A) 2600 km
B) 1600 km
C) 1400 km
D) 2000 km
Correct Answer: A
Solution :
Linear magnification \[\operatorname{m}=\frac{I}{O}=\frac{\operatorname{f}}{\operatorname{f}-u}~\Rightarrow m=\frac{-10}{15}=\frac{-2}{3}\] \[\operatorname{Axial} magnification = {{m}^{2}} = \frac{4}{9}\] \[\operatorname{Area} of image = \frac{4}{9} area of object\] \[= \frac{4}{9}\times {{\left( 3 \right)}^{2}}=4{{\operatorname{cm}}^{2}}\]You need to login to perform this action.
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