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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)

Can you answer this question?
 
 

1 Answer

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$\large\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
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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