A self-luminous object of height $h$ is $40 \mathrm{~cm}$ to the left of a converging lens with a focal length of $10 \mathrm{~cm}$. A second converging lens with a focal length of $20 \mathrm{~cm}$ is $30 \mathrm{~cm}$ to the right of the first lens.

Calculate the ratio of the height of the final image to the height $h$ of the object.
From $\frac{1}{f_{1}}=\frac{1}{u_{1}}+\frac{1}{v_{1}}$, where $f_{1}=10 \mathrm{~cm}, u_{1}=40 \mathrm{~cm}$, we obtain $v_{1}=13 \frac{1}{3} \mathrm{~cm}$.

From $\frac{1}{f_{2}}=\frac{1}{u_{2}}+\frac{1}{v_{2}}$, where $f_{2}=20 \mathrm{~cm}, u_{2}=\left(30-\frac{10}{3}\right) \mathrm{cm}$, we obtain $v_{2}=-100 \mathrm{~cm}$, i.e., the final image is $100 \mathrm{~cm}$ to the left of the second lens.The ratio of the final image height to the height $h$ of the object can be calculated from

$$
m=m_{2} \times m_{1}=\left(\frac{v_{2}}{u_{2}}\right)\left(\frac{v_{1}}{u_{1}}\right)=-2 .
$$

The minus sign signifies an inverted image.