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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-




Question Number 41753 by Necxx last updated on 12/Aug/18
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?
$${How}\:{far}\:{must}\:{a}\:{man}\:{stand}\:{in}\:{front} \\ $$$${of}\:{a}\:{concave}\:{mirror}\:{of}\:{radius}\:\mathrm{120}{cm} \\ $$$${in}\:{order}\:{to}\:{see}\:{an}\:{erect}\:{image}\:{of} \\ $$$${his}\:{face}\:{four}\:{times}\:{its}\:{actual}\:{size}? \\ $$
Answered by ajfour last updated on 12/Aug/18
(1/u)+(1/v)=(1/f)    and   m=−(v/u)  ⇒   (1/u)−(1/(mu)) =(1/f)  ⇒   u=f(1−(1/m)) = 60cm(1−(1/4))       u= 45cm .
$$\frac{\mathrm{1}}{{u}}+\frac{\mathrm{1}}{{v}}=\frac{\mathrm{1}}{{f}}\:\:\:\:{and}\:\:\:{m}=−\frac{{v}}{{u}} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{{u}}−\frac{\mathrm{1}}{{mu}}\:=\frac{\mathrm{1}}{{f}} \\ $$$$\Rightarrow\:\:\:{u}={f}\left(\mathrm{1}−\frac{\mathrm{1}}{{m}}\right)\:=\:\mathrm{60}{cm}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:{u}=\:\mathrm{45}{cm}\:. \\ $$
Commented by Necxx last updated on 12/Aug/18
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by alex041103 last updated on 12/Aug/18
(1/a)+(1/b)=(2/R)=(2/(1.2))=(5/3) ∣×a  ⇒1−(1/W)=(5/3)a  where W=−(b/a)  is the   linear magnification...  for W=+4 ⇒(3/4)=(5/3)a ⇒ a=(9/(20))=((45)/(100))=0.45m  ⇒The distance required is 45cm
$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}=\frac{\mathrm{2}}{{R}}=\frac{\mathrm{2}}{\mathrm{1}.\mathrm{2}}=\frac{\mathrm{5}}{\mathrm{3}}\:\mid×{a} \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{1}}{{W}}=\frac{\mathrm{5}}{\mathrm{3}}{a}\:\:{where}\:{W}=−\frac{{b}}{{a}}\:\:{is}\:{the}\: \\ $$$${linear}\:{magnification}… \\ $$$${for}\:{W}=+\mathrm{4}\:\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{3}}{a}\:\Rightarrow\:{a}=\frac{\mathrm{9}}{\mathrm{20}}=\frac{\mathrm{45}}{\mathrm{100}}=\mathrm{0}.\mathrm{45}{m} \\ $$$$\Rightarrow{The}\:{distance}\:{required}\:{is}\:\mathrm{45}{cm} \\ $$
Commented by Necxx last updated on 12/Aug/18
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$

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