If a concave mirror has a focal length of 10 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.

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Solution

f = -10cm
case 1: m=2 (Image is virtual and erect)
$m equals negative v over u 2 equals negative v over u rightwards double arrow v equals space minus 2 u 1 over v plus 1 over u equals 1 over f fraction numerator 1 over denominator negative 2 u end fraction plus 1 over u equals fraction numerator 1 over denominator negative 10 end fraction fraction numerator 1 over denominator 2 u end fraction equals fraction numerator negative 1 over denominator 10 end fraction u space equals space minus 5 c m$
object is placed 5cm infront of mirror.

Case 2: m=-2 (Image is real and inverted)
$m equals negative v over u minus 2 equals negative v over u rightwards double arrow v equals space 2 u 1 over v plus 1 over u equals 1 over f fraction numerator 1 over denominator 2 u end fraction plus 1 over u equals fraction numerator 1 over denominator negative 10 end fraction fraction numerator 3 over denominator 2 u end fraction equals fraction numerator negative 1 over denominator 10 end fraction u space equals space minus 15 c m$
object is placed 15cm infront of mirror .

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If a concave mirror has a focal length of 10 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.

f = -10cm case 1: m=2 (Image is virtual and erect) object is placed 5cm infront of mirror. Case 2: m=-2 (Image is real and inverted) object is placed 15cm infront of mirror .