/span><div id=\"target-id69e37b67be849\" class=\"collapseomatic_content \">\n<ol style=\"text-align: justify; color: #000000;\">\n<li type=\"a\">En d\u00e9finissant l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\beta<\/span><\/span> entre la normale du second miroir et le rayon lumineux r\u00e9fl\u00e9chi par le premier miroir, et en utilisant la loi de r\u00e9flexion dans les miroirs plans, nous pouvons compl\u00e9ter le sch\u00e9ma de la mani\u00e8re suivante :<br \/>\n<center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-zheSpWUbupU\/YSMU24o_WkI\/AAAAAAAAFdE\/iOtRNY_vBWMWaY24ycU8llcenhsBRVpDQCLcBGAsYHQ\/s0\/espejoplano2.PNG\" width=\"400\" height=\"100\" alt=\"angle entre les miroirs plans\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-zheSpWUbupU\/YSMU24o_WkI\/AAAAAAAAFdE\/iOtRNY_vBWMWaY24ycU8llcenhsBRVpDQCLcBGAsYHQ\/s0\/espejoplano2.PNG\" width=\"400\" height=\"100\" alt=\"angle entre les miroirs plans\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><br \/>\nEn gardant cela \u00e0 l&#8217;esprit, il est maintenant possible de faire le raisonnement suivant :<\/p>\n<table>\n<tbody>\n<tr>\n<td style=\"width: 50px;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(1)<\/span><\/span><\/td>\n<td style=\"width: 350px;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(90^o - \\alpha) + (90^o - \\beta) + \\theta = 180^o<\/span><\/span><\/td>\n<td>; Parce que la somme des angles int\u00e9rieurs d&#8217;un triangle est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">180^o<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv<\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\alpha + \\beta = \\theta <\/span><\/span><\/td>\n<td> <\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50px;\"><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\"> 2\\alpha +2\\beta + \\gamma = 180 <\/span><\/span><\/td>\n<td>; Parce que la somme des angles int\u00e9rieurs d&#8217;un triangle est <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">180^o<\/span><\/span><\/td>\n<\/tr>\n<tr>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">\\equiv<\/span><\/td>\n<td><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\gamma = 180 - 2(\\alpha + \\beta)<\/span><\/span><\/td>\n<td> <\/td>\n<\/tr>\n<tr>\n<td style=\"width: 50px;\"><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\">\\color{blue}{\\gamma = 180 - 2\\theta}<\/span><\/span><\/td>\n<td>; De (1,2)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Par cons\u00e9quent, il est d\u00e9duit que l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma<\/span><\/span> ne sera qu&#8217;une fonction de l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span><\/span> form\u00e9 par les miroirs et sa formule sera <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma(\\theta) = 180^0 - 2\\theta<\/span><\/span><\/li>\n<li type=\"a\">Bas\u00e9 sur le raisonnement de la partie pr\u00e9c\u00e9dente, nous avons que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma = 180^o - 2\\cdot 50^o = 80^o <\/span><\/span><\/li>\n<\/ol>\n<\/div>\n<p><a name=\"3\"><\/a><\/p>\n<h3>Examen des limites du raisonnement<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=sX--0tertWI&amp;t=464s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">L&#8217;exercice pr\u00e9c\u00e9dent a un probl\u00e8me d\u00e9licat<\/span><\/strong><\/a>. Si vous observez l&#8217;\u00e9nonc\u00e9, vous verrez qu&#8217;il est exig\u00e9 que le rayon lumineux ne doit rebondir qu&#8217;une seule fois sur chaque miroir ; cependant, toutes les valeurs de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span><\/span> ne conviennent pas. Trouvez les valeurs de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span><\/span> qui satisfont cette condition et, par cons\u00e9quent, permettent \u00e0 la formule obtenue dans l&#8217;exercice pr\u00e9c\u00e9dent d&#8217;\u00eatre valide.<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-FIE6EKNxu1Q\/YSM34HUa-NI\/AAAAAAAAFdM\/DVM38fUBrxggrGsXoJcAI4SgDC2u5gm1gCLcBGAsYHQ\/s0\/espejosplanos3.PNG\" width=\"1021\" height=\"485\" alt=\"Rayons lumineux rebondissant sur des miroirs plans\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-FIE6EKNxu1Q\/YSM34HUa-NI\/AAAAAAAAFdM\/DVM38fUBrxggrGsXoJcAI4SgDC2u5gm1gCLcBGAsYHQ\/s0\/espejosplanos3.PNG\" width=\"1021\" height=\"485\" alt=\"Rayons lumineux rebondissant sur des miroirs plans\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><span class=\"collapseomatic \" id=\"id69e37b67bea8e\"  tabindex=\"0\" title=\"SOLUTION\"    >SOLUTION<\/span><div id=\"target-id69e37b67bea8e\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">Nous avons que <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span><\/span> atteint la