Question

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

• A. 15 cm
• B. Infinity
• C. 45 cm
• D. 30 cm

Answer 1

The correct option is A 15 cm

For 1${}^{st}$ lens, $u_1=-30,f_1=+10cm,$

Formula of lens, $\frac{1}{v_1}+\frac{1}{30}=\frac{1}{10}$.

or $v_1=15cm$ at $I_1$ behind the lens.

The images $I_1$ serves as virtual object for concave lens.

For second lens, which is concave, $u_2=(15-5)=10cm.I_1$ acts as object. $f_2=-10cm$.

The rays will emerge parallel to axis as the virtual object is at focus of concave lens, as shown in the figure. Image of $I_1$ will be at infinity. The parallel rays are incident on the third lens viz the convex lens, $f_3=+30cm$. These parallel rays will be brought to convergence at the focus of the third lens.

$\therefore$ Image distance from third lens $=f_3=30cm$