Question

На дно акваріума,наповненого водою до висоти 20 см,помістили маленьку лампочку. Якого найменшого радіуса непрозорий круг треба покласти на воду,щоб жодний промінь не зміг вийти через поверхню води?​

Answer 1

Ответ: 22,8 см

Объяснение:  Див. додаток

Answer 2

I can answer your question in detail.

The problem you are asking is related to the phenomenon of total internal reflection, which occurs when light travels from a denser medium to a less dense medium and the angle of incidence is greater than the critical angle. The critical angle is the angle of incidence that produces an angle of refraction of 90 degrees.

To find the radius of the opaque circle that needs to be placed on the water surface, we need to find the critical angle for the water-air interface. Using Snell's law, we can write:

$$n_1 \sin \theta_1 = n_2 \sin \theta_2$$

where $n_1$ and $n_2$ are the refractive indices of water and air, respectively, and $\theta_1$ and $\theta_2$ are the angles of incidence and refraction, respectively. Since we want $\theta_2$ to be 90 degrees, we can write:

$$\sin \theta_2 = 1$$

and solve for $\theta_1$:

$$\theta_1 = \sin^{-1} \left( \frac{n_2}{n_1} \right)$$

Using the values of $n_1 = 1.33$ and $n_2 = 1$, we get:

$$\theta_1 = \sin^{-1} \left( \frac{1}{1.33} \right) \approx 48.8^{\circ}$$

This is the critical angle for the water-air interface. Any light ray that hits the surface at an angle greater than this will be totally reflected back into the water.

Now, we can use geometry to find the radius of the opaque circle. If we draw a right triangle with the height of the water as the vertical leg, the radius of the circle as the horizontal leg, and the hypotenuse as the path of the light ray, we can see that the angle of incidence is equal to the angle opposite to the radius. Therefore, we can write:

$$\sin \theta_1 = \frac{r}{h}$$

where $r$ is the radius of the circle and $h$ is the height of the water. Substituting the values of $\theta_1 = 48.8^{\circ}$ and $h = 20$ cm, we get:

$$r = h \sin \theta_1 \approx 20 \times 0.74 \approx 14.8 \text{ cm}$$

This is the radius of the opaque circle that needs to be placed on the water surface, so that no light can escape through the surface. The diameter of the circle is twice the radius, so it is about 29.6 cm.

I hope this answer helps you understand the problem. You can learn more about total internal reflection and its applications from [this article](https://edufuture.biz/index.php?title=%D0%9F%D0%BE%D0%B2%D0%BD%D0%B5_%D0%B2%D0%BD%D1%83%D1%82%D1%80%D1%96%D1%88%D0%BD%D1%94_%D0%B2%D1%96%D0%B4%D0%B1%D0%B8%D0%B2%D0%B0%D0%BD%D0%BD%D1%8F) or [this video](https://www.youtube.com/watch?v=7Y3mfAGVn1c