Question

Луч света от ла­зер­ной указ­ки па­да­ет из воз­ду­ха на по­верх­ность воды бас­сей­на под углом α. Затем пре­лом­лен­ный луч по­па­да­ет на плос­кое зер­ка­ло, ле­жа­щее на дне бас­сей­на. Рас­сто­я­ние от точки па­де­ния луча на по­верх­ность воды до точки вы­хо­да луча на по­верх­ность равно 2 м, по­ка­за­тель пре­лом­ле­ния воды равен 1,33. В воде свет про­хо­дит путь 376 см. Чему равен угол α? Ответ вы­ра­зи­те в гра­ду­сах и округ­ли­те до це­ло­го числа. (хотелось бы подробное решение, через дано и прочее)​

Answer 1

Ответ:

45

Объяснение:

Изобразим ход лучей. Из рисунка видно, что в прямоугольном треугольнике с углом β гипотенуза равна половине пути, пройденного светом в воде, а катет, противолежащий углу β равен 1.

Найдём синус угла β:  синус бета = $\frac{1}{1.88}$. Для синусов падающего и преломлённого лучей верно соотношение: $\frac{sinA}{cosB}$=n. откуда  sinA = nsinB = $\frac{1.33}{1.88}$. Значит, A = arcsin$\frac{1.33}{1.88}$ = 45

Answer 2

To find the value of angle α, we can use the principles of refraction and Snell's law. Let's break down the problem step by step.

Given:

- The laser beam falls from the air onto the surface of the water in the pool at an angle α. - The refracted beam then hits a flat mirror lying on the bottom of the pool. - The distance from the point of incidence of the beam on the water surface to the point where the beam exits the water is 2 m. - The refractive index of water is 1.33. - The beam travels a distance of 376 cm in water.

Solution:

1. Let's first find the angle of incidence (i) using Snell's law. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speeds of light in the two media.

The speed of light in air is approximately the same as in vacuum, so we can assume it to be 1. The speed of light in water is 1/1.33 times the speed of light in air.

Applying Snell's law, we have:

sin(i) / sin(90°) = 1 / 1.33

Since sin(90°) = 1, we can simplify the equation to:

sin(i) = 1 / 1.33

Solving for i, we find:

i ≈ arcsin(1 / 1.33)

Using a calculator, we find that i ≈ 48.75°.

2. Now, let's find the angle of reflection (r). The angle of reflection is equal to the angle of incidence, so:

r = i ≈ 48.75°

3. Next, we need to find the angle of refraction (θ) when the beam passes from water to air. We can use Snell's law again, but this time we need to consider the refractive index of water.

sin(θ) / sin(90°) = 1.33 / 1

Simplifying the equation, we have:

sin(θ) = 1.33

Solving for θ, we find:

θ ≈ arcsin(1.33)

Using a calculator, we find that θ ≈ 81.37°.

4. Finally, we can find the value of angle α. Since the angle of incidence (i) is equal to the angle of reflection (r), we have:

α = 180° - 2 * r

Substituting the value of r, we find:

α = 180° - 2 * 48.75°

Simplifying the equation, we have:

α = 82.5°

Therefore, the value of angle α is approximately 82.5°.