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

Solution. The extensions of the parallel rays intersect at the focus, so $F$ is the focus of the given lens.

![](https://cdn.mathpix.com/cropped/2024_05_06_7a4bc87492c06fc34cd6g-12.jpg?height=468&width=266&top_left_y=1488&top_left_x=975)

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

Therefore, $\frac{O F}{\sin 135^{\circ}}=\frac{O A}{\sin 15^{\circ}}$.

We obtain that the focal length:

$O C=O F \sin 30^{\circ}=O A \frac{\sin 135^{\circ}}{\sin 15^{\circ}} \sin 30^{\circ} \approx 13.7 \mathrm{~cm}$.