An object AB is placed 15 cm to the left of a convex lens P of focal length $$10 cm$$. Another convex lens Q is now placed $$15 cm$$ right of lens P . If the focal length of lens Q is 15 cm,  final image is______.

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$m_{\text{total}} = m_1 \times m_2 = \left(\frac{v_1}{u_1}\right) \times \left(\frac{v_2}{u_2}\right)$$

$$\frac{1}{v_1} - \frac{1}{-15} = \frac{1}{10}$$

$$\frac{1}{v_1} + \frac{1}{15} = \frac{1}{10} \implies \frac{1}{v_1} = \frac{1}{10} - \frac{1}{15}$$

$$\frac{1}{v_1} = \frac{3 - 2}{30} = \frac{1}{30} \implies v_1 = +30\text{ cm}$$

Lens $$P$$ forms a real image $$30\text{ cm}$$ to its right.

$$m_1 = \frac{v_1}{u_1} = \frac{30}{-15} = -2$$

Lens $$Q$$ is located $$15\text{ cm}$$ to the right of lens $$P$$.

Because the image from lens $$P$$ forms at $$30\text{ cm}$$ to the right of $$P$$, it ends up further down the path, exactly:

$$\text{Distance from Q} = 30\text{ cm} - 15\text{ cm} = 15\text{ cm} \text{ (to the right of } Q)$$

Since the light rays converge toward this position behind lens $$Q$$, it serves as a virtual object for lens $$Q$$:

Object distance for $$Q$$, $$u_2 = +15\text{ cm}$$

Focal length of $$Q$$, $$f_2 = +15\text{ cm}$$

$$\frac{1}{v_2} - \frac{1}{+15} = \frac{1}{15}$$

$$\frac{1}{v_2} = \frac{1}{15} + \frac{1}{15} = \frac{2}{15} \implies v_2 = \frac{15}{2} = +7.5\text{ cm}$$

$$m_2 = \frac{v_2}{u_2} = \frac{7.5}{15} = +\frac{1}{2}$$

$$m_{\text{total}} = m_1 \times m_2 = (-2) \times \left(+\frac{1}{2}\right) = -1$$

An absolute total magnification value of $$|m_{\text{total}}| = 1$$ means the final image has the exact same size as the original object $$AB$$ (the negative sign simply means it is inverted).