question_answer

A 3 cm tall object is placed 18 cm in front of a concave mirror of focal length 12 cm. At what distance from the mirror should a screen be placed to see a sharp image of the object on the screen. Also calculate the height of the image formed.

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         \[-2=\frac{hi}{3}\]

\[{{h}_{i}}=-6cm\]