{"id":87269,"date":"2025-06-01T06:39:01","date_gmt":"2025-06-01T06:39:01","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87269"},"modified":"2025-06-01T06:39:01","modified_gmt":"2025-06-01T06:39:01","slug":"an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/","title":{"rendered":"An object is placed 10 cm in front of a convex lens of focal length 15"},"content":{"rendered":"

An object is placed 10 cm in front of a convex lens of focal length 15 cm. The image produced will be<\/p>\n

[amp_mcq option1=”Real and magnified” option2=”Virtual and magnified” option3=”Virtual and reduced in size” option4=”Real and reduced in size” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2015<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nFor a convex lens, the focal length ($f$) is positive. Given $f = +15$ cm. The object is placed at $u = -10$ cm (negative sign indicates it’s in front of the lens). Using the lens formula $1\/f = 1\/v – 1\/u$:
\n$1\/15 = 1\/v – 1\/(-10)$
\n$1\/15 = 1\/v + 1\/10$
\n$1\/v = 1\/15 – 1\/10$
\n$1\/v = (2 – 3)\/30$
\n$1\/v = -1\/30$
\n$v = -30$ cm
\nThe image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image.
\nThe magnification ($m$) is given by $m = v\/u$:
\n$m = (-30) \/ (-10) = +3$
\nThe positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object).
\nTherefore, the image is virtual and magnified.
\n<\/section>\n
\nApplying the lens formula and magnification formula to determine the nature and size of the image formed by a convex lens for a given object position.
\n<\/section>\n
\nFor a convex lens, when the object is placed between the optical center and the focal point (i.e., $u < f$), a virtual, erect, and magnified image is formed on the same side as the object. In this case, $u=10$ cm and $f=15$ cm, so $u < f$, fitting this scenario.\n<\/section>\n","protected":false},"excerpt":{"rendered":"

An object is placed 10 cm in front of a convex lens of focal length 15 cm. The image produced will be [amp_mcq option1=”Real and magnified” option2=”Virtual and magnified” option3=”Virtual and reduced in size” option4=”Real and reduced in size” correct=”option2″] This question was previously asked in UPSC NDA-1 – 2015 Download PDFAttempt Online For a … <\/p>\n

Detailed SolutionAn object is placed 10 cm in front of a convex lens of focal length 15<\/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":[1093],"tags":[1443,1153,1128],"class_list":["post-87269","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1443","tag-optics","tag-physics","no-featured-image-padding"],"yoast_head":"\nAn object is placed 10 cm in front of a convex lens of focal length 15<\/title>\n<meta name=\"description\" content=\"For a convex lens, the focal length ($f$) is positive. Given $f = +15$ cm. The object is placed at $u = -10$ cm (negative sign indicates it's in front of the lens). Using the lens formula $1\/f = 1\/v - 1\/u$: $1\/15 = 1\/v - 1\/(-10)$ $1\/15 = 1\/v + 1\/10$ $1\/v = 1\/15 - 1\/10$ $1\/v = (2 - 3)\/30$ $1\/v = -1\/30$ $v = -30$ cm The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image. The magnification ($m$) is given by $m = v\/u$: $m = (-30) \/ (-10) = +3$ The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object). Therefore, the image is virtual and magnified. Applying the lens formula and magnification formula to determine the nature and size of the image formed by a convex lens for a given object position.\" \/>\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\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"An object is placed 10 cm in front of a convex lens of focal length 15\" \/>\n<meta property=\"og:description\" content=\"For a convex lens, the focal length ($f$) is positive. Given $f = +15$ cm. The object is placed at $u = -10$ cm (negative sign indicates it's in front of the lens). Using the lens formula $1\/f = 1\/v - 1\/u$: $1\/15 = 1\/v - 1\/(-10)$ $1\/15 = 1\/v + 1\/10$ $1\/v = 1\/15 - 1\/10$ $1\/v = (2 - 3)\/30$ $1\/v = -1\/30$ $v = -30$ cm The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image. The magnification ($m$) is given by $m = v\/u$: $m = (-30) \/ (-10) = +3$ The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object). Therefore, the image is virtual and magnified. Applying the lens formula and magnification formula to determine the nature and size of the image formed by a convex lens for a given object position.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:39:01+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":"An object is placed 10 cm in front of a convex lens of focal length 15","description":"For a convex lens, the focal length ($f$) is positive. Given $f = +15$ cm. The object is placed at $u = -10$ cm (negative sign indicates it's in front of the lens). Using the lens formula $1\/f = 1\/v - 1\/u$: $1\/15 = 1\/v - 1\/(-10)$ $1\/15 = 1\/v + 1\/10$ $1\/v = 1\/15 - 1\/10$ $1\/v = (2 - 3)\/30$ $1\/v = -1\/30$ $v = -30$ cm The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image. The magnification ($m$) is given by $m = v\/u$: $m = (-30) \/ (-10) = +3$ The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object). Therefore, the image is virtual and magnified. Applying the lens formula and magnification formula to determine the nature and size of the image formed by a convex lens for a given object position.","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\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/","og_locale":"en_US","og_type":"article","og_title":"An object is placed 10 cm in front of a convex lens of focal length 15","og_description":"For a convex lens, the focal length ($f$) is positive. Given $f = +15$ cm. The object is placed at $u = -10$ cm (negative sign indicates it's in front of the lens). Using the lens formula $1\/f = 1\/v - 1\/u$: $1\/15 = 1\/v - 1\/(-10)$ $1\/15 = 1\/v + 1\/10$ $1\/v = 1\/15 - 1\/10$ $1\/v = (2 - 3)\/30$ $1\/v = -1\/30$ $v = -30$ cm The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image. The magnification ($m$) is given by $m = v\/u$: $m = (-30) \/ (-10) = +3$ The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object). Therefore, the image is virtual and magnified. Applying the lens formula and magnification formula to determine the nature and size of the image formed by a convex lens for a given object position.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:39:01+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\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/","url":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/","name":"An object is placed 10 cm in front of a convex lens of focal length 15","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:39:01+00:00","dateModified":"2025-06-01T06:39:01+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"For a convex lens, the focal length ($f$) is positive. Given $f = +15$ cm. The object is placed at $u = -10$ cm (negative sign indicates it's in front of the lens). Using the lens formula $1\/f = 1\/v - 1\/u$: $1\/15 = 1\/v - 1\/(-10)$ $1\/15 = 1\/v + 1\/10$ $1\/v = 1\/15 - 1\/10$ $1\/v = (2 - 3)\/30$ $1\/v = -1\/30$ $v = -30$ cm The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image. The magnification ($m$) is given by $m = v\/u$: $m = (-30) \/ (-10) = +3$ The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object). Therefore, the image is virtual and magnified. Applying the lens formula and magnification formula to determine the nature and size of the image formed by a convex lens for a given object position.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-convex-lens-of-focal-length-15\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"An object is placed 10 cm in front of a convex lens of focal length 15"}]},{"@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\/87269","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=87269"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87269\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87269"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87269"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87269"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}