{"id":90567,"date":"2025-06-01T10:31:09","date_gmt":"2025-06-01T10:31:09","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=90567"},"modified":"2025-06-01T10:31:09","modified_gmt":"2025-06-01T10:31:09","slug":"an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/","title":{"rendered":"An object is placed 10 cm in front of a lens. The image formed is real"},"content":{"rendered":"

An object is placed 10 cm in front of a lens. The image formed is real, inverted and of same size as the object. What is the focal length and nature of the lens?<\/p>\n

[amp_mcq option1=”5 cm, converging” option2=”10 cm, diverging” option3=”20 cm, converging” option4=”20 cm, diverging” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2021<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nAn object is placed 10 cm in front of a lens. So, object distance u = 10 cm.
\nThe image formed is real, inverted, and of the same size as the object.
\nA real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images.
\nFor a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v\/u for erect images and m = -v\/u for inverted images.
\nSince the image is inverted, m = -v\/u.
\nThe image is of the same size as the object, so the magnitude of magnification |m| = 1.
\nThus, m = -1.
\n-v\/u = -1 => v = u.
\nGiven u = 10 cm, so v = 10 cm.
\nUsing the lens formula (thin lens equation): 1\/f = 1\/v – 1\/u.
\nApplying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm.
\n1\/f = 1\/(+10 cm) – 1\/(-10 cm)
\n1\/f = 1\/10 + 1\/10
\n1\/f = 2\/10
\n1\/f = 1\/5
\nf = 5 cm.
\nThe focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens).
\nAlternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f.
\nGiven u = 10 cm, 10 cm = 2f => f = 5 cm.
\nThe nature of the lens is converging.
\n<\/section>\n
\n– Identify the nature of the lens based on the image characteristics (real, inverted implies converging).
\n– Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u).
\n– Apply the lens formula (1\/f = 1\/v – 1\/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f).
\n– Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).
\n<\/section>\n
\nFor a converging lens (convex lens):
\n– Object at infinity: real, inverted, point image at F.
\n– Object beyond 2F: real, inverted, diminished image between F and 2F.
\n– Object at 2F: real, inverted, same size image at 2F.
\n– Object between F and 2F: real, inverted, magnified image beyond 2F.
\n– Object at F: real, inverted, image at infinity.
\n– Object between the lens and F: virtual, erect, magnified image on the same side as the object.
\nThe condition “real, inverted, and of same size” is unique to the object being placed at 2F for a converging lens.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

An object is placed 10 cm in front of a lens. The image formed is real, inverted and of same size as the object. What is the focal length and nature of the lens? [amp_mcq option1=”5 cm, converging” option2=”10 cm, diverging” option3=”20 cm, converging” option4=”20 cm, diverging” correct=”option1″] This question was previously asked in UPSC … <\/p>\n

