{"id":87964,"date":"2025-06-01T06:58:43","date_gmt":"2025-06-01T06:58:43","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87964"},"modified":"2025-06-01T06:58:43","modified_gmt":"2025-06-01T06:58:43","slug":"a-rectangle-abcd-is-kept-in-front-of-a-concave-mirror-of-focal-length","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-rectangle-abcd-is-kept-in-front-of-a-concave-mirror-of-focal-length\/","title":{"rendered":"A rectangle ABCD is kept in front of a concave mirror of focal length"},"content":{"rendered":"

A rectangle ABCD is kept in front of a concave mirror of focal length f with its corners A and B being, respectively, at distances 2f and 3f from the mirror with AB along the principal axis as shown in the figure. It forms an image A’B’C’D’ in front of the mirror. What is the ratio of B’C’ to A’D’ ?<\/p>\n

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This question was previously asked in<\/div>\n
UPSC NDA-1 – 2023<\/div>\n<\/div>\n
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The ratio of B’C’ to A’D’ is 1\/2.<\/section>\n
For a concave mirror, the magnification (m) is given by m = -v\/u, where v is the image distance and u is the object distance. The size of the image perpendicular to the axis (like B’C’ and A’D’) is related to the object size (BC and AD) by the magnification: Image size = |m| * Object size. Since ABCD is a rectangle, AD = BC.<\/section>\n
Using the mirror formula 1\/f = 1\/u + 1\/v:
\nFor point A at u_A = -2f (assuming standard sign convention where f is negative for concave mirror, but distances are given as 2f and 3f. Let’s assume f > 0 is magnitude, and u is negative): u_A = -2f. 1\/f = 1\/(-2f) + 1\/v_A => 1\/v_A = 1\/f + 1\/(2f) = 3\/(2f) => v_A = 2f\/3. No, if A is at 2f from the mirror, and f is the focal length, then 2f is the radius of curvature C. Object at C forms image at C. So u_A = -2f, then v_A = -2f. Magnification m_A = -v_A\/u_A = -(-2f)\/(-2f) = -1. A’D’ = |m_A| * AD = AD.
\nFor point B at u_B = -3f: 1\/f = 1\/(-3f) + 1\/v_B => 1\/v_B = 1\/f + 1\/(3f) = 4\/(3f) => v_B = 3f\/4. Wait, the standard formula for concave mirror with positive f is 1\/f = 1\/u + 1\/v. Let’s use this and assume distances are positive. u_A = 2f, u_B = 3f, f=f. 1\/v_A = 1\/f – 1\/u_A = 1\/f – 1\/(2f) = (2-1)\/(2f) = 1\/(2f) => v_A = 2f. Magnification m_A = -v_A\/u_A = -(2f)\/(2f) = -1. A’D’ = |-1| * AD = AD.
\n1\/v_B = 1\/f – 1\/u_B = 1\/f – 1\/(3f) = (3-1)\/(3f) = 2\/(3f) => v_B = 3f\/2. Magnification m_B = -v_B\/u_B = -(3f\/2)\/(3f) = -1\/2. B’C’ = |-1\/2| * BC = 1\/2 * BC.
\nSince AD = BC, the ratio B’C’ \/ A’D’ = (1\/2 * BC) \/ AD = 1\/2 * (BC\/AD) = 1\/2 * 1 = 1\/2.<\/section>\n","protected":false},"excerpt":{"rendered":"

A rectangle ABCD is kept in front of a concave mirror of focal length f with its corners A and B being, respectively, at distances 2f and 3f from the mirror with AB along the principal axis as shown in the figure. It forms an image A’B’C’D’ in front of the mirror. What is the … <\/p>\n

Detailed SolutionA rectangle ABCD is kept in front of a concave mirror of focal length<\/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":[1105,1153,1128],"class_list":["post-87964","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1105","tag-optics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA rectangle ABCD is kept in front of a concave mirror of focal length<\/title>\n<meta name=\"description\" content=\"The ratio of B'C' to A'D' is 1\/2. For a concave mirror, the magnification (m) is given by m = -v\/u, where v is the image distance and u is the object distance. The size of the image perpendicular to the axis (like B'C' and A'D') is related to the object size (BC and AD) by the magnification: Image size = |m| * Object size. Since ABCD is a rectangle, AD = BC.\" \/>\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-rectangle-abcd-is-kept-in-front-of-a-concave-mirror-of-focal-length\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A rectangle ABCD is kept in front of a concave mirror of focal length\" \/>\n<meta property=\"og:description\" content=\"The ratio of B'C' to A'D' is 1\/2. For a concave mirror, the magnification (m) is given by m = -v\/u, where v is the image distance and u is the object distance. The size of the image perpendicular to the axis (like B'C' and A'D') is related to the object size (BC and AD) by the magnification: Image size = |m| * Object size. 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