valeur \u00abcritique\u00bb lorsqu&#8217;il fait <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\beta=0^o;<\/span><\/span> et lorsque cela se produit, nous pouvons prendre un angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">x<\/span><\/span> qui permet le raisonnement suivant :<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-b_rcYFH-moo\/YSM--pA_6sI\/AAAAAAAAFdU\/VQRBmlh7I-oAhjuWi83GOphA1JKQHrrbACLcBGAsYHQ\/s0\/espejoplano4.PNG\" width=\"288\" height=\"319\" alt=\"rayon rebondissant sur des miroirs plans avec un angle critique\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-b_rcYFH-moo\/YSM--pA_6sI\/AAAAAAAAFdU\/VQRBmlh7I-oAhjuWi83GOphA1JKQHrrbACLcBGAsYHQ\/s0\/espejoplano4.PNG\" width=\"288\" height=\"319\" alt=\"rayon rebondissant sur des miroirs plans avec un angle critique\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify; color: #000000;\">Les deux \u00e9quations suivantes doivent \u00eatre satisfaites :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha + x = 90^o<\/span><\/span><\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta + x = 90^o<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Et cela n&#8217;est possible que si :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha = \\theta<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Autrement dit : la valeur <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha=\\theta<\/span><\/span> est la valeur de l&#8217;angle d&#8217;incidence critique telle que, si elle est d\u00e9pass\u00e9e, alors le rayon rebondira plus de deux fois sur un miroir et, par cons\u00e9quent, rendra la formule obtenue dans l&#8217;exercice pr\u00e9c\u00e9dent invalide. \u00c0 partir de ces r\u00e9sultats, nous pouvons corriger le r\u00e9sultat de l&#8217;exercice pr\u00e9c\u00e9dent en \u00e9crivant :<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\gamma(\\theta, \\alpha) = 180^0 - 2\\theta \\;\\;\\;\\; ; \\;\\;\\;\\; \\alpha \\in ]0,\\theta[ <\/span><\/span><\/p>\n<\/div>\n<p><a name=\"4\"><\/a><\/p>\n<h3>Angles de retour<\/h3>\n<p style=\"text-align: justify; color: #000000;\"><a href=\"https:\/\/www.youtube.com\/watch?v=sX--0tertWI&amp;t=809s\" target=\"_blank\" rel=\"noopener\"><strong><span style=\"color: #ff0000;\">\u00c0 partir de ces r\u00e9sultats, nous pouvons voir que<\/span><\/strong><\/a>, pour certains angles d&#8217;incidence, le rayon lumineux retourne sur lui-m\u00eame. Cela se produit lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha = 0^o<\/span><\/span> ou lorsque <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha = \\theta,<\/span><\/span> o\u00f9 <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta<\/span><\/span> est l&#8217;angle form\u00e9 entre les deux miroirs plans. Existe-t-il d&#8217;autres angles de retour? et s&#8217;ils existent, comment peuvent-ils \u00eatre calcul\u00e9s?<\/p>\n<span class=\"collapseomatic \" id=\"id69e37b67beb83\"  tabindex=\"0\" title=\"SOLUTION\"    >SOLUTION<\/span><div id=\"target-id69e37b67beb83\" class=\"collapseomatic_content \">\n<p style=\"text-align: justify; color: #000000;\">Pour r\u00e9soudre ce probl\u00e8me, nous devons imaginer la situation qui se produit lorsque le rayon lumineux tombe sur le premier miroir avec un angle par rapport \u00e0 la normale <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha\\in ]\\theta, 180^o[<\/span><\/span>. Lorsque cela se produit, nous avons une situation comme celle montr\u00e9e dans la figure suivante :<\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-T3bmPeT9SHU\/YSNnIfiDnzI\/AAAAAAAAFdc\/68ireIOvkuYdVvLw0dZD4oqeP2YL8Ml3QCLcBGAsYHQ\/s0\/espejo5.PNG\" width=\"650\" height=\"294\" alt=\"rayon contre des miroirs plans\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-T3bmPeT9SHU\/YSNnIfiDnzI\/AAAAAAAAFdc\/68ireIOvkuYdVvLw0dZD4oqeP2YL8Ml3QCLcBGAsYHQ\/s0\/espejo5.PNG\" width=\"650\" height=\"294\" alt=\"rayon contre des miroirs plans\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<p style=\"text-align: justify; color: #000000;\">Comme la somme des angles int\u00e9rieurs d&#8217;un triangle est de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">180^o<\/span><\/span>:<\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">(90^o - \\alpha) + (90^o + \\beta) + \\theta = 180<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">En simplifiant cette relation, nous pouvons obtenir l&#8217;angle <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\beta<\/span><\/span> en termes de <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span><\/span> et <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\theta.