A ray of light of vacuum wavelength 550 nm
traveling in air enters a slab of transparent
material. The incoming ray makes an angle of
37.4◦
with the normal, and the refracted ray
makes an angle of 22.0◦
with the normal.
Find the index of refraction of the transparent material. (Assume that the index of
refraction of air for light of wavelength 550
nm is 1.00.)
disregard
To find the index of refraction of the transparent material, we can use Snell's law, which relates the angle of incidence of a ray of light to the angle of refraction and the indices of refraction of the two materials.
Snell's Law is given by:
n1 sin(θ1) = n2 sin(θ2)
Where:
n1 is the index of refraction of the first material (in this case, air),
θ1 is the angle of incidence,
n2 is the index of refraction of the second material (the transparent material),
θ2 is the angle of refraction.
Given in the problem:
Wavelength in air, λ = 550 nm = 550 x 10^(-9) m.
Index of refraction of air, n1 = 1.00.
Angle of incidence, θ1 = 37.4°.
Angle of refraction, θ2 = 22.0°.
To find n2, we need to rearrange Snell's law:
n2 = n1 * (sin(θ1) / sin(θ2))
Now, we can substitute the known values and solve for n2:
n2 = 1.00 * (sin(37.4°) / sin(22.0°))
Using a calculator:
n2 ≈ 1.49
Therefore, the index of refraction of the transparent material is approximately 1.49.