A light beam in air enters a block of transparent material at an angle of 28.0˚ with respect to the normal. After refraction it makes an angle of 19.0˚ with respect to the normal.
a) Find the index of refraction of the material.
b) In a second setting, the light actually originates within the transparent material. It enters the surface at an angle of 38.9˚ with respect to the normal before emerging out in the air. What angle would it make with the surface of the material once it emerges out in the air?
a) To find the index of refraction of the material, we can use Snell's Law, which relates the angles of incidence and refraction to the index of refraction of the material.
Snell's Law is given by:
n1 * sin(θ1) = n2 * sin(θ2)
where n1 is the index of refraction of the initial medium (in this case, air), θ1 is the angle of incidence with respect to the normal, n2 is the index of refraction of the material, and θ2 is the angle of refraction with respect to the normal.
Given: θ1 = 28.0˚ and θ2 = 19.0˚.
Since the light beam is entering the material from air, we can assume the index of refraction of air to be approximately 1.
Using Snell's Law, we can rearrange the equation to solve for n2:
n2 = (n1 * sin(θ1)) / sin(θ2)
Plugging in the values, we get:
n2 = (1 * sin(28.0˚)) / sin(19.0˚)
Using a scientific calculator or an online tool to evaluate the sines of these angles, we can calculate n2.
b) Similarly, in the second setting, we can also use Snell's Law to determine the angle at which the light emerges from the material and enters the air.
Using the same Snell's Law equation:
n1 * sin(θ1) = n2 * sin(θ2)
where n1 is the index of refraction of the material, θ1 is the angle of incidence with respect to the normal within the material, n2 is the index of refraction of the second medium (in this case, air), and θ2 is the angle of refraction with respect to the normal in air.
Given: θ1 = 38.9˚
Using Snell's Law, we rearrange the equation to solve for θ2:
θ2 = arcsin((n1 * sin(θ1)) / n2)
Plugging in the values, we can calculate θ2 using the index of refraction of the material determined in part a (n1) and the index of refraction of air (n2), which is approximately 1.
a) To find the index of refraction of the material, we can use Snell's law. Snell's law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the reciprocal of the index of refraction:
n₁sinθ₁ = n₂sinθ₂
where n₁ is the index of refraction of the initial medium (air), θ₁ is the angle of incidence, n₂ is the index of refraction of the material, and θ₂ is the angle of refraction.
Given:
θ₁ = 28.0˚
θ₂ = 19.0˚
Since we are moving from air to a transparent material, n₁ = 1 (approximately).
Using the formula and solving for n₂:
1 x sin(28.0˚) = n₂ x sin(19.0˚)
n₂ = sin(28.0˚) / sin(19.0˚)
n₂ ≈ 1.517
Therefore, the index of refraction of the material is approximately 1.517.
b) In this case, we need to find the angle at which the light beam emerges out of the material and enters the air. To do that, we can again use Snell's law:
n₁sinθ₁ = n₂sinθ₂
Given:
n₁ = 1 (air)
θ₁ = 38.9˚ (angle of incidence)
n₂ = 1.517 (index of refraction of the material)
We want to find θ₂, the angle at which the light beam emerges out of the material.
Using the formula:
1 x sin(38.9˚) = 1.517 x sin(θ₂)
sin(θ₂) = sin(38.9˚) / 1.517
θ₂ = arcsin(sin(38.9˚) / 1.517)
θ₂ ≈ 26.4˚
Therefore, once the light emerges out in the air, it makes an angle of approximately 26.4˚ with the surface of the material.