<\/span><\/span><\/p>\n<p style=\"text-align: center; color: #000000;\"><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\beta=\\alpha - \\theta<\/span><\/span><\/p>\n<p style=\"text-align: justify; color: #000000;\">Cette expression est importante car si <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\beta=\\theta,<\/span><\/span> alors, selon le raisonnement de l&#8217;exercice pr\u00e9c\u00e9dent, le rayon devrait revenir sur lui-m\u00eame lors de la prochaine r\u00e9flexion. Ainsi <span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha=2\\theta.<\/span><\/span> Par cons\u00e9quent, ce raisonnement peut \u00eatre \u00e9tendu de mani\u00e8re inductive \u00e0 travers :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_0 = 0^o<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_1 = \\theta<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_{n-1} = \\alpha_n - \\theta<\/span><\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">Et, \u00e0 partir de cela, nous avons la s\u00e9quence d&#8217;angles de retour :<\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li style=\"list-style-type: none;\">\n<ul style=\"text-align: justify; color: #000000;\">\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_0 = 0^o<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_1 = \\theta<\/span><\/span><\/li>\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_{2} = 2\\theta<\/span><\/span><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vdots<\/span><\/span><\/p>\n<ul style=\"text-align: justify; color: #000000;\">\n<li><span dir=\"ltr\"><span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_{n} = n\\theta<\/span><\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify; color: #000000;\">De plus, nous devons noter que l&#8217;angle entre les miroirs plans ainsi que chaque angle d&#8217;incidence doivent \u00eatre aigus.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Miroirs Plans, Probl\u00e8mes R\u00e9solus R\u00e9sum\u00e9 : Dans cette le\u00e7on, nous examinerons quelques probl\u00e8mes r\u00e9solus des miroirs plans. L&#8217;angle de r\u00e9flexion est d\u00e9termin\u00e9 en fonction de l&#8217;angle entre deux miroirs plans reli\u00e9s par une charni\u00e8re, et des exemples sp\u00e9cifiques sont calcul\u00e9s. Les valeurs critiques de sont examin\u00e9es pour que le rayon rebondisse une fois sur chaque [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":27598,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":16,"footnotes":""},"categories":[847,647],"tags":[],"class_list":["post-27620","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>Probl\u00e8mes R\u00e9solus des Miroirs Plans - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Probl\u00e8mes R\u00e9solus des Miroirs Plans : D\u00e9terminer les Angles de R\u00e9flexion, Examiner les Valeurs Critiques avec des Exemples Pratiques\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Probl\u00e8mes R\u00e9solus des Miroirs Plans\" \/>\n<meta property=\"og:description\" content=\"Probl\u00e8mes R\u00e9solus des Miroirs Plans : D\u00e9terminer les Angles de R\u00e9flexion, Examiner les Valeurs Critiques avec des Exemples Pratiques\" \/>\n<meta property=\"og:url\" content=\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2021-08-26T13:00:48+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-08-07T09:39:20+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/08\/espejosplanos-1024x342.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Probl\u00e8mes R\u00e9solus des Miroirs Plans\" \/>\n<meta name=\"twitter:description\" content=\"Probl\u00e8mes R\u00e9solus des Miroirs Plans : D\u00e9terminer les Angles de R\u00e9flexion, Examiner les Valeurs Critiques avec des Exemples Pratiques\" \/>\n<meta name=\"twitter:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/08\/espejosplanos.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minuto\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/#article\",\"isPartOf\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"http:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Probl\u00e8mes R\u00e9solus des Miroirs Plans\",\"datePublished\":\"2021-08-26T13:00:48+00:00\",\"dateModified\":\"2024-08-07T09:39:20+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/\"},\"wordCount\":1147,\"commentCount\":0,\"publisher\":{\"@id\":\"http:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/#primaryimage\"},\"thumbnailUrl\":\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2021\/08\/espejosplanos.jpg\",\"articleSection\":[\"Optique G\u00e9om\u00e9trique\",\"Physique\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/\",\"url\":\"http:\/\/toposuranos.com\/material\/fr\/problemes-resolus-des-miroirs-plans\/\",\"name\":\"Probl\u00e8mes R\u00e9solus des Miroirs Plans - 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