{"id":180381,"date":"2025-01-06T05:36:51","date_gmt":"2025-01-06T05:36:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=180381"},"modified":"2025-01-06T05:36:53","modified_gmt":"2025-01-06T05:36:53","slug":"a-ray-of-monochromatic-light-in-air-falls-on-the-upper-surface-of-two-parallel-sided-slabs-of-transparent-material-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/06\/a-ray-of-monochromatic-light-in-air-falls-on-the-upper-surface-of-two-parallel-sided-slabs-of-transparent-material-2\/","title":{"rendered":"A ray of monochromatic light in air falls on the upper surface of two parallel-sided slabs of transparent material"},"content":{"rendered":"\n<p>A ray of monochromatic light in air falls on the upper surface of two parallel-sided slabs of transparent material, as shown in the diagram. What is the angle of refraction f in the lower slab? 60\u00b0 n1- 2.30 n2-1.73<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-41.png\" alt=\"\" class=\"wp-image-180382\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>To find the angle of refraction ((f)) in the lower slab, we need to use <strong>Snell&#8217;s Law<\/strong>, which is expressed as:<\/p>\n\n\n\n<p>[<br>n_1 \\sin \\theta_1 = n_2 \\sin \\theta_2<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(n_1) and (n_2) are the refractive indices of the materials,<\/li>\n\n\n\n<li>(\\theta_1) and (\\theta_2) are the angles of incidence and refraction, respectively.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Refraction at the air-upper slab interface<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Angle of incidence ((\\theta_1)) = 60\u00b0,<\/li>\n\n\n\n<li>Refractive index of air ((n_{\\text{air}})) = 1,<\/li>\n\n\n\n<li>Refractive index of the upper slab ((n_1)) = 2.30.<\/li>\n<\/ul>\n\n\n\n<p>Using Snell&#8217;s Law:<br>[<br>n_{\\text{air}} \\sin 60^\\circ = n_1 \\sin \\theta_2<br>]<br>[<br>1 \\cdot \\sin 60^\\circ = 2.30 \\cdot \\sin \\theta_2<br>]<br>[<br>\\sin \\theta_2 = \\frac{\\sin 60^\\circ}{2.30}<br>]<br>[<br>\\sin \\theta_2 = \\frac{0.866}{2.30} = 0.376<br>]<br>[<br>\\theta_2 = \\arcsin(0.376) \\approx 22.1^\\circ<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Refraction at the upper-lower slab interface<\/h3>\n\n\n\n<p>The light passes into the second slab. Here:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Angle of incidence ((\\theta_2&#8242;)) = (22.1^\\circ),<\/li>\n\n\n\n<li>Refractive index of upper slab ((n_1)) = 2.30,<\/li>\n\n\n\n<li>Refractive index of lower slab ((n_2)) = 1.73.<\/li>\n<\/ul>\n\n\n\n<p>Using Snell&#8217;s Law:<br>[<br>n_1 \\sin \\theta_2&#8242; = n_2 \\sin f<br>]<br>[<br>2.30 \\cdot \\sin 22.1^\\circ = 1.73 \\cdot \\sin f<br>]<br>[<br>\\sin f = \\frac{2.30 \\cdot \\sin 22.1^\\circ}{1.73}<br>]<br>[<br>\\sin f = \\frac{2.30 \\cdot 0.376}{1.73} = \\frac{0.865}{1.73} \\approx 0.500<br>]<br>[<br>f = \\arcsin(0.500) = 30^\\circ<br>]<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p>The angle of refraction ((f)) in the lower slab is <strong>30\u00b0<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Snell&#8217;s Law governs how light bends when moving between materials with different refractive indices.<\/li>\n\n\n\n<li>In the first interface, light bends toward the normal because it moves into a denser medium ((n_1 = 2.30)).<\/li>\n\n\n\n<li>At the second interface, the light bends away from the normal because it enters a less dense medium ((n_2 = 1.73)).<\/li>\n\n\n\n<li>By systematically applying Snell&#8217;s Law at each boundary, the final angle of refraction ((f)) is found to be 30\u00b0.<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>A ray of monochromatic light in air falls on the upper surface of two parallel-sided slabs of transparent material, as shown in the diagram. What is the angle of refraction f in the lower slab? 60\u00b0 n1- 2.30 n2-1.73 The Correct Answer and Explanation is : To find the angle of refraction ((f)) in the [&hellip;]<\/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-180381","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180381","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=180381"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180381\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=180381"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=180381"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=180381"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}