Light is incident from air onto a flat transparent slab of material, at an angle of 62.2° with the normal. The reflected ray and the refracted ray are perpendicular to each other.
What is the index of refraction of the transparent substance?
1.65
To find the index of refraction of the transparent substance, we can use Snell's Law which relates the angles of incidence and refraction to the indices of refraction of the two media involved.
Snell's Law is given by:
n₁ * sinθ₁ = n₂ * sinθ₂
Where:
- n₁ is the refractive index of the initial medium (in this case, air)
- θ₁ is the angle of incidence
- n₂ is the refractive index of the medium the light enters (in this case, the transparent substance)
- θ₂ is the angle of refraction
In this problem, it is given that the refracted ray and the reflected ray are perpendicular to each other. This implies that the angle of refraction (θ₂) is 90°.
Let's substitute the given values into Snell's Law:
n₁ * sin(θ₁) = n₂ * sin(90°)
Since sin(90°) = 1, the equation becomes:
n₁ * sin(θ₁) = n₂ * 1
n₁ * sin(θ₁) = n₂
We can rearrange the equation to solve for n₂:
n₂ = n₁ * sin(θ₁)
Now, plug in the known values:
- n₁ is the refractive index of air, which is approximately 1 (since air's refractive index is close to 1)
- θ₁ is the angle of incidence, given as 62.2°
n₂ = 1 * sin(62.2°)
Using a calculator:
n₂ ≈ 0.895
Therefore, the index of refraction of the transparent substance is approximately 0.895.