{"id":182161,"date":"2025-01-13T13:04:55","date_gmt":"2025-01-13T13:04:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182161"},"modified":"2025-01-13T13:04:57","modified_gmt":"2025-01-13T13:04:57","slug":"you-have-been-asked-to-measure-the-width-of-a-slit-in-a-piece-of-paper","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/you-have-been-asked-to-measure-the-width-of-a-slit-in-a-piece-of-paper\/","title":{"rendered":"You have been asked to measure the width of a slit in a piece of paper"},"content":{"rendered":"\n<p>You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80.0 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance between them to be 17.9millimeters. (Figure 1)<\/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-147.png\" alt=\"\" class=\"wp-image-182162\"\/><\/figure>\n\n\n\n<p>Part A<br>What is the width a of the slit?<br>Express your answer in micrometers, to three significant figures.<\/p>\n\n\n\n<p>Part B<br>If the entire apparatus were submerged in water, would the width of the central peak change?<br>If the entire apparatus were submerged in water, would the width of the central peak change?<br>The width would increase.<br>The width would decrease.<br>The width would not change.<\/p>\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><strong>Part A: Determining the Slit Width<\/strong><\/p>\n\n\n\n<p>\ue203In single-slit diffraction, the angular positions of intensity minima are given by:\ue204\ue206<\/p>\n\n\n\n<p>\ue203asin\u2061\u03b8m=m\u03bba \\sin \\theta_m = m \\lambda\ue204\ue206<\/p>\n\n\n\n<p>\ue203where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>aa is the slit width,<\/li>\n\n\n\n<li>\u03b8m\\theta_m is the angle to the mm-th minimum,<\/li>\n\n\n\n<li>mm is the order of the minimum (with m=1m = 1 for the first minimum),<\/li>\n\n\n\n<li>\u03bb\\lambda is the wavelength of the light.\ue204\ue206<\/li>\n<\/ul>\n\n\n\n<p>\ue203The distance between two adjacent minima on the screen is related to the angle between them by:\ue204\ue206<\/p>\n\n\n\n<p>\ue203\u0394y=L(\u03b8m+1\u2212\u03b8m)\\Delta y = L (\\theta_{m+1} &#8211; \\theta_m)\ue204\ue206<\/p>\n\n\n\n<p>\ue203where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u0394y\\Delta y is the distance between adjacent minima,<\/li>\n\n\n\n<li>LL is the distance from the slit to the screen.\ue204\ue206<\/li>\n<\/ul>\n\n\n\n<p>\ue203For small angles, sin\u2061\u03b8\u2248\u03b8\\sin \\theta \\approx \\theta, so the angular separation between adjacent minima is approximately:\ue204\ue206<\/p>\n\n\n\n<p>\ue203\u03b8m+1\u2212\u03b8m\u2248\u03bba\\theta_{m+1} &#8211; \\theta_m \\approx \\frac{\\lambda}{a}\ue204\ue206<\/p>\n\n\n\n<p>\ue203Thus, the distance between adjacent minima is:\ue204\ue206<\/p>\n\n\n\n<p>\ue203\u0394y=L\u03bba\\Delta y = L \\frac{\\lambda}{a}\ue204\ue206<\/p>\n\n\n\n<p>\ue203Rearranging to solve for aa:\ue204\ue206<\/p>\n\n\n\n<p>\ue203a=L\u03bb\u0394ya = \\frac{L \\lambda}{\\Delta y}\ue204\ue206<\/p>\n\n\n\n<p><strong>Given:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\ue203L=80.0\u2009cm=0.800\u2009mL = 80.0 \\, \\text{cm} = 0.800 \\, \\text{m}\ue204\ue206<\/li>\n\n\n\n<li>\ue203\u03bb=633\u2009nm=633\u00d710\u22129\u2009m\\lambda = 633 \\, \\text{nm} = 633 \\times 