Detailed SolutionAn object is placed 10 cm in front of a lens. The image formed is real<\/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":[1085],"tags":[1110,1153,1128],"class_list":["post-90567","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1110","tag-optics","tag-physics","no-featured-image-padding"],"yoast_head":"\nAn object is placed 10 cm in front of a lens. The image formed is real<\/title>\n<meta name=\"description\" content=\"An object is placed 10 cm in front of a lens. So, object distance u = 10 cm. The image formed is real, inverted, and of the same size as the object. A real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images. For a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v\/u for erect images and m = -v\/u for inverted images. Since the image is inverted, m = -v\/u. The image is of the same size as the object, so the magnitude of magnification |m| = 1. Thus, m = -1. -v\/u = -1 => v = u. Given u = 10 cm, so v = 10 cm. Using the lens formula (thin lens equation): 1\/f = 1\/v - 1\/u. Applying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm. 1\/f = 1\/(+10 cm) - 1\/(-10 cm) 1\/f = 1\/10 + 1\/10 1\/f = 2\/10 1\/f = 1\/5 f = 5 cm. The focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens). Alternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f. Given u = 10 cm, 10 cm = 2f => f = 5 cm. The nature of the lens is converging. - Identify the nature of the lens based on the image characteristics (real, inverted implies converging). - Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u). - Apply the lens formula (1\/f = 1\/v - 1\/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f). - Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).\" \/>\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-lens-the-image-formed-is-real\/\" \/>\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 lens. The image formed is real\" \/>\n<meta property=\"og:description\" content=\"An object is placed 10 cm in front of a lens. So, object distance u = 10 cm. The image formed is real, inverted, and of the same size as the object. A real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images. For a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v\/u for erect images and m = -v\/u for inverted images. Since the image is inverted, m = -v\/u. The image is of the same size as the object, so the magnitude of magnification |m| = 1. Thus, m = -1. -v\/u = -1 => v = u. Given u = 10 cm, so v = 10 cm. Using the lens formula (thin lens equation): 1\/f = 1\/v - 1\/u. Applying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm. 1\/f = 1\/(+10 cm) - 1\/(-10 cm) 1\/f = 1\/10 + 1\/10 1\/f = 2\/10 1\/f = 1\/5 f = 5 cm. The focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens). Alternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f. Given u = 10 cm, 10 cm = 2f => f = 5 cm. The nature of the lens is converging. - Identify the nature of the lens based on the image characteristics (real, inverted implies converging). - Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u). - Apply the lens formula (1\/f = 1\/v - 1\/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f). - Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:31:09+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=\"3 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"An object is placed 10 cm in front of a lens. The image formed is real","description":"An object is placed 10 cm in front of a lens. So, object distance u = 10 cm. The image formed is real, inverted, and of the same size as the object. A real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images. For a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v\/u for erect images and m = -v\/u for inverted images. Since the image is inverted, m = -v\/u. The image is of the same size as the object, so the magnitude of magnification |m| = 1. Thus, m = -1. -v\/u = -1 => v = u. Given u = 10 cm, so v = 10 cm. Using the lens formula (thin lens equation): 1\/f = 1\/v - 1\/u. Applying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm. 1\/f = 1\/(+10 cm) - 1\/(-10 cm) 1\/f = 1\/10 + 1\/10 1\/f = 2\/10 1\/f = 1\/5 f = 5 cm. The focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens). Alternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f. Given u = 10 cm, 10 cm = 2f => f = 5 cm. The nature of the lens is converging. - Identify the nature of the lens based on the image characteristics (real, inverted implies converging). - Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u). - Apply the lens formula (1\/f = 1\/v - 1\/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f). - Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).","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-lens-the-image-formed-is-real\/","og_locale":"en_US","og_type":"article","og_title":"An object is placed 10 cm in front of a lens. The image formed is real","og_description":"An object is placed 10 cm in front of a lens. So, object distance u = 10 cm. The image formed is real, inverted, and of the same size as the object. A real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images. For a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v\/u for erect images and m = -v\/u for inverted images. Since the image is inverted, m = -v\/u. The image is of the same size as the object, so the magnitude of magnification |m| = 1. Thus, m = -1. -v\/u = -1 => v = u. Given u = 10 cm, so v = 10 cm. Using the lens formula (thin lens equation): 1\/f = 1\/v - 1\/u. Applying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm. 1\/f = 1\/(+10 cm) - 1\/(-10 cm) 1\/f = 1\/10 + 1\/10 1\/f = 2\/10 1\/f = 1\/5 f = 5 cm. The focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens). Alternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f. Given u = 10 cm, 10 cm = 2f => f = 5 cm. The nature of the lens is converging. - Identify the nature of the lens based on the image characteristics (real, inverted implies converging). - Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u). - Apply the lens formula (1\/f = 1\/v - 1\/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f). - Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:31:09+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"3 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/","url":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/","name":"An object is placed 10 cm in front of a lens. The image formed is real","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:31:09+00:00","dateModified":"2025-06-01T10:31:09+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"An object is placed 10 cm in front of a lens. So, object distance u = 10 cm. The image formed is real, inverted, and of the same size as the object. A real and inverted image is formed by a converging lens (convex lens). Diverging lenses (concave lenses) always produce virtual, erect, and diminished images. For a lens, the magnification (m) is given by the ratio of image distance (v) to object distance (u), with a sign convention: m = v\/u for erect images and m = -v\/u for inverted images. Since the image is inverted, m = -v\/u. The image is of the same size as the object, so the magnitude of magnification |m| = 1. Thus, m = -1. -v\/u = -1 => v = u. Given u = 10 cm, so v = 10 cm. Using the lens formula (thin lens equation): 1\/f = 1\/v - 1\/u. Applying the sign convention: object distance u is typically taken as negative when placed in front of the lens, so u = -10 cm. The image is real and formed on the opposite side of the lens from the object, so image distance v is positive, v = +10 cm. 1\/f = 1\/(+10 cm) - 1\/(-10 cm) 1\/f = 1\/10 + 1\/10 1\/f = 2\/10 1\/f = 1\/5 f = 5 cm. The focal length is positive (f > 0), which confirms that the lens is a converging lens (convex lens). Alternatively, for a converging lens, a real image of the same size as the object is formed only when the object is placed at a distance of 2f from the lens, and the image is also formed at 2f on the other side. So, u = 2f. Given u = 10 cm, 10 cm = 2f => f = 5 cm. The nature of the lens is converging. - Identify the nature of the lens based on the image characteristics (real, inverted implies converging). - Use the magnification information (same size implies |m|=1) and image type (inverted implies m=-1) to relate object and image distances (v=u). - Apply the lens formula (1\/f = 1\/v - 1\/u) with appropriate sign conventions or use the special case rule for an image of the same size formed by a converging lens (object at 2f). - Determine the sign of the focal length to confirm the nature of the lens (positive f for converging).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/an-object-is-placed-10-cm-in-front-of-a-lens-the-image-formed-is-real\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC CAPF","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-capf\/"},{"@type":"ListItem","position":3,"name":"An object is placed 10 cm in front of a lens. The image formed is real"}]},{"@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\/90567","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=90567"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/90567\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=90567"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=90567"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=90567"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}