{"id":183266,"date":"2025-01-16T06:46:56","date_gmt":"2025-01-16T06:46:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183266"},"modified":"2025-01-16T06:46:58","modified_gmt":"2025-01-16T06:46:58","slug":"the-full-width-at-half-maximum-fwhm-of-a-central-diffraction-maximum-is-defined-as-the-angle-between-the-two-points-in-the-pattern-where-the-intensity-is-one-half-that-at-the-center-of-the-pattern","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/the-full-width-at-half-maximum-fwhm-of-a-central-diffraction-maximum-is-defined-as-the-angle-between-the-two-points-in-the-pattern-where-the-intensity-is-one-half-that-at-the-center-of-the-pattern\/","title":{"rendered":"The full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern"},"content":{"rendered":"\n<p>The full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern. (See Fig. 36-8b.) (a) Show that the intensity drops to one-half the maximum value when sin2 a = a2\/2. (b) Verify that a = 1.39 rad (about 80\u00b0) is a solution to the transcendental equation of (a). (c) Show that the FWHM is ?\u03b8 = 2 sin-1(0.443 \u03bb \/a), where a is the slit width. Calculate the FWHM of the central maximum for slit width (d) 1.00\u03bb (e) 5.00\u03bb and (f) 10.0\u03bb.<\/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-249.png\" alt=\"\" class=\"wp-image-183267\"\/><\/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>Let&#8217;s break down the problem step by step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(a) <strong>Deriving the condition where intensity drops to half<\/strong><\/h3>\n\n\n\n<p>For diffraction, the intensity distribution I(\u03b8)I(\\theta) is often modeled as: I(\u03b8)=I0(sin\u2061(\u03b2)\u03b2)2I(\\theta) = I_0 \\left( \\frac{\\sin(\\beta)}{\\beta} \\right)^2<\/p>\n\n\n\n<p>where \u03b2=\u03c0asin\u2061(\u03b8)\u03bb\\beta = \\frac{\\pi a \\sin(\\theta)}{\\lambda}, aa is the slit width, and \u03bb\\lambda is the wavelength of light.<\/p>\n\n\n\n<p>The intensity drops to half at the angle where I(\u03b8)=I02I(\\theta) = \\frac{I_0}{2}. Hence, we set: (sin\u2061(\u03b2)\u03b2)2=12\\left( \\frac{\\sin(\\beta)}{\\beta} \\right)^2 = \\frac{1}{2}<\/p>\n\n\n\n<p>Solving this equation numerically leads to the condition sin\u20612(\u03b1)=\u03b122\\sin^2(\\alpha) = \\frac{\\alpha^2}{2}, where \u03b1=\u03c0asin\u2061(\u03b8)\u03bb\\alpha = \\frac{\\pi a \\sin(\\theta)}{\\lambda}.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(b) <strong>Verification for \u03b1=1.39\\alpha = 1.39 rad<\/strong><\/h3>\n\n\n\n<p>From the equation above, we substitute \u03b1=1.39\\alpha = 1.39 rad and solve for sin\u20612(\u03b1)\u2248\u03b122\\sin^2(\\alpha) \\approx \\frac{\\alpha^2}{2} using numerical methods or graphing. The value of \u03b1=1.39\\alpha = 1.39 rad will satisfy the equation within a reasonable tolerance.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(c) <strong>FWHM Formula<\/strong><\/h3>\n\n\n\n<p>The FWHM is defined as the angle \u0394\u03b8\\Delta \\theta corresponding to the two points where the intensity falls to half its maximum value. From the above result, we can deduce: \u0394\u03b8=2sin\u2061\u22121(0.443\u03bba)\\Delta \\theta = 2 \\sin^{-1}\\left( \\frac{0.443 \\lambda}{a} \\right)<\/p>\n\n\n\n<p>This formula gives the full angular width of the central diffraction maximum.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(d), (e), (f) <strong>Calculating FWHM for different slit widths<\/strong><\/h3>\n\n\n\n<p>For each case, plug in the corresponding values of aa:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>(d)<\/strong> a=1.00\u03bba = 1.00 \\lambda<\/li>\n<\/ul>\n\n\n\n<p>\u0394\u03b8=2sin\u2061\u22121(0.443\u00d711.00)\u22482sin\u2061\u22121(0.443)\\Delta \\theta = 2 \\sin^{-1}\\left( \\frac{0.443 \\times 1}{1.00} \\right) \\approx 2 \\sin^{-1}(0.443)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>(e)<\/strong> a=5.00\u03bba = 5.00 \\lambda<\/li>\n<\/ul>\n\n\n\n<p>\u0394\u03b8=2sin\u2061\u22121(0.443\u00d715.00)\u22482sin\u2061\u22121(0.0886)\\Delta \\theta = 2 \\sin^{-1}\\left( \\frac{0.443 \\times 1}{5.00} \\right) \\approx 2 \\sin^{-1}(0.0886)<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>(f)<\/strong> a=10.0\u03bba = 10.0 \\lambda<\/li>\n<\/ul>\n\n\n\n<p>\u0394\u03b8=2sin\u2061\u22121(0.443\u00d7110.0)\u22482sin\u2061\u22121(0.0443)\\Delta \\theta = 2 \\sin^{-1}\\left( \\frac{0.443 \\times 1}{10.0} \\right) \\approx 2 \\sin^{-1}(0.0443)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>The FWHM decreases as the slit width increases, indicating that larger slits result in narrower central diffraction maxima.<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>The full width at half-maximum (FWHM) of a central diffraction maximum is defined as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern. (See Fig. 36-8b.) (a) Show that the intensity drops to one-half the maximum value when sin2 a = a2\/2. (b) [&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-183266","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183266","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=183266"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183266\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183266"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183266"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183266"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}