The drawing shows a ray of light traveling through three materials whose surfaces are parallel to each other. The refracted rays (but not the reflected rays) are shown as the light passes through each material. A ray of light strikes the a–b interface at a 50.0˚ angle of incidence. The index of refraction of material a is na = 1.20. The angles of refraction in materials b and c are, respectively, 41.8˚ and 56.1˚. Find the indices of refraction in these two media.
To find the indices of refraction in materials b and c, we can use Snell's law, which relates the angles of incidence and refraction of a light ray as it passes from one medium to another.
Snell's law is given by the equation: n1 * sin(θ1) = n2 * sin(θ2)
Where:
n1 and n2 are the indices of refraction of the two media
θ1 and θ2 are the angles of incidence and refraction
In this case, we know the angle of incidence (θ1) and the angle of refraction in material b (θ2). We only need to find the index of refraction of material b (nb).
Step 1: Find the index of refraction in material b (nb)
Using Snell's law, we have:
na * sin(θ1) = nb * sin(θ2)
Plugging in the values we know:
1.20 * sin(50.0˚) = nb * sin(41.8˚)
Solving for nb:
nb = (1.20 * sin(50.0˚)) / sin(41.8˚)
nb ≈ 1.514
Step 2: Find the index of refraction in material c (nc)
We know the angle of refraction in material c (θ3), which is 56.1˚. We can use Snell's law again, but this time the equation will be:
nb * sin(θ2) = nc * sin(θ3)
Plugging in the values we know:
1.514 * sin(41.8˚) = nc * sin(56.1˚)
Solving for nc:
nc = (1.514 * sin(41.8˚)) / sin(56.1˚)
nc ≈ 1.332
Therefore, the indices of refraction in materials b and c are approximately 1.514 and 1.332, respectively.
To find the indices of refraction in materials b and c, we can use the laws of refraction.
The law of refraction, also known as 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 ratio of the velocities of light in the two materials. This law can be expressed as:
sin(θ1) / sin(θ2) = n2 / n1
Where:
θ1 = angle of incidence
θ2 = angle of refraction
n1 = index of refraction of the first material
n2 = index of refraction of the second material
First, let's find the index of refraction in material b. Given that the angle of incidence at the a-b interface is 50.0˚ and the angle of refraction in material b is 41.8˚, we can use the above equation to solve for n2.
sin(50.0˚) / sin(41.8˚) = n2 / 1.20
Now, we can rearrange the equation to solve for n2:
n2 = 1.20 * (sin(50.0˚) / sin(41.8˚))
Using a calculator, we find that n2 ≈ 1.44.
Now, let's find the index of refraction in material c. Given that the angle of refraction in material c is 56.1˚, we can apply Snell's law once again to solve for n3.
sin(41.8˚) / sin(56.1˚) = n3 / 1.44
Rearranging the equation, we can solve for n3:
n3 = 1.44 * (sin(41.8˚) / sin(56.1˚))
Using a calculator, we find that n3 ≈ 1.07.
Therefore, the indices of refraction in materials b and c are approximately 1.44 and 1.07, respectively.