question_answer

    For a given lens, the magnification was found to be twice as large as when the object was 0.15 m distant from It as when the distance was 0.2 m. The focal length of the lens is

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, \[\frac{1}{f}=\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}=\frac{1}{15}+\frac{1}{30}=\frac{3}{30}\] \[\therefore \]  \[f=10\,cm=0.10\,m\]