A) 1.5m
B) 0.20m
C) 0.10m
D) 0.05m
Correct Answer: C
Solution :
Let as shown, 1 and 2 are positions of objects and images in two different situations. Object It-is given \[\left| \frac{{{v}_{1}}}{{{u}_{1}}} \right|=2\left| \frac{{{v}_{2}}}{{{u}_{2}}} \right|\] Here, \[{{u}_{1}}=-15\,cm,{{u}_{2}}=-20\,cm\] \[\therefore \] \[{{v}_{1}}=2{{v}_{2}}\times \frac{{{u}_{1}}}{{{u}_{2}}}=2{{v}_{2}}\times \frac{15}{20}=\frac{3}{2}{{v}_{2}}\] Now \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\] \[\therefore \] \[\frac{1}{f}=\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}\]and \[\frac{1}{f}=\frac{1}{{{v}_{2}}}-\frac{1}{{{u}_{2}}}\] So, \[\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}=\frac{1}{{{v}_{2}}}-\frac{1}{{{u}_{2}}}\] \[\Rightarrow \] \[\frac{2}{3{{v}_{2}}}+\frac{1}{15}=\frac{1}{{{v}_{2}}}+\frac{1}{20}\] \[\Rightarrow \] \[v=20\,cm\] \[\therefore \] \[\frac{{{v}_{1}}}{{{u}_{1}}}=2\frac{{{v}_{2}}}{{{u}_{2}}}=2\times \frac{20}{20}=2\] \[\Rightarrow \] \[{{v}_{1}}=2{{u}_{1}}-2\times 15=30\,cm\] There fore, \[\frac{1}{f}=\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}=\frac{1}{15}+\frac{1}{30}=\frac{3}{30}\] \[\therefore \] \[f=10\,cm=0.10\,m\]You need to login to perform this action.
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