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(B) Given: Height of object $(h_o)$ = $10\, mm = 1\, cm$. Height of image $(h_i)$ = $-5\, mm = -0.5\, cm$ (since the image is real and inverted,it is

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$-20$

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(B) Given: Height of object $(h_o)$ = $10\, mm = 1\, cm$. Height of image $(h_i)$ = $-5\, mm = -0.5\, cm$ (since the image is real and inverted,it is formed in front of the mirror). Image distance $(v)$ = $-30\, cm$.Magnification $(m)$ = $h_i / h_o = -v / u$.Substituting the values: $-0.5 / 1 = -(-30) / u$.$-0.5 = 30 / u$.$u = 30 / -0.5 = -60\, cm$.Using the mirror formula: $1/f = 1/v + 1/u$.$1/f = 1/(-30) + 1/(-60)$.$1/f = (-2 - 1) / 60 = -3 / 60$.$1/f = -1 / 20$.Therefore,$f = -20\, cm$.

$A$ $10\, mm$ long awl pin is placed vertically in front of a concave mirror. $A$ $5\, mm$ long image of the awl pin is formed at $30\, cm$ in front of the mirror. The focal length of this mirror is ......... $cm$.

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