A) \[4c{{m}^{2}}\]
B) \[6c{{m}^{2}}\]
C) \[16c{{m}^{2}}\]
D) \[36c{{m}^{2}}\]
Correct Answer: A
Solution :
Linear magnification, \[m=\frac{I}{O}=\frac{f}{u-f}\] \[=\frac{10}{25-10}\] \[m=\frac{10}{15}=\frac{2}{3}\] Areal magnification \[={{m}^{2}}=\frac{4}{9}\] Area of image\[=\frac{4}{9}\times \]area of object \[=\frac{4}{9}\times {{(3)}^{2}}=4c{{m}^{2}}\]
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