Answer:
Using mirror equation: \[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\] \[\Rightarrow \] \[\frac{1}{v}-\frac{1}{18}=\frac{1}{12}\] \[\frac{1}{v}=-\frac{1}{12}+\frac{1}{18}\] \[\Rightarrow \] \[\frac{1}{v}=\frac{-3+2}{36}=\frac{-1}{36}\] \[V=-36cm\] Thus, to obtain a sharp image of the object, the screen should be placed at a distance of 36 cm, in front of the mirror. Now, \[m=\frac{-v}{u}=\frac{{{h}_{i}}}{{{h}_{o}}}\] \[m=-\left( \frac{-36}{-18} \right)=-2\] or \[-2=\frac{hi}{3}\] \[{{h}_{i}}=-6cm\]
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