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Where should an object be placed in from of a concave mirror of focal length 30 cm so that the image size is 5 times the object size?

(a) 24 cm in front of mirror

(b) 36 cm in front of mirror

(c) 48 cm in front of mirror

(d) (a) and (b)

1 Answer

$\large\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
or $\large\frac{u}{v}$$+1=\frac{u}{f}$
Since $m =\large\frac{-v}{u}$
$\therefore m= \large\frac{f}{f-u}$
Here $f= -30\;cm$ and $m=\pm 5$ because image can be real or virtual .
for real image $m=-5$
$\therefore -5= \large\frac{-30}{-30 -u}$
or $u= -36 \;cm$
for virtual image $m=+5$
$\therefore 5=\large\frac{-30}{-30-u}$
or $u= -24 \;cm$
Hence the object must be placed at 24 cm or 36 cm in front of concave mirror.
Hence d is the correct answer.
answered Feb 11, 2014 by meena.p
 

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