{"id":269569,"date":"2025-07-24T16:38:10","date_gmt":"2025-07-24T16:38:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=269569"},"modified":"2025-07-24T16:38:12","modified_gmt":"2025-07-24T16:38:12","slug":"a-plane-monochromatic-electromagnetic-wave-travels-from-one-medium-refractive-index-1-1-to-another-refractive-index-2-2-with-electric-field-oscillating-within-the-plane-of-incidence","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/24\/a-plane-monochromatic-electromagnetic-wave-travels-from-one-medium-refractive-index-1-1-to-another-refractive-index-2-2-with-electric-field-oscillating-within-the-plane-of-incidence\/","title":{"rendered":"A plane monochromatic electromagnetic wave travels from one medium (refractive index = 1.1) to another (refractive index = 2.2) with electric field oscillating within the plane of incidence."},"content":{"rendered":"\n

A plane monochromatic electromagnetic wave travels from one medium (refractive index = 1.1) to another (refractive index = 2.2) with electric field oscillating within the plane of incidence. What will be the reflection and transmission coefficients for the system if incident wave makes an angle 60\u00b0 with the normal to the interface?<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To calculate the reflection and transmission coefficients for the given system, we will use the Fresnel equations<\/strong>, which describe the behavior of light when it hits an interface between two media.<\/p>\n\n\n\n

Given:<\/h3>\n\n\n\n
    \n
  • Refractive index of medium 1 (incident medium), n1=1.1n_1 = 1.1n1\u200b=1.1<\/li>\n\n\n\n
  • Refractive index of medium 2 (transmitting medium), n2=2.2n_2 = 2.2n2\u200b=2.2<\/li>\n\n\n\n
  • Angle of incidence \u03b81=60\u2218\\theta_1 = 60^\\circ\u03b81\u200b=60\u2218<\/li>\n\n\n\n
  • The electric field oscillates within the plane of incidence, which corresponds to the p-polarized<\/strong> (parallel) case.<\/li>\n<\/ul>\n\n\n\n

    Step 1: Calculate the angle of refraction<\/h3>\n\n\n\n

    We can use Snell’s law<\/strong> to find the angle of refraction \u03b82\\theta_2\u03b82\u200b:n1sin\u2061(\u03b81)=n2sin\u2061(\u03b82)n_1 \\sin(\\theta_1) = n_2 \\sin(\\theta_2)n1\u200bsin(\u03b81\u200b)=n2\u200bsin(\u03b82\u200b)<\/p>\n\n\n\n

    Substitute the given values:1.1sin\u2061(60\u2218)=2.2sin\u2061(\u03b82)1.1 \\sin(60^\\circ) = 2.2 \\sin(\\theta_2)1.1sin(60\u2218)=2.2sin(\u03b82\u200b)sin\u2061(\u03b82)=1.1sin\u2061(60\u2218)2.2=1.1\u00d70.8662.2\u22480.431\\sin(\\theta_2) = \\frac{1.1 \\sin(60^\\circ)}{2.2} = \\frac{1.1 \\times 0.866}{2.2} \\approx 0.431sin(\u03b82\u200b)=2.21.1sin(60\u2218)\u200b=2.21.1\u00d70.866\u200b\u22480.431\u03b82\u2248sin\u2061\u22121(0.431)\u224825.7\u2218\\theta_2 \\approx \\sin^{-1}(0.431) \\approx 25.7^\\circ\u03b82\u200b\u2248sin\u22121(0.431)\u224825.7\u2218<\/p>\n\n\n\n

    Step 2: Calculate the reflection and transmission coefficients<\/h3>\n\n\n\n

    For p-polarization<\/strong>, the Fresnel equations for reflection RRR and transmission TTT are given by:R=\u2223n1cos\u2061(\u03b81)\u2212n2cos\u2061(\u03b82)n1cos\u2061(\u03b81)+n2cos\u2061(\u03b82)\u22232R = \\left| \\frac{n_1 \\cos(\\theta_1) – n_2 \\cos(\\theta_2)}{n_1 \\cos(\\theta_1) + n_2 \\cos(\\theta_2)} \\right|^2R=\u200bn1\u200bcos(\u03b81\u200b)+n2\u200bcos(\u03b82\u200b)n1\u200bcos(\u03b81\u200b)\u2212n2\u200bcos(\u03b82\u200b)\u200b\u200b2T=\u22232n1cos\u2061(\u03b81)n1cos\u2061(\u03b81)+n2cos\u2061(\u03b82)\u22232\u00d7n2n1T = \\left| \\frac{2 n_1 \\cos(\\theta_1)}{n_1 \\cos(\\theta_1) + n_2 \\cos(\\theta_2)} \\right|^2 \\times \\frac{n_2}{n_1}T=\u200bn1\u200bcos(\u03b81\u200b)+n2\u200bcos(\u03b82\u200b)2n1\u200bcos(\u03b81\u200b)\u200b\u200b2\u00d7n1\u200bn2\u200b\u200b<\/p>\n\n\n\n

