Question

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

Answer 1

Answer: (D)

30cm

Answer 2

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 $({{L}_{2}})$ .