question_answer

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

A)  0.10 m                 

B)  0.20 m

C)  0.35 m                 

D)  0.40 m

Correct Answer: A

Solution :

                It is given \[\left| \frac{{{v}_{1}}}{{{u}_{1}}} \right|=2\left| \frac{{{v}_{2}}}{{{u}_{2}}} \right|\] \[{{u}_{1}}=-15\,cm,\]\[{{u}_{2}}=-20\,cm\] \[{{v}_{1}}=2{{v}_{2}}\times \frac{{{u}_{1}}}{{{u}_{2}}}=2{{v}_{2}}\times \frac{15}{20}=\frac{3}{2}{{v}_{2}}\] \[\frac{1}{f}=\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}\] and \[\frac{1}{f}=\frac{1}{{{v}_{2}}}-\frac{1}{{{u}_{2}}}\] \[\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}=\frac{1}{{{v}_{2}}}-\frac{1}{{{u}_{2}}}\] \[\frac{2}{3{{v}_{2}}}+\frac{1}{15}=\frac{1}{{{v}_{2}}}+\frac{1}{20}\] \[{{v}_{2}}=20\,cm\] \[\frac{{{v}_{1}}}{{{u}_{2}}}=2\frac{{{v}_{2}}}{{{u}_{2}}}=2\times \frac{20}{20}=2\] \[{{v}_{1}}=2{{u}_{1}}=2\times 15=30\,cm\] \[\frac{1}{f}=\frac{1}{{{v}_{1}}}-\frac{1}{{{u}_{1}}}=\frac{1}{15}+\frac{1}{30}=\frac{3}{30}\] \[f=10\,cm=0.10\,m\]