Video Transcript
Light rays travel through a layer of kerosene floating on the surface of water that has a refractive index of 1.33. Light rays that are incident on the interface of kerosene and water at angles of 16.9 degrees from the surface or less are totally internally reflected. What is the refractive index of the kerosene? Give your answer to two decimal places.
Let’s say that this is our interface between kerosene and water. Light rays travel through the kerosene until they reach this boundary. And we’re told that as long as this angle here is less than or equal to 16.9 degrees, then the ray of light will be totally reflected from the water surface. The reason the ray of light is reflected here is that the index of refraction of kerosene here is different to the index of refraction of water here.
Let’s label the index of refraction of kerosene 𝑛 sub k and that of water 𝑛 sub w. In our problem statement, we’re told that 𝑛 sub w is 1.33. We don’t know, however, the index of refraction of kerosene. That’s what we want to solve for. To start doing this, let’s clear some space on screen and remember a law of optics called Snell’s law. This law says that if we have an interface between materials of different index of refraction, 𝑛 sub i and 𝑛 sub r, then the index of refraction of the material the ray of light originally travels through, 𝑛 sub i, multiplied by the angle of incidence, 𝜃 sub i, equals the refractive index of the material on the other side of the boundary, 𝑛 sub r, times the sin of 𝜃 sub r, the angle of refraction.
An important thing to note here with these angles, 𝜃 sub i and 𝜃 sub r, is that they’re defined with respect to the line normal to the interface. We bring this up because in our original sketch, the angle we have of 16.9 degrees is defined with respect to the interface surface itself. We wouldn’t say then that this is equal to the angle of incidence 𝜃 sub i as it appears in Snell’s law. That angle is actually defined like this. If we had a case where our incoming ray of light was exactly at the critical angle so that it reflects like this, then our actual angle of incidence would be 90 degrees minus 16.9 degrees. This angle, the angle of incidence, plus this angle equals 90 degrees together. Using the symbols in our Snell’s law equation, we can say that 𝜃 sub i is 90 degrees minus 16.9 degrees, or 73.1 degrees.
Therefore, if we apply Snell’s law to our scenario, 𝑛 sub i, the index of refraction of the material the light ray originally travels through, is 𝑛 sub k, the index of refraction of kerosene. 𝜃 sub i, the angle of incidence of this ray onto the interface, is 73.1 degrees. 𝑛 sub r, the index of refraction of the material the light ray is refracted into or reflected from, is 1.33. That’s the index of refraction of water. Lastly, because this angle of incidence equals the critical angle for this interface, the angle of refraction is 90 degrees. This is how we describe the reflected ray traveling along the kerosene–water interface.
We can recall now that in this entire expression, it’s 𝑛 sub k that we want to solve for. If we divide both sides of the equation by the sin of 73.1 degrees, then that factor cancels out on the left. The index of refraction of kerosene equals the index of refraction of water times the sin of 90 degrees divided by the sin of 73.1 degrees. Before we calculate this fraction, we can simplify it a bit further by recognizing that the sin of 90 degrees is equal to one. Our entire numerator then is 1.33. And calculating this fraction to two decimal places, we get 1.39. This is the index of refraction of the kerosene layer in this scenario.
