A candle flame of 3 , cm is placed at 300 , c | vedclass

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(C) The height of the object is $h_o = 3 \, cm$ and the height of the real image formed on the wall is $h_i = -9 \, cm$ (since it is inverted).Magnifi

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$450$

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(C) The height of the object is $h_o = 3 \, cm$ and the height of the real image formed on the wall is $h_i = -9 \, cm$ (since it is inverted).Magnification $m = \frac{h_i}{h_o} = \frac{-9}{3} = -3$.We know that $m = -\frac{v}{u}$,so $-3 = -\frac{v}{u} \implies v = 3u$.From the geometry of the setup,the distance of the image from the mirror is $v = x$,and the distance of the object from the mirror is $u = x - 300$.Substituting these into the magnification equation: $x = 3(x - 300)$.$x = 3x - 900$.$2x = 900$.$x = 450 \, cm$.

$A$ candle flame of $3 \, cm$ is placed at $300 \, cm$ from a wall. $A$ concave mirror is kept at distance $x$ from the wall in such a way that the image of the flame on the wall is $9 \, cm$. Then $x$ is......$cm$.

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