Answer:
Given: \[u=-15\text{ }cm,\text{ }f=-10\text{ }cm,\,\,{{h}_{i}}=4\,cm\]
Using the mirror formula
\[\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\]
\[\frac{1}{v}-\frac{1}{15}=\frac{-1}{10}\]
\[\frac{1}{v}=\frac{-1}{10}+\frac{1}{15}\]
\[\frac{1}{v}=\frac{-15+10}{150}=\frac{-5}{150}\]
\[v=-30\,cm\]
Thus, to obtain a sharp image of the object, the screen should be placed in front of the mirror at a distance of 30 cm.
Now, \[m=\frac{-v}{u}=\frac{\text{height of image}}{\text{height of object}}\]
\[m=-\left( \frac{-30}{-15} \right)\]\[\Rightarrow m=-2\]
\[-2=\frac{{{h}_{1}}}{4}\] \[\Rightarrow {{h}_{1}}=-8\,cm\]
‹ Height of the image is 8 cm.
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