Light is incident from water onto a flat transparent slab of material, at an angle of 48.8° 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?
The index of water is 1.33.
That happens to be the Brewster angle in this case, since the reflected and refracted rays are perendicular.
tan48.8 = n2/n1 = n2/1.33
n2 = 1.52
http://webphysics.davidson.edu/physlet_resources/bu_semester2/c27_brewster.html
To find the index of refraction of the transparent substance, we can use Snell's law, which relates the angles of incidence and refraction with the indices of refraction of the two media involved.
Let's assume the incident medium is water, and the transparent substance has an unknown index of refraction "n". The angle of incidence is 48.8°, and the angle of refraction is 90° (since the reflected ray and the refracted ray are perpendicular to each other).
Snell's law states:
n₁sinθ₁ = n₂sinθ₂
Where:
n₁ = index of refraction of the incident medium (water)
θ₁ = angle of incidence
n₂ = index of refraction of the transparent substance (unknown)
θ₂ = angle of refraction
In this case, θ₁ = 48.8° and θ₂ = 90°.
Let's substitute the values into Snell's law:
water index of refraction (1.33) * sin(48.8°) = n * sin(90°)
1.33 * sin(48.8°) = n * sin(90°)
sin(48.8°) = n * sin(90°)
(sin(90°) = 1)
1.33 * sin(48.8°) = n
Now, let's calculate the value:
n = 1.33 * sin(48.8°)
Using a calculator:
n ≈ 1.33 * 0.7431
n ≈ 0.9851
Therefore, the index of refraction of the transparent substance is approximately 0.9851.
To determine the index of refraction of the transparent substance, we can make use of Snell's law and the condition that the reflected and refracted rays are perpendicular to each other.
Snell's law states that the ratio of the sine of the angle of incidence (θ₁) to the sine of the angle of refraction (θ₂) is equal to the ratio of the speed of light in the incident medium (v₁) to the speed of light in the refracted medium (v₂), which can be expressed as:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Where:
n₁ = index of refraction of the incident medium (water)
n₂ = index of refraction of the refracted medium (transparent substance)
Since the problem states that the reflected and refracted rays are perpendicular to each other, the angle of refraction can be found using the fact that the angle of reflection (θᵣ) is equal to the angle of incidence.
θᵣ = θ₁
With this information, we can set up the equation using Snell's law:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
n₁ * sin(θ₁) = n₂ * sin(θ₁)
Since sin(θ) = sin(90° - θ), we can substitute the equivalent value:
n₁ * sin(θ₁) = n₂ * sin(90° - θ₁)
Now, substitute the given values:
n₁ * sin(48.8°) = n₂ * sin(90° - 48.8°)
Using a scientific calculator or trigonometric table, we can calculate the values:
n₁ * 0.7471 = n₂ * 0.6710
Now, divide both sides of the equation by the value of 0.6710 to isolate n₂:
n₂ = (n₁ * 0.7471) / 0.6710
To obtain the index of refraction of the transparent substance, we need to know the index of refraction of water, which is approximately 1.33.
Substituting the value of n₁ = 1.33 into the equation:
n₂ = (1.33 * 0.7471) / 0.6710
Evaluating this expression:
n₂ ≈ 1.48
Therefore, the index of refraction of the transparent substance is approximately 1.48.