A concave mirror forms a real image three tim | vedclass

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(A) For the first case,magnification $m_1 = -3$ (real image). So,$v_1 = 3u_1$. Let $u_1 = u$,then $v_1 = 3u$.Using mirror formula $\frac{1}{v} + \frac

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$36 \ cm, 36 \ cm$

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(A) For the first case,magnification $m_1 = -3$ (real image). So,$v_1 = 3u_1$. Let $u_1 = u$,then $v_1 = 3u$.Using mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{3u} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{4}{3u} = \frac{1}{f} \Rightarrow f = \frac{3u}{4}$.For the second case,magnification $m_2 = -2$. The object is shifted by $6 \ cm$,so $u_2 = u + 6$. Then $v_2 = 2(u + 6)$.Using mirror formula $\frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f} \Rightarrow \frac{1}{2(u+6)} + \frac{1}{u+6} = \frac{1}{f} \Rightarrow \frac{3}{2(u+6)} = \frac{1}{f}$.Equating the two expressions for $f$: $\frac{3u}{4} = \frac{2(u+6)}{3} \Rightarrow 9u = 8u + 48 \Rightarrow u = 48 \ cm$.Then $f = \frac{3(48)}{4} = 36 \ cm$.Initial screen position $v_1 = 3(48) = 144 \ cm$. Final screen position $v_2 = 2(48+6) = 108 \ cm$.Shift of screen = $v_1 - v_2 = 144 - 108 = 36 \ cm$.

$A$ concave mirror forms a real image three times larger than the object on a screen. The object and screen are moved until the image becomes twice the size of the object. If the shift of the object is $6 \ cm$,find the shift of the screen and the focal length of the mirror.

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