Question

Луч света падает из воды н стекло под углом 30°. Определить синус ушла приломления, если показатель преломления воды 1.33, а показатель преломления стекла 1,5

Answer 1

Sina/sinb=n2/n1
sinb=sina*n1/n2=0,5*1,33/1,5=0,443
b=26,5 градусов

Answer 2

Problem Statement

A beam of light passes from water to glass at an angle of 30°. The refractive index of water is 1.33, and the refractive index of glass is 1.5. We need to determine the sine of the angle of refraction.

Solution

To solve this problem, we can use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. Snell's law is given by:

\[n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\]

where: - \(n_1\) is the refractive index of the first medium (water in this case) - \(\theta_1\) is the angle of incidence - \(n_2\) is the refractive index of the second medium (glass in this case) - \(\theta_2\) is the angle of refraction

We are given that the angle of incidence (\(\theta_1\)) is 30°, the refractive index of water (\(n_1\)) is 1.33, and the refractive index of glass (\(n_2\)) is 1.5. We need to find the sine of the angle of refraction (\(\theta_2\)).

Using Snell's law, we can rearrange the equation to solve for \(\sin(\theta_2)\):

\[\sin(\theta_2) = \frac{n_1}{n_2} \sin(\theta_1)\]

Substituting the given values, we have:

\[\sin(\theta_2) = \frac{1.33}{1.5} \sin(30°)\]

Now we can calculate the value of \(\sin(\theta_2)\):

\[\sin(\theta_2) = \frac{1.33}{1.5} \times 0.5\]

Simplifying the expression, we get:

\[\sin(\theta_2) = 0.4433\]

Therefore, the sine of the angle of refraction is approximately 0.4433.