A narrow ray of yellow light from glowing sodium (λ0 = 589 nm) traveling in air strikes a smooth surface of water at an angle of θi = 40.5°. Determine the angle of refraction θr.

To determine the angle of refraction (θr), we can use Snell's Law, which relates the angles of incidence and refraction to the indices of refraction of the two media.

Snell's Law is given by:

n1 * sin(θi) = n2 * sin(θr)

Where:
- n1 is the refractive index of the medium the light is coming from (in this case air)
- θi is the angle of incidence
- n2 is the refractive index of the medium the light is entering (in this case water)
- θr is the angle of refraction

In this case, the light is coming from air and entering water. The refractive index of air is very close to 1, while the refractive index of water is approximately 1.33.

Using Snell's Law, we can rearrange the equation to solve for θr:

sin(θr) = (n1/n2) * sin(θi)

Let's plug in the values:

n1 = 1 (refractive index of air)
n2 = 1.33 (refractive index of water)
θi = 40.5°

sin(θr) = (1/1.33) * sin(40.5°)

Now, let's calculate:

sin(θr) ≈ (0.7519) * (0.643)

sin(θr) ≈ 0.4837

Now, we can find the angle of refraction using the inverse sine (arcsine) function:

θr ≈ arcsin(0.4837)

θr ≈ 29.16°

Therefore, the angle of refraction (θr) is approximately 29.16°.

To determine the angle of refraction, we can use Snell's Law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the indices of refraction of the two media.

Snell's Law: n1 * sin(θi) = n2 * sin(θr)

In this scenario, the initial medium is air with an index of refraction of approximately 1 (since the index of refraction of air is very close to 1). The second medium is water with an index of refraction of approximately 1.33.

Plugging in the given values into Snell's Law, we get:

1 * sin(40.5°) = 1.33 * sin(θr)

Now, we can solve for the angle of refraction θr:

sin(θr) = (sin(40.5°) / 1.33)

Using a scientific calculator or trigonometric tables, we can find:

sin(θr) ≈ 0.6564

To find the angle of refraction θr, we can take the inverse sine (arcsine) of the value:

θr ≈ sin^(-1)(0.6564)

Calculating this value, we find:

θr ≈ 41.78°

Therefore, the angle of refraction, θr, is approximately 41.78°.

so, what is the index of refraction of water?

Then use Snell's law