    Now, calculate RRR:R=\u22231.1cos\u2061(60\u2218)\u22122.2cos\u2061(25.7\u2218)1.1cos\u2061(60\u2218)+2.2cos\u2061(25.7\u2218)\u22232R = \\left| \\frac{1.1 \\cos(60^\\circ) – 2.2 \\cos(25.7^\\circ)}{1.1 \\cos(60^\\circ) + 2.2 \\cos(25.7^\\circ)} \\right|^2R=\u200b1.1cos(60\u2218)+2.2cos(25.7\u2218)1.1cos(60\u2218)\u22122.2cos(25.7\u2218)\u200b\u200b2cos\u2061(60\u2218)=0.5,cos\u2061(25.7\u2218)\u22480.899\\cos(60^\\circ) = 0.5, \\quad \\cos(25.7^\\circ) \\approx 0.899cos(60\u2218)=0.5,cos(25.7\u2218)\u22480.899<\/p>\n\n\n\n

    Substitute the values:R=\u22231.1\u00d70.5\u22122.2\u00d70.8991.1\u00d70.5+2.2\u00d70.899\u22232R = \\left| \\frac{1.1 \\times 0.5 – 2.2 \\times 0.899}{1.1 \\times 0.5 + 2.2 \\times 0.899} \\right|^2R=\u200b1.1\u00d70.5+2.2\u00d70.8991.1\u00d70.5\u22122.2\u00d70.899\u200b\u200b2R=\u22230.55\u22121.9780.55+1.978\u22232=\u2223\u22121.4282.528\u22232\u22480.324R = \\left| \\frac{0.55 – 1.978}{0.55 + 1.978} \\right|^2 = \\left| \\frac{-1.428}{2.528} \\right|^2 \\approx 0.324R=\u200b0.55+1.9780.55\u22121.978\u200b\u200b2=\u200b2.528\u22121.428\u200b\u200b2\u22480.324<\/p>\n\n\n\n

    Now, calculate TTT:T=\u22232\u00d71.1\u00d70.51.1\u00d70.5+2.2\u00d70.899\u22232\u00d72.21.1T = \\left| \\frac{2 \\times 1.1 \\times 0.5}{1.1 \\times 0.5 + 2.2 \\times 0.899} \\right|^2 \\times \\frac{2.2}{1.1}T=\u200b1.1\u00d70.5+2.2\u00d70.8992\u00d71.1\u00d70.5\u200b\u200b2\u00d71.12.2\u200bT=\u22231.12.528\u22232\u00d72=(0.435)2\u00d72\u22480.379T = \\left| \\frac{1.1}{2.528} \\right|^2 \\times 2 = \\left( 0.435 \\right)^2 \\times 2 \\approx 0.379T=\u200b2.5281.1\u200b\u200b2\u00d72=(0.435)2\u00d72\u22480.379<\/p>\n\n\n\n

    Final Results:<\/h3>\n\n\n\n
      \n
    • Reflection coefficient R\u22480.324R \\approx 0.324R\u22480.324<\/li>\n\n\n\n
    • Transmission coefficient T\u22480.379T \\approx 0.379T\u22480.379<\/li>\n<\/ul>\n\n\n\n

      Explanation:<\/h3>\n\n\n\n

      The reflection and transmission coefficients give us the fraction of the incident wave that is reflected and transmitted, respectively. The reflection coefficient RRR is the ratio of the reflected intensity to the incident intensity, and similarly, TTT is the ratio of the transmitted intensity to the incident intensity.<\/p>\n\n\n\n

      These values depend on the refractive indices of the media and the angle of incidence. Since the refractive index of the second medium is higher, the transmitted wave bends towards the normal (as expected from Snell’s law), and part of the wave is reflected back into the first medium.<\/p>\n\n\n\n

      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

      A plane monochromatic electromagnetic wave travels from one medium (refractive index = 1.1) to another (refractive index = 2.2) with electric field oscillating within the plane of incidence. What will be the reflection and transmission coefficients for the system if incident wave makes an angle 60\u00b0 with the normal to the interface? The Correct Answer […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-269569","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/269569","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=269569"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/269569\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=269569"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=269569"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=269569"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}