{"id":88707,"date":"2025-06-01T07:18:46","date_gmt":"2025-06-01T07:18:46","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88707"},"modified":"2025-06-01T07:18:46","modified_gmt":"2025-06-01T07:18:46","slug":"light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/","title":{"rendered":"Light of uniform intensity impinges perpendicularly on a totally refle"},"content":{"rendered":"

Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become<\/p>\n

[amp_mcq option1=”double” option2=”half” option3=”four times” option4=”one fourth” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2020<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe radiation force (F) on a totally reflecting surface is given by F = 2IA\/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A\u2082 = A\u2081\/2. The new radiation force F\u2082 will be F\u2082 = 2IA\u2082\/c = 2I(A\u2081\/2)\/c = (2IA\u2081\/c) \/ 2 = F\u2081\/2. Thus, the radiation force on it will become half.
\n<\/section>\n
\nRadiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.
\n<\/section>\n
\nThe radiation pressure is defined as the force per unit area (P = F\/A). For a totally reflecting surface, the radiation pressure is P = 2I\/c. The force is the product of pressure and area: F = P * A = (2I\/c) * A.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become [amp_mcq option1=”double” option2=”half” option3=”four times” option4=”one fourth” correct=”option2″] This question was previously asked in UPSC NDA-2 – 2020 Download PDFAttempt Online The radiation force (F) on a totally reflecting … <\/p>\n

Detailed SolutionLight of uniform intensity impinges perpendicularly on a totally refle<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1288,1153,1128],"class_list":["post-88707","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1288","tag-optics","tag-physics","no-featured-image-padding"],"yoast_head":"\nLight of uniform intensity impinges perpendicularly on a totally refle<\/title>\n<meta name=\"description\" content=\"The radiation force (F) on a totally reflecting surface is given by F = 2IA\/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A\u2082 = A\u2081\/2. The new radiation force F\u2082 will be F\u2082 = 2IA\u2082\/c = 2I(A\u2081\/2)\/c = (2IA\u2081\/c) \/ 2 = F\u2081\/2. Thus, the radiation force on it will become half. Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Light of uniform intensity impinges perpendicularly on a totally refle\" \/>\n<meta property=\"og:description\" content=\"The radiation force (F) on a totally reflecting surface is given by F = 2IA\/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A\u2082 = A\u2081\/2. The new radiation force F\u2082 will be F\u2082 = 2IA\u2082\/c = 2I(A\u2081\/2)\/c = (2IA\u2081\/c) \/ 2 = F\u2081\/2. Thus, the radiation force on it will become half. Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:18:46+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Light of uniform intensity impinges perpendicularly on a totally refle","description":"The radiation force (F) on a totally reflecting surface is given by F = 2IA\/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A\u2082 = A\u2081\/2. The new radiation force F\u2082 will be F\u2082 = 2IA\u2082\/c = 2I(A\u2081\/2)\/c = (2IA\u2081\/c) \/ 2 = F\u2081\/2. Thus, the radiation force on it will become half. Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/","og_locale":"en_US","og_type":"article","og_title":"Light of uniform intensity impinges perpendicularly on a totally refle","og_description":"The radiation force (F) on a totally reflecting surface is given by F = 2IA\/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A\u2082 = A\u2081\/2. The new radiation force F\u2082 will be F\u2082 = 2IA\u2082\/c = 2I(A\u2081\/2)\/c = (2IA\u2081\/c) \/ 2 = F\u2081\/2. Thus, the radiation force on it will become half. Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:18:46+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/","url":"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/","name":"Light of uniform intensity impinges perpendicularly on a totally refle","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:18:46+00:00","dateModified":"2025-06-01T07:18:46+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The radiation force (F) on a totally reflecting surface is given by F = 2IA\/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A\u2082 = A\u2081\/2. The new radiation force F\u2082 will be F\u2082 = 2IA\u2082\/c = 2I(A\u2081\/2)\/c = (2IA\u2081\/c) \/ 2 = F\u2081\/2. Thus, the radiation force on it will become half. Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/light-of-uniform-intensity-impinges-perpendicularly-on-a-totally-refle\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"Light of uniform intensity impinges perpendicularly on a totally refle"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88707","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=88707"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/88707\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=88707"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=88707"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=88707"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}