A convergent lens is placed $40 \mathrm{~cm}$ to the right of a diverging lens of focal length $15…

A convergent lens is placed $40 \mathrm{~cm}$ to the right of a diverging lens of focal length $15 \mathrm{~cm}$. A parallel beam of light enters the divergent lens from the left and the beam is again parallel when it emerges from the convergent lens. The focal length of the convergent lens is
  1. $40 \mathrm{~cm}$
  2. $25 \mathrm{~cm}$
  3. $55 \mathrm{~cm}$
  4. $27.5 \mathrm{~cm}$

Solution

We are given with following arrangement,
From diagram it is clear that focus of lens 1, point $F_1$ must be focus of lens 2 also. In this way, a parallel beam, of light becomes again paralkel after passing through system of lenses. Hence, distance $F_2 L_2$, $F_2 L_2=15+40=55 \mathrm{~cm}$ So, focal length of convex lens is $55 \mathrm{~cm}$.

Asked in: AP EAMCET 2022 (08 Jul Shift 2)

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