8. (15 points) A thin ray of light falls on a thin converging lens at a distance of $x=10$ cm from its optical center. The angle between the incident ray and the plane of the lens $\alpha=45^{\circ}$, and the angle between the refracted ray and the plane of the lens $\beta=30^{\circ}$. Determine its focal length.
Answer: $\approx 13.7$ cm.

Solution. Parallel rays intersect at the focus, so $F$ is the focus of the given lens.

![](https://cdn.mathpix.com/cropped/2024_05_06_7a4bc87492c06fc34cd6g-06.jpg?height=480&width=302&top_left_y=1502&top_left_x=957)

In triangle $O A F$: angle $F A O=30^{\circ}$, angle $O F A=15^{\circ}$, angle $A O F=135^{\circ}$.

Therefore, $\frac{O F}{\sin 30^{\circ}}=\frac{A O}{\sin 15^{\circ}}$.

We obtain that the focal length:

$F C=O F \sin 45^{\circ}=A O \frac{\sin 30^{\circ}}{\sin 15^{\circ}} \sin 45^{\circ} \approx 13.7 \mathrm{~cm}$.

## Multidisciplinary Engineering Olympiad "Star" in Natural Sciences

10th grade

Final Round

$2022-2023$

Version 2

Problems, answers, and evaluation criteria