The spectral distribution of the radiation emitted by a diffuse surface may be approximated as follows. HW 2 Q2 <\/p>\n\n\n\n
(a) What is the total emissive power? <\/p>\n\n\n\n
(b) What is the total intensity of the radiation emitted in the normal direction and at an angle of 30\u00b0 from the normal? <\/p>\n\n\n\n
(c) Determine the fraction of the emissive power leaving the surface in the directions \u03c0\/4 \u2264 \u03b8 \u2264 \u03c0\/2.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To solve this problem, we need to analyze the radiation emitted by a diffuse surface based on the spectral distribution provided. Let\u2019s denote the spectral emissive power by ( E(\\lambda) ) and assume that it is given in a specific form (e.g., Planck’s law, or a similar distribution).<\/p>\n\n\n\n The total emissive power ( E ) of a surface can be calculated by integrating the spectral emissive power over all wavelengths:<\/p>\n\n\n\n [ For a blackbody, this can be evaluated using Planck\u2019s law, resulting in the Stefan-Boltzmann law:<\/p>\n\n\n\n [ where ( \\sigma ) is the Stefan-Boltzmann constant and ( T ) is the absolute temperature of the surface. If the surface is not a perfect blackbody, it is necessary to multiply by the emissivity ( \\epsilon ) of the surface:<\/p>\n\n\n\n [ The intensity ( I ) of radiation emitted in a specific direction is given by:<\/p>\n\n\n\n [ where ( \\theta ) is the angle from the normal. To find the intensity in the normal direction (( \\theta = 0^\\circ )) and at ( \\theta = 30^\\circ ):<\/p>\n\n\n\n [ [ To determine the fraction of emissive power leaving the surface in the range ( \\frac{\\pi}{4} \\leq \\theta \\leq \\frac{\\pi}{2} ), we need to integrate the intensity over this angle and divide by the total emissive power:<\/p>\n\n\n\n [ Using ( I(\\theta) = \\frac{E}{\\pi} \\cos(\\theta) ):<\/p>\n\n\n\n [ Calculating the integral:<\/p>\n\n\n\n [ Thus,<\/p>\n\n\n\n [ So the fraction of the emissive power is:<\/p>\n\n\n\n [ This represents the fraction of the total emissive power leaving the surface within the specified angle range.<\/p>\n","protected":false},"excerpt":{"rendered":" The spectral distribution of the radiation emitted by a diffuse surface may be approximated as follows. HW 2 Q2 (a) What is the total emissive power? (b) What is the total intensity of the radiation emitted in the normal direction and at an angle of 30\u00b0 from the normal? (c) Determine the fraction of the […]<\/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-153929","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153929","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=153929"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/153929\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=153929"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=153929"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=153929"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Total Emissive Power<\/h3>\n\n\n\n
E = \\int_0^\\infty E(\\lambda) \\, d\\lambda
]<\/p>\n\n\n\n
E = \\sigma T^4
]<\/p>\n\n\n\n
E = \\epsilon \\sigma T^4
]<\/p>\n\n\n\n(b) Total Intensity of Radiation Emitted<\/h3>\n\n\n\n
I(\\theta) = \\frac{E}{\\pi} \\cos(\\theta)
]<\/p>\n\n\n\n\n
I(0) = \\frac{E}{\\pi} \\cos(0) = \\frac{E}{\\pi}
]<\/p>\n\n\n\n\n
I(30) = \\frac{E}{\\pi} \\cos(30) = \\frac{E}{\\pi} \\cdot \\frac{\\sqrt{3}}{2} = \\frac{\\sqrt{3} E}{2\\pi}
]<\/p>\n\n\n\n(c) Fraction of Emissive Power in Specific Directions<\/h3>\n\n\n\n
\\text{Fraction} = \\frac{\\int_{\\pi\/4}^{\\pi\/2} I(\\theta) \\, d\\theta}{E}
]<\/p>\n\n\n\n
\\text{Fraction} = \\frac{1}{E} \\int_{\\pi\/4}^{\\pi\/2} \\frac{E}{\\pi} \\cos(\\theta) \\, d\\theta = \\frac{1}{\\pi} \\int_{\\pi\/4}^{\\pi\/2} \\cos(\\theta) \\, d\\theta
]<\/p>\n\n\n\n
\\int \\cos(\\theta) \\, d\\theta = \\sin(\\theta)
]<\/p>\n\n\n\n
\\int_{\\pi\/4}^{\\pi\/2} \\cos(\\theta) \\, d\\theta = \\sin\\left(\\frac{\\pi}{2}\\right) – \\sin\\left(\\frac{\\pi}{4}\\right) = 1 – \\frac{\\sqrt{2}}{2}
]<\/p>\n\n\n\n
\\text{Fraction} = \\frac{1 – \\frac{\\sqrt{2}}{2}}{\\pi}
]<\/p>\n\n\n\n