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The size of the image, if an object of 2.5 m height is placed at a distance of 10 cm from a concave mirror is:(Take radius of curvature of concave mirror =

30 cm)


We have height of object ,h=2.5 m and distance of object from mirror u=-10 and focal length,f=-15 now 1/f=1/u+1/v 1/f-1/u=1/v -1/15-(-1/10)=1/v -1/15+1/10=1/v 1/v=1/30 v=30 now h'/h=-v/u h'/2.5=30/10 h'=7.5 m------------------------ans
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Ht of the object ho=2.5m distance of the object u=-10cm radius of curvature=2x focallength=30cm so  focallengthf=-15cm  using the equn 1/f=i/u+1/v  1/v=1/f-1/u sub the values  1/v=-1/15- _1/10=-1/15 +1/101/v=1/10-1/15=0.1-.066=.034 orv=1/.034=29cm             m the magnification=-v/u=ht of theimage/ht of the object  .subthe values ,-29/-10=hi/250 2.9=hi/250 or hi=2.9x250=725cm =7.25m ie ht of the object=7.25m and magnification=-v/u=-29/-10=2.9
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