InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
12cmf mixxas then Find pos |
| Answer» <p>an object is placed at i)10cm, ii)5cm in front of a concave mirror of radius of curvature 15 cm. find the position, nature, and magnifiaction of the image in each case.</p> <p>Given:</p><p>the radius of curvature of theconcave mirror(R) = 15 cmtherefore,focal lengthof the mirror = -7.5cm</p><p>(as focal length(f) = 1/2 * radius of curvature of the mirror(R))i)Object distance(u) = -10cmLet image distance from pole = v cmusingmirror formula: - 1/v + 1/u =1/f1/v = 1/f - 1/u1/v = -1/7.5 - (- 1/10)1/v = -1/7.5 + 1/101/v = (-100 + 75)/7501/v = -25/7501/v = -1/30therefore v = -30cmtherefore image is formed at 30cm from the pole of the mirror in front of the mirror.</p><p>magnification(m) = -v/um = [-(-30)]/-10m = -3here minus sign indicates that the image is inverted in nature.therefore the image is formed at 30cm from the pole of the mirror in front of the mirrorit is inverted in nature and Magnification = -3</p><p>ii) Object distance= -5cm</p><p>Let the image distance from pole= v cm</p><p>Using the mirror formula : 1/v +1/u = 1/f</p><p>1/v = 1/f -1/u</p><p>1/v = -1/7.5 –(-1/5)</p><p>1/v= - 10/75 + 1/5</p><p>1/v= 5/375</p><p>1/v= 1/15 cm</p><p>v= 15cm</p><p>therefore the image is formed at 15cm from the pole of the mirror</p><p>magnification (m)= -v/u</p><p>m=-(15)/-5</p><p>m=3</p><p>therefore the image is formed 15cm from the pole it is erect n magnification=3cm</p> <p>This was solution to a similar question, kindly follow the methodology.</p> | |