By what length is a light ray displaced after passing through a 4.7 cm thick sheet of material (n=1.38) with an incident angle of theta = 33.5°.
Draw yourself a figure and apply Snell's law of refraction. Someone will gladly critique your work.
By the way, Snell did not derive the law that is commonly named for him. He didn't even spell his name that way.. It was Snel. I only learned that recently.
To calculate the displacement of a light ray passing through a material, we can use Snell's law and the concept of refraction. Snell's law states that the ratio of the sines of the angles of incidence (θ1) and refraction (θ2) is equal to the ratio of the velocities of light in the two mediums:
n1 * sin(θ1) = n2 * sin(θ2)
In this case, the incident angle (θ1) is given as 33.5°, the thickness of the material (d) is given as 4.7 cm, and the refractive index (n) of the material is given as 1.38.
First, let's calculate the refracted angle (θ2) using Snell's law:
n1 * sin(θ1) = n2 * sin(θ2)
1 * sin(33.5°) = 1.38 * sin(θ2)
sin(θ2) = (1 * sin(33.5°)) / 1.38
θ2 ≈ 22.29°
Next, we can calculate the horizontal displacement of the light ray using trigonometry. Since the incident and refracted angles are different, the ray will be displaced when it passes through the material. The horizontal displacement is given by:
Displacement = d * tan(θ1) - d * tan(θ2)
Substituting the given values:
Displacement = 4.7 cm * tan(33.5°) - 4.7 cm * tan(22.29°)
Displacement ≈ 2.796 cm
Therefore, the light ray will be displaced by approximately 2.796 cm after passing through the 4.7 cm thick sheet of material.