Question

На поверхности водоема плавает плотик, под центром которого находится точечный источник света. Если глубина водоёма 1,3 м, то каким должен быть радиус плотика, чтобы свет от источника не был виден. Показатель преломления воды принять 4/3.

Answer 1

Дано:
n = 4/3
H = 1,3 м
__________
R - ?

Решение:

1)
Находим угол полного внутреннего отражения:
α₀ = arcsin (1/n) = arcsin (3/4) = arcsin (0,75) ≈ 48,6°

2)
R = H·tg (α₀) = 1,3·tg (48,6°) ≈ 1,3·1,13 ≈ 1,47 м

Answer 2

Understanding the Problem

The problem states that there is a small floating object on the surface of a body of water, with a point light source located at the center of the object. The depth of the water is given as 1.3 meters. We need to determine the radius of the object such that the light from the source is not visible.

Solution

To solve this problem, we need to consider the principles of refraction and total internal reflection.

When light passes from one medium to another, it changes direction due to the change in speed. This phenomenon is called refraction. The amount of bending depends on the refractive indices of the two media involved.

In this case, light is passing from air (where the point light source is located) to water. The refractive index of water is given as 4/3.

To ensure that the light from the point source is not visible, we need to ensure that the light rays entering the water are refracted away from the line of sight of an observer above the water. This can be achieved by using the principle of total internal reflection.

Total internal reflection occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index, and the angle of incidence is greater than the critical angle. The critical angle can be calculated using the formula:

critical angle = arcsin(n2/n1)

Where n1 is the refractive index of the medium the light is coming from (in this case, air) and n2 is the refractive index of the medium the light is entering (in this case, water).

Once we have the critical angle, we can determine the maximum angle at which the light rays can enter the water without being visible. This angle can be calculated as:

maximum angle = 90 degrees - critical angle

Finally, we can use trigonometry to calculate the radius of the object. The radius can be calculated as:

radius = depth * tan(maximum angle)

Substituting the given values, we can calculate the radius of the object.

Calculation

Given: - Depth of the water (d) = 1.3 meters - Refractive index of water (n2) = 4/3

To calculate the critical angle: critical angle = arcsin(n2/n1) Substituting the values: critical angle = arcsin(4/3/1) = arcsin(4/3) ≈ 53.13 degrees

To calculate the maximum angle: maximum angle = 90 degrees - critical angle Substituting the values: maximum angle = 90 degrees - 53.13 degrees ≈ 36.87 degrees

To calculate the radius of the object: radius = depth * tan(maximum angle) Substituting the values: radius = 1.3 meters * tan(36.87 degrees) ≈ 1.3 meters * 0.75 ≈ 0.975 meters

Therefore, the radius of the object should be approximately 0.975 meters to ensure that the light from the point source is not visible.

Please note that the calculations provided above are based on the given information and assumptions.