{"id":86140,"date":"2025-06-01T03:38:33","date_gmt":"2025-06-01T03:38:33","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=86140"},"modified":"2025-06-01T03:38:33","modified_gmt":"2025-06-01T03:38:33","slug":"a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/","title":{"rendered":"A luminous object is placed at a distance of 40 cm from a converging l"},"content":{"rendered":"

A luminous object is placed at a distance of 40 cm from a converging lens of focal length 25 cm. The image obtained in the screen is<\/p>\n

[amp_mcq option1=”erect and magnified” option2=”erect and smaller” option3=”inverted and magnified” option4=”inverted and smaller” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CDS-2 – 2020<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nFor a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1\/f = 1\/v – 1\/u to find the image distance (v).
\n1\/25 = 1\/v – 1\/(-40)
\n1\/25 = 1\/v + 1\/40
\n1\/v = 1\/25 – 1\/40 = (8 – 5) \/ 200 = 3 \/ 200
\nv = 200\/3 cm \u2248 +66.7 cm.
\nSince v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted.
\nThe magnification is given by m = v\/u = (200\/3) \/ (-40) = -200 \/ 120 = -5\/3 \u2248 -1.67.
\nSince |m| = 5\/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified.
\n<\/section>\n
\nFor a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).
\n<\/section>\n
\nIf the object were placed beyond 2f, the image would be real, inverted, and diminished. If the object were at 2f, the image would be real, inverted, and of the same size at 2f on the other side. If the object were between the optical center and f, the image would be virtual, erect, and magnified, formed on the same side as the object.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A luminous object is placed at a distance of 40 cm from a converging lens of focal length 25 cm. The image obtained in the screen is [amp_mcq option1=”erect and magnified” option2=”erect and smaller” option3=”inverted and magnified” option4=”inverted and smaller” correct=”option3″] This question was previously asked in UPSC CDS-2 – 2020 Download PDFAttempt Online For … <\/p>\n

Detailed SolutionA luminous object is placed at a distance of 40 cm from a converging l<\/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":[1088],"tags":[1288,1153,1128],"class_list":["post-86140","post","type-post","status-publish","format-standard","hentry","category-upsc-cds-2","tag-1288","tag-optics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA luminous object is placed at a distance of 40 cm from a converging l<\/title>\n<meta name=\"description\" content=\"For a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1\/f = 1\/v - 1\/u to find the image distance (v). 1\/25 = 1\/v - 1\/(-40) 1\/25 = 1\/v + 1\/40 1\/v = 1\/25 - 1\/40 = (8 - 5) \/ 200 = 3 \/ 200 v = 200\/3 cm \u2248 +66.7 cm. Since v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted. The magnification is given by m = v\/u = (200\/3) \/ (-40) = -200 \/ 120 = -5\/3 \u2248 -1.67. Since |m| = 5\/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified. For a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).\" \/>\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\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A luminous object is placed at a distance of 40 cm from a converging l\" \/>\n<meta property=\"og:description\" content=\"For a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1\/f = 1\/v - 1\/u to find the image distance (v). 1\/25 = 1\/v - 1\/(-40) 1\/25 = 1\/v + 1\/40 1\/v = 1\/25 - 1\/40 = (8 - 5) \/ 200 = 3 \/ 200 v = 200\/3 cm \u2248 +66.7 cm. Since v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted. The magnification is given by m = v\/u = (200\/3) \/ (-40) = -200 \/ 120 = -5\/3 \u2248 -1.67. Since |m| = 5\/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified. For a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T03:38:33+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":"A luminous object is placed at a distance of 40 cm from a converging l","description":"For a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1\/f = 1\/v - 1\/u to find the image distance (v). 1\/25 = 1\/v - 1\/(-40) 1\/25 = 1\/v + 1\/40 1\/v = 1\/25 - 1\/40 = (8 - 5) \/ 200 = 3 \/ 200 v = 200\/3 cm \u2248 +66.7 cm. Since v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted. The magnification is given by m = v\/u = (200\/3) \/ (-40) = -200 \/ 120 = -5\/3 \u2248 -1.67. Since |m| = 5\/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified. For a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).","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\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/","og_locale":"en_US","og_type":"article","og_title":"A luminous object is placed at a distance of 40 cm from a converging l","og_description":"For a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1\/f = 1\/v - 1\/u to find the image distance (v). 1\/25 = 1\/v - 1\/(-40) 1\/25 = 1\/v + 1\/40 1\/v = 1\/25 - 1\/40 = (8 - 5) \/ 200 = 3 \/ 200 v = 200\/3 cm \u2248 +66.7 cm. Since v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted. The magnification is given by m = v\/u = (200\/3) \/ (-40) = -200 \/ 120 = -5\/3 \u2248 -1.67. Since |m| = 5\/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified. For a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T03:38:33+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\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/","url":"https:\/\/exam.pscnotes.com\/mcq\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/","name":"A luminous object is placed at a distance of 40 cm from a converging l","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T03:38:33+00:00","dateModified":"2025-06-01T03:38:33+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"For a converging lens, the focal length is positive, f = +25 cm. The luminous object is placed at a distance u = -40 cm (object distance is negative by convention for real objects). We can use the lens formula 1\/f = 1\/v - 1\/u to find the image distance (v). 1\/25 = 1\/v - 1\/(-40) 1\/25 = 1\/v + 1\/40 1\/v = 1\/25 - 1\/40 = (8 - 5) \/ 200 = 3 \/ 200 v = 200\/3 cm \u2248 +66.7 cm. Since v is positive, the image is real and formed on the opposite side of the lens from the object, which can be obtained on a screen. Real images formed by a single converging lens are always inverted. The magnification is given by m = v\/u = (200\/3) \/ (-40) = -200 \/ 120 = -5\/3 \u2248 -1.67. Since |m| = 5\/3 > 1, the image is magnified. The negative sign indicates that the image is inverted. Thus, the image obtained on the screen is inverted and magnified. For a converging lens, when the object is placed between f and 2f, a real, inverted, and magnified image is formed beyond 2f. In this case, u = 40 cm, which is between f (25 cm) and 2f (50 cm).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/a-luminous-object-is-placed-at-a-distance-of-40-cm-from-a-converging-l\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CDS-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-cds-2\/"},{"@type":"ListItem","position":3,"name":"A luminous object is placed at a distance of 40 cm from a converging l"}]},{"@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\/86140","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=86140"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/86140\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=86140"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=86140"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=86140"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}