An object 5 cm in lenght is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm.Find the position of the image,its nature and size.

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Solution

$G i v e n h o = 5 c m u = - 20 c m f = 30 / 2 = 15 c m 1 / f = 1 / v + 1 / u 1 / 15 = 1 / v - 1 / 20 v = 60 / 7 c m m = - v / u = 0.428 H e i g h t o f i m a g e = m * h o = 2.142 c m V i r t a u l, e r e c t a n d d i m i n i s h e d$

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An object 5 cm in lenght is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm.Find the position of the image,its nature and size.

Given ho=5cmu=-20 cmf=30/2=15 cm1/f=1/v+1/u1/15 = 1/v-1/20 v=60/7 cmm= -v/u = 0.428Height of image = m* ho=2.142 cmVirtaul, erect and diminished

$G i v e n h o = 5 c m u = - 20 c m f = 30 / 2 = 15 c m 1 / f = 1 / v + 1 / u 1 / 15 = 1 / v - 1 / 20 v = 60 / 7 c m m = - v / u = 0.428 H e i g h t o f i m a g e = m * h o = 2.142 c m V i r t a u l, e r e c t a n d d i m i n i s h e d$