A) 4 \[c{{m}^{2}}\]
B) 6 \[c{{m}^{2}}\]
C) 16 \[c{{m}^{2}}\]
D) 36 \[c{{m}^{2}}\]
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 \[=\frac{4}{9}\times {{(3)}^{2}}=4\,c{{m}^{2}}\]
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