A) \[-4\]
B) \[-8\]
C) \[+8\]
D) \[-2\]
Correct Answer: D
Solution :
Given: \[{{f}_{0}}=50cm,\,{{f}_{e}}=5\,cm\] \[d=25cm,\,{{u}_{0}}=-200\,cm\] Magnification M =? As \[\frac{1}{{{v}_{0}}}=\frac{1}{{{f}_{0}}}+\frac{1}{{{u}_{0}}}=\frac{1}{50}-\frac{1}{200}=\frac{4-1}{200}=\frac{3}{200}\] or \[=\frac{200}{3}\,cm\] Now \[{{v}_{e}}=d=-25\,cm\] From, \[\frac{1}{{{v}_{e}}}-\frac{1}{{{u}_{e}}}=\frac{1}{{{f}_{e}}}\] \[-\frac{1}{{{u}_{e}}}=\frac{1}{{{f}_{e}}}-\frac{1}{{{v}_{e}}}\] \[=\frac{1}{5}+\frac{1}{25}=\frac{6}{25}\] or, \[{{v}_{e}}=\frac{-25}{6}\,cm\] Magnification \[M={{M}_{0}}\times {{M}_{e}}\] \[=\frac{{{v}_{0}}}{{{u}_{0}}}\times \frac{{{v}_{e}}}{{{u}_{e}}}=\frac{-200/3}{200}\times \frac{-25}{-25/6}\] \[=-\frac{1}{3}\times 6=-2\]You need to login to perform this action.
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