question_answer

The position of final image formed by the given lens combination from the third lens will be at a distance of \[[{{f}_{1}}=+10\,cm,\,{{f}_{2}}=-10\,cm,{{f}_{3}}=30\,cm]\]            30cm   5cm     10cm

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 \[{{L}_{3}}\]\[{{v}_{2}}={{f}_{3}}=30\,cm\]