Solution to Question 21
1. Analysis of Positions
Let the optical center of the lens be the origin $O(0,0)$. The sources $S_1, S_2$ and image $I_1$ are on the principal axis. We are given that distances $S_2I_1 = I_1S_1 = S_1O$. Let this common distance be $d$. Since $I_1$ is the image of $S_1$ and $S_1$ is a real source, for a single lens to form an image further away on the same side, the image must be virtual and the lens must be convex. The order of points from left to right is $S_2, I_1, S_1, O$.
- Position of $S_1$: $u_1 = -d$
- Position of $I_1$: $v_1 = -2d$ (Since $I_1O = I_1S_1 + S_1O = d + d = 2d$)
- Position of $S_2$: $u_2 = -3d$ (Since $S_2O = S_2I_1 + I_1O = d + 2d = 3d$)
2. Finding Focal Length
Using the lens formula $\frac{1}{v} – \frac{1}{u} = \frac{1}{f}$ for source $S_1$: $$\frac{1}{-2d} – \frac{1}{-d} = \frac{1}{f}$$ $$-\frac{1}{2d} + \frac{1}{d} = \frac{1}{f} \implies \frac{1}{2d} = \frac{1}{f} \implies f = 2d$$
3. Finding Image of S₂
Now, we find the image $I_2$ for the source $S_2$ located at $u_2 = -3d$. $$\frac{1}{v_2} – \frac{1}{u_2} = \frac{1}{f}$$ $$\frac{1}{v_2} – \frac{1}{-3d} = \frac{1}{2d}$$ $$\frac{1}{v_2} = \frac{1}{2d} – \frac{1}{3d} = \frac{3-2}{6d} = \frac{1}{6d}$$ $$v_2 = +6d$$ This means the image $I_2$ is real and formed at a distance $6d$ from the optical center on the other side.
Figure: Ray diagram showing the formation of virtual image $I_1$ and real image $I_2$.
4. Conclusion
The distance of the image $I_2$ from the optical center is $6d$. Comparing this to the distance of source $S_1$ (which is $d$): $$OI_2 = 6 \times OS_1$$ Using the notation from the diagram where $OS_2 = 3 OS_1$: $$OI_2 = 2 \times OS_2$$