How far must a man stand in front of a concave mirror of radius 120cm in order to see an erect image of his face four times its actual size?


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Question Number 41753 by Necxx last updated on 12/Aug/18

$H o w f a r m u s t a m a n s t a n d i n f r o n t$ $o f a c o n c a v e m i r r o r o f r a d i u s 120 c m$ $i n o r d e r t o s e e a n e r e c t i m a g e o f$ $h i s f a c e f o u r t i m e s i t s a c t u a l s i z e ?$
	Answered by ajfour last updated on 12/Aug/18

	$\frac{1}{u} + \frac{1}{v} = \frac{1}{f} a n d m = - \frac{v}{u}$ $\Rightarrow \frac{1}{u} - \frac{1}{m u} = \frac{1}{f}$ $\Rightarrow u = f (1 - \frac{1}{m}) = 60 c m (1 - \frac{1}{4})$ $u = 45 c m .$
	Commented by Necxx last updated on 12/Aug/18

	$t h a n k y o u s i r$
	Answered by alex041103 last updated on 12/Aug/18

	$\frac{1}{a} + \frac{1}{b} = \frac{2}{R} = \frac{2}{1.2} = \frac{5}{3} ∣ \times a$ $\Rightarrow 1 - \frac{1}{W} = \frac{5}{3} a w h e r e W = - \frac{b}{a} i s t h e$ $l i n e a r m a g n i f i c a t i o n . . .$ $f o r W = + 4 \Rightarrow \frac{3}{4} = \frac{5}{3} a \Rightarrow a = \frac{9}{20} = \frac{45}{100} = 0.45 m$ $\Rightarrow T h e d i s t a n c e r e q u i r e d i s 45 c m$
	Commented by Necxx last updated on 12/Aug/18

	$T h a n k y o u s i r$
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