InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
prove that 1/f=1/u+1/v |
| Answer» <p>The formula is a relation between the object distance u , inmage distance v and the focal length from the pole of the concave mirror. The formula is valid for the images in convex mirror and even for the images in lens.</p><p>We consider the image formed by aconcave mirror whose focal length is f and whose radius of curvature is r = 2f.</p><p>Let P be the pole of the concave mirror. Let P, F , C be the pole, focucal point , and centre of curvature along principal axis . So, PC = 2PF , as r = 2f.Let AB be a vertically standinding object beyond C on the principal axis.</p><p>Then the ray starting from B parallel to principal axis incident on the mirror atD reflects through the focus F. Let the reflected ray be CFB' .</p><p>The another ray starting from B through the centre C incident on the mirror atE retracesits path by reflection being normal to the mirror.Now BE and DF produced meet at B'.</p><p>Now drop the perpendicular from B' to PC to meet at A'.</p><p>Drop the perpendicular from D to PC to meet at G.</p><p>NowPF = f , the focal length. PA = u object distance from the mirror. PA' = v the image distance.</p><p>Now consider the similar triangles ABC and A'BC.</p><p>AB/AB' = AC/ A'C =( PU-PC)(PC-PA') = (u-2f)/((2f-v).....(1)</p><p>Consider the similar triangles DFG and A'B'F.</p><p>DG/A'B' = PF/PA' PF/(PA'-PF)= f/(v-f)... (2)</p><p>DG = AB. So (2) could be rewritten as:</p><p>AB/A'B' + f/v ....................(3).</p><p>From (2) and (3), LHS being same , we can equate right sides.</p><p>(u-2f)/(2f-v) = f/(v-f).</p><p>(u-2f)(v-f) = (2f-v)f.uv-2fv -fu +2f^2 = 2f^2 -fv</p><p>uv = fu +fv</p><p>Dvide by uvf;</p><p>1/f = 1/v+1/u.</p> <p>1/v+1/u is the correct answer</p> | |