5. (20 points) A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=20 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its image.
Answer: 5 Dpt $i \approx 14.1$ cm

Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{20}{10}=20 \mathrm{~cm} . \quad$ The optical power of the lens: $D=\frac{1}{d}-\frac{1}{f}=\frac{1}{0.1}-\frac{1}{0.2}=5$ Dpt. The distance between the source and its image: $s=\sqrt{(x-y)^{2}+(d-f)^{2}}=\sqrt{200} \approx 14.1 \mathrm{~cm}$.