Question

Водолаз ростом 1.7м стоит на дне горизонтального водоема глубиной 10м. на каком минимальном расстоянии от водолаза находятся те части дна которые он может увидеть отраженными от поверхности воды? Преломление воды 1,33

Answer 1

Найдем предельный угол полного внутреннего отражения:
sina=1/n=1/1,33=0,752
a=arcsin(0,762)=49 градусов
s/h=tga    s=h*tga=1,7*1,15=1,96 м

Answer 2

Calculation of the Minimum Distance

To determine the minimum distance from the diver to the parts of the bottom that he can see reflected from the water's surface, we need to consider the concept of refraction. Refraction is the bending of light as it passes from one medium to another, such as from water to air. The refractive index of water is 1.33.

When light passes from a denser medium (water) to a less dense medium (air), it bends away from the normal (a line perpendicular to the surface of the water). This bending of light is what allows us to see objects underwater.

To calculate the minimum distance, we can use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media:

n1 * sin(theta1) = n2 * sin(theta2)

Where: - n1 is the refractive index of the first medium (water) - theta1 is the angle of incidence (the angle between the incident ray and the normal) - n2 is the refractive index of the second medium (air) - theta2 is the angle of refraction (the angle between the refracted ray and the normal)

In this case, the incident ray is the light coming from the bottom of the water, and the refracted ray is the light that reaches the diver's eyes after passing through the water and air.

Since we want to find the minimum distance, we can assume that the light ray travels parallel to the surface of the water after refraction. This means that the angle of refraction (theta2) is 90 degrees.

By rearranging Snell's law, we can solve for theta1:

theta1 = arcsin((n2 / n1) * sin(theta2))

In this case, n1 is 1.33 (the refractive index of water) and theta2 is 90 degrees. Plugging in these values, we can calculate theta1.

Calculation:

Using the given information: - Height of the diver (h) = 1.7 m - Depth of the water (d) = 10 m - Refractive index of water (n1) = 1.33 - Angle of refraction (theta2) = 90 degrees

We can calculate the angle of incidence (theta1) using the formula:

theta1 = arcsin((n2 / n1) * sin(theta2))

Substituting the values: theta1 = arcsin((1 / 1.33) * sin(90))

Calculating the value of theta1: theta1 ≈ 48.75 degrees

To find the minimum distance (x) from the diver to the parts of the bottom that he can see reflected from the water's surface, we can use trigonometry. The tangent of theta1 is equal to the opposite side (x) divided by the adjacent side (d):

tan(theta1) = x / d

Rearranging the equation to solve for x: x = d * tan(theta1)

Substituting the values: x = 10 * tan(48.75)

Calculating the value of x: x ≈ 10 * 1.19 ≈ 11.9 meters

Therefore, the minimum distance from the diver to the parts of the bottom that he can see reflected from the water's surface is approximately 11.9 meters.

Please note that this calculation assumes ideal conditions and does not take into account factors such as the clarity of the water or the diver's field of vision.