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)  1.5 m

B)  0.20 m

C)  0.10 m

D)  0.05 m

Correct Answer: C

Solution :

Let as shown, 1 and 2 are positions of objects and images in two different situations. It is given Object \[\left| \frac{{{v}_{1}}}{{{u}_{1}}} \right|=2\left| \frac{{{v}_{2}}}{{{u}_{2}}} \right|\] Here,\[{{u}_{1}}=-15cm,{{u}_{2}}=-20cm\] \[\therefore \]\[{{v}_{1}}=2{{v}_{2}}\times \frac{{{u}_{1}}}{{{u}_{2}}}=2{{v}_{2}}\times \frac{15}{20}=\frac{3}{2}{{v}_{2}}\] 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, \[\frac{1}{{{v}_{1}}}=\frac{1}{{{u}_{1}}}=\frac{1}{{{v}_{2}}}-\frac{1}{{{u}_{2}}}\] \[\Rightarrow \] \[\frac{2}{3{{v}_{2}}}=\frac{1}{15}=\frac{1}{{{v}_{2}}}+\frac{1}{20}\] \[\Rightarrow \] \[v=20cm\] \[\therefore \] \[\frac{{{v}_{1}}}{{{u}_{1}}}=2\frac{{{v}_{2}}}{{{u}_{2}}}=2\times \frac{20}{20}=2\] \[\Rightarrow \] \[{{v}_{1}}=2{{u}_{1}}=2\times 15=30cm\] Therefore,\[\frac{1}{f}=\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}=\frac{1}{15}+\frac{1}{30}=\frac{3}{30}\] \[\therefore \] \[f=10cm=0.10m\]