A) 15 cm
B) infinity
C) 45 cm
D) 30 cm
Correct Answer: D
Solution :
For first lens \[{{u}_{1}}=-30\,\,cm,\,\,{{f}_{1}}=10\,cm\] We have, \[\frac{1}{f}=\frac{1}{v}\,=\frac{1}{u}\Rightarrow \,\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\] or \[\frac{1}{v}=\frac{1}{10}-\frac{1}{30}\,=\frac{1}{15}\Rightarrow \,v=15\,cm\] Therefore, image formed by convex lens \[({{L}_{1}})\] is at point \[{{I}_{1}}\] and acts as virtual object for concave lens \[({{L}_{2}})\].You need to login to perform this action.
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