Question

Луч, падающий из воздуха под углом 30° к поверхности жидкости, после преломления отклоняется от первоначаль-ного направления на 15°. Найти показатель преломления жидкости.МНЕ НУЖНО РЕШЕНИЕ НЕ ОТВЕТ РЕШЕНИЕ !!!!ПОЖАЛУЙСТА​

Answer 1

Ответ:

Объяснение:

Угол падения:

α = 30°

Угол преломления

β = α - 15 = 15°

n = sin α / sin β = sin 30° / sin 15° = 15*√2 / 15 = √30 ≈ 5,5  

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Answer 2

Problem Analysis

We are given that a light ray is incident on the surface of a liquid at an angle of 30° and after refraction, it deviates from its original direction by 15°. We need to find the refractive index of the liquid.

Solution

Let's denote the angle of incidence as i and the angle of refraction as r. According to Snell's law, the relationship between the angles of incidence and refraction and the refractive indices of the two media is given by:

n1 * sin(i) = n2 * sin(r)

where n1 is the refractive index of the medium from which the light is coming (in this case, air) and n2 is the refractive index of the liquid.

We are given that the angle of incidence is 30° and the angle of refraction is 15°. Substituting these values into Snell's law, we get:

n1 * sin(30°) = n2 * sin(15°)

We know that the refractive index of air is approximately 1. So the equation becomes:

1 * sin(30°) = n2 * sin(15°)

Now we can solve for n2:

n2 = sin(30°) / sin(15°)

Using a calculator, we can find that sin(30°) ≈ 0.5 and sin(15°) ≈ 0.259. Substituting these values, we get:

n2 ≈ 0.5 / 0.259

Simplifying the expression, we find:

n2 ≈ 1.932

Therefore, the refractive index of the liquid is approximately 1.932.

Answer

The refractive index of the liquid is approximately 1.932.