Light passes from air into water at an angle of

incidence of 33.5�.degrees
Find the angle of refraction in the water.
Answer in units of �degrees

By Snell's law,

n(incident)sin(theta(incident)) = n(refracted)sin(theta(refracted)) or n(1)sin(theta(1)) = n(2)sin(theta(2))

Solving for theta(2)

sin(theta(2)) = [n(1)sin(theta(1))]/n(2)

Theta(2) = arcsin [n(1)sin(theta(1))]/n(2)

Theta(2) = arcsin ((1)(sin(33.5))/1.33 or whatever your instructor says to use for the indexes of refraction for air and water

To find the angle of refraction in water, you can use Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the two media.

The formula for Snell's Law is:
n₁ * sin(θ₁) = n₂ * sin(θ₂)

Where:
n₁ = refractive index of the initial medium (air)
θ₁ = angle of incidence in the initial medium
n₂ = refractive index of the second medium (water)
θ₂ = angle of refraction in the second medium (water)

In this case, the initial medium is air, which has a refractive index of approximately 1.0003, and the second medium is water, which has a refractive index of approximately 1.33.

Given that the angle of incidence (θ₁) is 33.5˚, you can substitute the values into Snell's Law to solve for the angle of refraction:

1.0003 * sin(33.5˚) = 1.33 * sin(θ₂)

To find θ₂, isolate it on one side of the equation:

sin(θ₂) = (1.0003 * sin(33.5˚)) / 1.33

Now, take the inverse sine (also known as arcsine) of both sides:

θ₂ = arcsin((1.0003 * sin(33.5˚)) / 1.33)

Using a calculator, evaluate the right side of the equation:

θ₂ ≈ 22.77˚

Therefore, the angle of refraction in the water is approximately 22.77˚.