# Question Video: Calculating the Refractive Index of Air from Incident and Refracted Angle Physics • 9th Grade

A light ray traveling in air is incident on the flat surface of a plastic block, hitting the surface at an angle of 45 degrees from the line normal to it. The refracted ray in the block travels at an angle of 33 degrees to the line normal to the surface. What is the refractive index of the plastic? Give your answer to one decimal place.

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### Video Transcript

A light ray traveling in air is incident on the flat surface of a plastic block, hitting the surface at an angle of 45 degrees from the line normal to it. The refracted ray in the block travels at an angle of 33 degrees to the line normal to the surface. What is the refractive index of the plastic? Give your answer to one decimal place.

In this question, we’re dealing with a ray of light initially traveling in air. The ray is incident on the surface of a plastic block, and we want to determine the refractive index of the plastic. We will start by drawing out this problem.

So let’s begin by saying that this is the initial ray of light. We’ve been told that this ray is incident on the flat surface of a plastic block. So let’s say that this is the plastic block. To make things clear, let’s label the air and plastic block as follows. Now, additionally, we’ve been told that this ray of light hits the surface of the plastic block at an angle of 45 degrees from the line normal to the surface. So, if this is the surface of the plastic block, then this is the line normal to that surface, because the normal line is at 90 degrees to the surface, perpendicular to the surface. That’s what it means for it to be normal. And so the angle between the ray and the normal line here is 45 degrees. Let’s call this angle 𝜃 sub one.

Now, because this ray of light is traveling from air to plastic, in other words, from a medium with a certain refractive index to a medium with a different refractive index, this ray of light is going to refract or change direction in the plastic block. We’ve specifically been told that the refracted ray in the block travels at an angle of 33 degrees to the line normal to the surface. And so if we imagine that this is the ray of light now traveling through the plastic block, then we can say that the angle between that ray of light and the normal line, which we will call here 𝜃 sub two, is equal to 33 degrees as we’ve been told in the question.

Now that we have drawn out the problem, we need to work out the refractive index of the plastic that results in this specific refraction. To do this, we need to recall Snell’s law. This is given by the equation 𝑛 sub 𝑖 multiplied by sin 𝜃 sub 𝑖 equals 𝑛 sub 𝑟 multiplied by sin 𝜃 sub 𝑟, where 𝑛 sub 𝑖 is the refractive index of the light ray’s initial material. 𝑛 sub 𝑟 is the refractive index of the light ray’s final material. 𝜃 sub 𝑖 is the angle of incident. And 𝜃 sub 𝑟 is the angle of refraction.

We want to calculate the refractive index of the plastic, which is the light ray’s final material. So we need to rearrange this equation to make 𝑛 sub 𝑟 the subject. We can do this by dividing both sides of the equation by sin 𝜃 sub 𝑟, which will leave us with 𝑛 sub 𝑟 equals 𝑛 sub 𝑖 sin 𝜃 sub 𝑖 over sin 𝜃 sub 𝑟.

Now, we are told in the question that the light ray’s initial material is air. So 𝑛 sub 𝑖 is the refractive index of air. We can recall that the refractive index of air is very close to one and can be approximated as equal to exactly one. So 𝑛 sub 𝑖 equals one. We also know that the angle of incidence is this angle here. So 𝜃 sub 𝑖 is equal to 45 degrees. And the angle of refraction is this angle here. So 𝜃 sub 𝑟 equals 33 degrees. Substituting these values into this equation and grabbing a calculator, we find that 𝑛 sub 𝑟 is equal to 1.298 and so on.

We need to give this answer to one decimal place. So, if we round this up, we find that 𝑛 sub 𝑟 is equal to 1.3. And we have arrived at the final answer. The refractive index of the plastic is 1.3.