Question

The position of final image formed by the given lens combination from the third lens will be at a distance of [ $f_1=+10cm,$ $f_2=-10cm,$ $f_3=+30cm$ ]

• A. 15 cm
• B. infinity
• C. 45cm
• D. 30cm
• E. 35 cm

Answer 1

Answer: (D)

For first lens, $u_1=-30cm,f_1=10cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ or $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
Or $\frac{1}{v}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15}$
Or $v=15cm$
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 $L_2$
$u_2=15-5=10cm,f_2=-10cm.$
These parallel rays are incident on the third convex lens $(L_3)$ and will be brought to convergence at the focus of the lens 1, 3. Hence, distance of final image from third lens $L_3,$ $v_2=f_3=30cm$