Example 9.3 An object is placed at 0 10 cm. h) 5 em in front of aconcave mirror of radius of curvature 15 cm. Find the position, natureand magnification of the image in each case

Example 9.3 An object is placed at 0 10 cm. h) 5 em in front of aconca

1.	Example 9.3 An object is placed at 0 10 cm. h) 5 em in front of aconcave mirror of radius of curvature 15 cm. Find the position, natureand magnification of the image in each case
Answer» <p>1) Object distance = 10 cmObject distance =u = -10 cmRadius of curvature = R = -15 cm</p><p>Focal length = ff = 2R</p><p>so,f = R/2 = -15/2 cm</p><p>We know that :-1/v + 1/u = 1/f</p><p>v = image distance</p><p>1/v = 1/f - 1/u = 1 / (-15/2) - (-1/10) = -2/15 + 1/10 = -4/30 + 3/30 = -1/30</p><p>1/v = -1/30v = -30 cmImage distance = v = -30 cm</p><p>The position of image is 30 cm in front of mirror.</p><p>we know that ,magnification , m = -v/u = -(-30)/-10 = -3</p><p>Nature :-This negative sign indicates that the image is real ,and inverted.</p><p>ii) Object distance = 5 cm</p><p>Object distance = u = -5cmFocal length = -15/2 cm</p><p>1/v = 1/f - 1/u = -2/15 - (-1/5) = -2/15 + 1/5 = 1/15</p><p>v = 15 cmImage distance = v = 15 cm</p><p>we know ,m = -v/u = -15/-5 = 3</p><p>Nature of image :- The positive sign of magnification indicates that the image is virtual , erect and magnified</p> <p>right answer </p>