10^{-9} \\, \\text{m}\ue204\ue206<\/li>\n\n\n\n<li>\ue203\u0394y=17.9\u2009mm=17.9\u00d710\u22123\u2009m\\Delta y = 17.9 \\, \\text{mm} = 17.9 \\times 10^{-3} \\, \\text{m}\ue204\ue206<\/li>\n<\/ul>\n\n\n\n<p>\ue203Substituting these values:\ue204\ue206<\/p>\n\n\n\n<p>\ue203a=0.800\u00d7633\u00d710\u2212917.9\u00d710\u22123a = \\frac{0.800 \\times 633 \\times 10^{-9}}{17.9 \\times 10^{-3}}\ue204\ue206<\/p>\n\n\n\n<p>\ue203a\u224835.7\u00d710\u22126\u2009m=35.7\u2009\u03bcma \\approx 35.7 \\times 10^{-6} \\, \\text{m} = 35.7 \\, \\mu\\text{m}\ue204\ue206<\/p>\n\n\n\n<p>\ue203Therefore, the slit width is approximately <strong>35.7 micrometers<\/strong>.\ue204\ue206<\/p>\n\n\n\n<p><strong>Part B: Effect of Submerging the Apparatus in Water<\/strong><\/p>\n\n\n\n<p>\ue203When the apparatus is submerged in water, the wavelength of the laser light decreases due to the change in the refractive index. The wavelength in a medium is given by:\ue204\ue206<\/p>\n\n\n\n<p>\ue203\u03bbmedium=\u03bbvacuumn\\lambda_{\\text{medium}} = \\frac{\\lambda_{\\text{vacuum}}}{n}\ue204\ue206<\/p>\n\n\n\n<p>\ue203where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03bbvacuum\\lambda_{\\text{vacuum}} is the wavelength in vacuum,<\/li>\n\n\n\n<li>nn is the refractive index of the medium.\ue204\ue206<\/li>\n<\/ul>\n\n\n\n<p>\ue203For water, n\u22481.33n \\approx 1.33. Therefore, the wavelength in water is:\ue204\ue206<\/p>\n\n\n\n<p>\ue203\u03bbwater=633\u00d710\u221291.33\u2248476.7\u00d710\u22129\u2009m\\lambda_{\\text{water}} = \\frac{633 \\times 10^{-9}}{1.33} \\approx 476.7 \\times 10^{-9} \\, \\text{m}\ue204\ue206<\/p>\n\n\n\n<p>\ue203The angular position of the first minimum in single-slit diffraction is given by:\ue204\ue206<\/p>\n\n\n\n<p>\ue203asin\u2061\u03b81=\u03bbmediuma \\sin \\theta_1 = \\lambda_{\\text{medium}}\ue204\ue206<\/p>\n\n\n\n<p>\ue203Since aa remains constant and \u03bbwater\\lambda_{\\text{water}} is smaller than \u03bbvacuum\\lambda_{\\text{vacuum}}, the angle \u03b81\\theta_1 decreases. For small angles, sin\u2061\u03b81\u2248\u03b81\\sin \\theta_1 \\approx \\theta_1, so the position of the first minimum on the screen, y1y_1, is:\ue204\ue206<\/p>\n\n\n\n<p>\ue203y1=L\u03b81=L\u03bbmediumay_1 = L \\theta_1 = L \\frac{\\lambda_{\\text{medium}}}{a}\ue204\ue206<\/p>\n\n\n\n<p>\ue203Since \u03bbmedium\\lambda_{\\text{medium}} decreases and aa remains constant, y1y_1 decreases. Therefore, the width of the central peak, defined as the distance between the first minima on either side, decreases.\ue204\ue206<\/p>\n\n\n\n<p>\ue203In conclusion, if the entire apparatus were submerged in water, the width of the central peak would decrease.\ue204<\/p>\n","protected":false},"excerpt":{"rendered":"<p>You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80.0 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance [&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-182161","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182161","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=182161"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182161\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182161"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182161"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}