question_answer

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

A)  15 cm   

B)                         infinity

C)  45cm                   

D)         30cm

E)  35 cm

Correct Answer: D

Solution :

For first lens, \[{{u}_{1}}=-30\,cm,{{f}_{1}}=10\,cm\] \[\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\]or \[\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\] Or           \[\frac{1}{v}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15}\] Or           \[v=15\text{ }cm\] 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 concave lens (as shown) and so emergent rays will be parallel to the principal axis. For lens\[{{L}_{2}}\] \[{{u}_{2}}=15-5=10\,cm,{{f}_{2}}=-10\,cm.\]These parallel rays are incident on the third convex lens \[({{L}_{3}})\]and will be brought to convergence at the focus of the lens 1, 3. Hence, distance of final image from third lens \[{{L}_{3}},\] \[{{v}_{2}}={{f}_{3}}=30cm\]