A) \[\frac{mx}{{{\left( m+1 \right)}^{2}}}\]
B) \[\frac{mx}{{{\left( m-1 \right)}^{2}}}\]
C) \[\frac{{{\left( m+1 \right)}^{2}}}{m}x\]
D) \[\frac{{{\left( m-1 \right)}^{2}}}{m}x\]
Correct Answer: A
Solution :
\[\operatorname{x}=u+v\] ?.(i) \[\operatorname{m}=\frac{f}{f-u}\,\,\,\,\,\Rightarrow \frac{f-v}{f}\] Image is real, magnification is negative \[-m=\frac{f}{f-u}\] \[\operatorname{u}=\left( \frac{m+1}{m} \right)\operatorname{f}\] \[-m=\frac{f-v}{\operatorname{f}}\] \[\operatorname{x}=\left[ \frac{m+1}{m} \right]f+\left( m+1 \right)f\Rightarrow f\left( m+1 \right)\left[ \frac{1}{m}+1 \right]\] \[\operatorname{f}=\frac{mx}{{{\left( m+1 \right)}^{2}}}\]
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