# 1. A ray of light passes from air into water at an angle of 30.0 degrees. Find the angle of refraction. (22.1 degrees)

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### 2 Answers

To find the angle of refraction, use Snell's Law.

Here, `theta_1` is the angle of incidence (angle the ray makes with the normal to the surface) and `theta_2` is the angle the refracted ray makes with the normal to the surface - angle of refraction.

`sintheta_1/sintheta_2 = n_2/n_1`

The values of `n_1` and `n_2` , indices of refraction, are `n_1 = 1` for air and `n_2 = 1.33` for water. Plugging in these values and also `sintheta_1 = sin30=1/2` , and using the property of proportion, solve for `sintheta_2` :

`sintheta_2 = (n_1sintheta_1)/n_2=(1/2)/1.33=0.38`

From here, `theta_2 = 22.1` degrees.

**Sources:**

Refraction is the phenomenon of the "bending" of rays of light as they pass the boundary between two media, for example, air and water. It happens because the light propagates with different speeds in different media.

In air, the speed of light is very close to the speed of light in the vacuum, `3*10^8 m/s` . In water, the speed of light decreases by a factor of `n_w = 1.33` .

This number is called *index of refraction *of water.

The relationship between the angle of incidence and the angle of refraction is described by Snell's Law:

`n_isin(theta_i) = n_rsin(theta_r)`

Here, `n_i` is the index of refraction of incident medium, which is in our case

air: `n_1 = 1` , `theta_i` is the angle the incident light ray makes with the normal (perpendicular line) to the boundary between the media, `n_r ` is the index of refraction of the refractive media (water), and `theta_r` is the angle the refracted ray makes with the normal.

For the values in the given problem,

`1*sin(30) = 1.33*sin(theta_r)`

From here, `sin(theta_r) = (sin(30)/1.33) =0.376 `

and `theta_r = 22.1` degrees.

**The angle of refraction is 22.1 degrees.**

**Sources:**