An erect image 3 times the size of the object is obtained with a concave mirror of radius of curvature 36cm. What is the position of the object?

15cm

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20cm

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12cm

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30cm

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Solution

The correct option is C 12cm
Radius of curvature = 36cm
Since mirror is concave
$f = \frac{R}{2} = \frac{- 36}{2} = - 18 c m m a g n i f i c a t i o n = + 3 (G i v e n) We know that, m = \frac{- v}{u} \Rightarrow v = - m u \Rightarrow v = - 3 u From mirror formula, \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{- 3 u} + \frac{1}{u} = \frac{2}{3 u} \Rightarrow f = \frac{3 u}{2} \Rightarrow u = \frac{2 f}{3} u = \frac{2 (- 18)}{3} u = - 12 c m [Infront of the mirror]$

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An erect image 3 times the size of the object is obtained with a concave mirror of radius of curvature 36cm. What is the position of the object?

The correct option is C 12cm Radius of curvature = 36cm Since mirror is concave f=R2=−362=−18 cmmagnification=+3 ( Given)We know that,m=−vu⇒v=−mu⇒v=−3uFrom mirror formula,1f=1v+1u⇒1f=1−3u+1u=23u⇒f=3u2⇒u=2f3u=2(−18)3u=−12 cm [Infront of the mirror]