Question

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

Answer 1

$$\begin{gathered} Letinitiallyobjectdis\tan ce=a \\ imagedis\tan ce=b \\ Finallyobjectdis\tan ce=c \\ imagediatnce=d \\ f=focallength \\ magnificationforthemirror \\ m=\frac{-v}{u} \\ \frac{h_{image}}{h_{oject}}=\frac{-v}{u} \\ initially \\ \frac{-3h_{object}}{h_{object}}=\frac{-(-b)}{-a} \\ b=3a \\ finally \\ \frac{-2h_{object}}{h_{oject}}=\frac{-(-d)}{-c} \\ d=2a \\ mirrorformula \\ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ initially \\ \frac{1}{f}=\frac{-1}{a}-\frac{1}{b} \\ \frac{1}{f}=\frac{-1}{a}-\frac{1}{3a}--\left(1\right) \\ finally \\ \frac{1}{f}=\frac{-1}{d}-\frac{1}{3c}---\left(2\right) \\ ONSolving\left(1\right)and\left(2\right)weget \\ 8c=9a \\ whenweshiftheobjectandscreen \\ objectdis\tan cefinal-objectdi\tan cebefore=6 \\ -c+a=6 \\ -8c+8a=4 \\ a=-48cm \\ c=-54cm \\ b=-144cm \\ focallength= \\ \frac{1}{f}=\frac{-1}{144}-\frac{1}{48} \\ f=-36cm \end{gathered}$$