The focal length of a concave mirror is 12; c | vedclass

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(C) Given: Focal length of the concave mirror $f = -12\; cm$ (sign convention).Object height $h = 4\; cm$.Image height $h' = -1\; cm$ (since the image

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$60$

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(C) Given: Focal length of the concave mirror $f = -12\; cm$ (sign convention).Object height $h = 4\; cm$.Image height $h' = -1\; cm$ (since the image is real,it is inverted).Magnification $m = \frac{h'}{h} = \frac{-1}{4} = -0.25$.We know that magnification $m = \frac{-v}{u}$,so $\frac{-v}{u} = -0.25$,which gives $v = 0.25u$.Using the mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the values: $\frac{1}{-12} = \frac{1}{0.25u} + \frac{1}{u}$.$\frac{1}{-12} = \frac{4}{u} + \frac{1}{u} = \frac{5}{u}$.$u = -12 	imes 5 = -60\; cm$.Thus,the object should be placed at a distance of $60\; cm$ in front of the mirror.

The focal length of a concave mirror is $12\; cm$. Where should an object of length $4\; cm$ be placed,so that a real image of $1\; cm$ length is formed (in $; cm$)?

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