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 \[({{L}_{2}})\]. The image \[{{I}_{1}}\] is formed at focus of concanve lens (as shown) and so emergent rays will be parallel to the principal axis, For lens \[{{I}_{2}},\,{{u}_{2}}=15-5=10\,cm,\,\,{{f}_{2}}=-10cm\] These parallel rays are incident on the third convex lens\[({{L}_{3}})\] and will be brought to convergence at the focus of the lens \[{{L}_{3}}\] Hence ,distance of final image from third lens \[{{L}_{3}}\]\[{{v}_{2}}={{f}_{3}}=30\,cm\]You need to login to perform this action.
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