{"id":233296,"date":"2025-06-13T04:10:23","date_gmt":"2025-06-13T04:10:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233296"},"modified":"2025-06-13T04:10:25","modified_gmt":"2025-06-13T04:10:25","slug":"light-of-wavelength-729-nm-and-observe-the-resulting-diffraction-pattern-on-a-screen-that-is-situated-2-43-m-from-the-slit","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/light-of-wavelength-729-nm-and-observe-the-resulting-diffraction-pattern-on-a-screen-that-is-situated-2-43-m-from-the-slit\/","title":{"rendered":"light of wavelength 729 nm and observe the resulting diffraction pattern on a screen that is situated 2.43 m from the slit"},"content":{"rendered":"\n<p>light of wavelength 729 nm and observe the resulting diffraction pattern on a screen that is situated 2.43 m from the slit. What is the width w, in centimeters, of the pattern&#8217;s central maximum?<br>1 point Problem 04: You illuminate a slit with a width of 73.5 \u00b5m with a light of wavelength 729 nm and observe the resulting diffraction pattern on a screen that is situated 2.43 m from the slit. What is the width w, in centimeters, of the pattern&#8217;s central maximum?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To determine the width of the central maximum in a single-slit diffraction pattern, we use the formula derived from the diffraction condition:w=2L\u22c5tan\u2061(\u03b8)\u22482L\u22c5sin\u2061(\u03b8)(for&nbsp;small&nbsp;angles)w = 2L \\cdot \\tan(\\theta) \\approx 2L \\cdot \\sin(\\theta) \\quad \\text{(for small angles)}w=2L\u22c5tan(\u03b8)\u22482L\u22c5sin(\u03b8)(for&nbsp;small&nbsp;angles)<\/p>\n\n\n\n<p>For a single slit, the angle \u03b8\\theta\u03b8 to the first minimum is given by the condition:a\u22c5sin\u2061(\u03b8)=\u03bba \\cdot \\sin(\\theta) = \\lambdaa\u22c5sin(\u03b8)=\u03bb<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>aaa = slit width = 73.5 \u00b5m = 73.5\u00d710\u2212673.5 \\times 10^{-6}73.5\u00d710\u22126 m<\/li>\n\n\n\n<li>\u03bb\\lambda\u03bb = wavelength of light = 729 nm = 729\u00d710\u22129729 \\times 10^{-9}729\u00d710\u22129 m<\/li>\n\n\n\n<li>LLL = distance from slit to screen = 2.43 m<\/li>\n\n\n\n<li>www = total width of central maximum = distance between first minima on both sides = 2y2y2y<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 1: Calculate sin\u2061(\u03b8)\\sin(\\theta)sin(\u03b8):<\/strong>sin\u2061(\u03b8)=\u03bba=729\u00d710\u2212973.5\u00d710\u22126=0.00991837\\sin(\\theta) = \\frac{\\lambda}{a} = \\frac{729 \\times 10^{-9}}{73.5 \\times 10^{-6}} = 0.00991837sin(\u03b8)=a\u03bb\u200b=73.5\u00d710\u22126729\u00d710\u22129\u200b=0.00991837<\/p>\n\n\n\n<p><strong>Step 2: Use small-angle approximation (sin\u2061(\u03b8)\u2248tan\u2061(\u03b8)\u2248\u03b8\\sin(\\theta) \\approx \\tan(\\theta) \\approx \\thetasin(\u03b8)\u2248tan(\u03b8)\u2248\u03b8)<\/strong>tan\u2061(\u03b8)\u2248\u03b8\u22480.00991837&nbsp;radians\\tan(\\theta) \\approx \\theta \\approx 0.00991837 \\ \\text{radians}tan(\u03b8)\u2248\u03b8\u22480.00991837&nbsp;radians<\/p>\n\n\n\n<p><strong>Step 3: Compute half-width y=L\u22c5tan\u2061(\u03b8)y = L \\cdot \\tan(\\theta)y=L\u22c5tan(\u03b8)<\/strong>y=2.43\u22c50.00991837=0.0240936&nbsp;my = 2.43 \\cdot 0.00991837 = 0.0240936 \\ \\text{m}y=2.43\u22c50.00991837=0.0240936&nbsp;m<\/p>\n\n\n\n<p><strong>Step 4: Compute full width of central maximum:<\/strong>w=2y=2\u22c50.0240936=0.0481872&nbsp;m=4.82&nbsp;cmw = 2y = 2 \\cdot 0.0240936 = 0.0481872 \\ \\text{m} = 4.82 \\ \\text{cm}w=2y=2\u22c50.0240936=0.0481872&nbsp;m=4.82&nbsp;cm<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Final Answer:<\/strong>4.82&nbsp;cm\\boxed{4.82 \\ \\text{cm}}4.82&nbsp;cm\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Explanation:<\/strong><br>In single-slit diffraction, the central maximum is the brightest and widest part of the diffraction pattern. It spans the distance between the first minima on either side of the center. The position of these minima depends on the wavelength of the light and the width of the slit. The relationship asin\u2061\u03b8=\u03bba \\sin\\theta = \\lambdaasin\u03b8=\u03bb describes where the first minimum occurs. Since the angle is typically small, we can apply the small-angle approximation, replacing tan\u2061\u03b8\\tan\\thetatan\u03b8 with sin\u2061\u03b8\\sin\\thetasin\u03b8, which simplifies calculations. The total width on the screen is then calculated using the geometry of the triangle formed by the slit, screen, and first minimum. After converting all units to SI, applying the formulas systematically yields the final width in meters, which is then converted to centimeters.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-439.jpeg\" alt=\"\" class=\"wp-image-233297\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>light of wavelength 729 nm and observe the resulting diffraction pattern on a screen that is situated 2.43 m from the slit. What is the width w, in centimeters, of the pattern&#8217;s central maximum?1 point Problem 04: You illuminate a slit with a width of 73.5 \u00b5m with a light of wavelength 729 nm and [&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-233296","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233296","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=233296"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233296\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233296"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233296"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233296"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}