A) \[{{u}_{1}}+\frac{{{u}_{2}}}{2}\]
B) \[\frac{{{u}_{1}}-{{u}_{2}}}{2}\]
C) \[\frac{{{u}_{1}}+{{u}_{2}}}{2}\]
D) \[{{u}_{1}}+{{u}_{2}}\]
Correct Answer: C
Solution :
: For a convex lens when real image is formed,\[{{u}_{1}}\]is negative,\[{{v}_{1}}\]is positive,\[f\]is positive. Lens formula is \[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\] \[\therefore \]For real image, \[\frac{1}{{{v}_{1}}}+\frac{1}{{{u}_{1}}}=\frac{1}{f}\] \[\therefore \] \[\frac{{{u}_{1}}}{{{v}_{1}}}+\frac{{{u}_{1}}}{{{u}_{1}}}=\frac{{{u}_{1}}}{f}\] \[\therefore \] \[\frac{{{u}_{1}}}{{{v}_{1}}}=\left( \frac{{{u}_{1}}}{f}-1 \right)\] ?.(i) When virtual image is formed,\[{{u}_{2}}\]is negative, \[{{v}_{2}}\]is also negative,\[f\]is positive \[\therefore \]For virtual image, \[\frac{1}{-{{v}_{2}}}+\frac{1}{{{u}_{2}}}=\frac{1}{f}\] Or \[\frac{{{u}_{2}}}{-{{v}_{2}}}+\frac{{{u}_{2}}}{{{u}_{2}}}=\frac{{{u}_{2}}}{f}\] or \[\frac{{{u}_{2}}}{{{v}_{2}}}-1=-\frac{{{u}_{2}}}{f}\] Or \[\frac{{{u}_{2}}}{{{v}_{2}}}=1-\frac{{{u}_{2}}}{f}\] ...(ii) Since magnification is same in the two cases. \[\frac{{{v}_{2}}}{{{u}_{2}}}=\frac{{{v}_{2}}}{{{u}_{1}}}\] or\[\frac{{{u}_{2}}}{{{v}_{2}}}=\frac{{{u}_{2}}}{{{v}_{1}}}\].Substitute for these from (i) and (ii). Or \[\left( 1-\frac{{{u}_{2}}}{f} \right)=\left( \frac{{{u}_{1}}}{f}-1 \right)\]or \[1+1=\frac{{{u}_{1}}}{f}+\frac{{{u}_{2}}}{f}\] Or \[2=\frac{({{u}_{1}}+{{u}_{2}})}{f}\]or \[f=\frac{{{u}_{1}}+{{u}_{2}}}{2}\]
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