Question Video: Solving for the Difference between Angles of Incidence of Two Light Rays Physics

Video Transcript

Light rays follow the paths shown in the diagram. Find the difference between the angles 𝜃 one and 𝜃 two. Answer to the nearest degree.

We can see these two angles, 𝜃 one and 𝜃 two, representing the angles between rays of light. Here’s one ray represented by the dotted line, and the other ray represented by a solid line and these two dashed lines that are normal or perpendicular to the surface. Our question asks us to solve for the difference between these angles. And we see that they described rays that move upward through the substance, called substance one, and then meet at the interface between substance one and substance two at the same point.

From there, the solid black line representing the ray associated with the angle 𝜃 two travels horizontally along the boundary between substance one and substance two. The dotted black line, though, representing the ray associated with the angle 𝜃 one refracts or bends as it crosses into substance two. And then when it reaches the interface between substance two and substance three here is refracted at an angle of 90 degrees, so that it travels along the boundary between these substances.

Note that we’re told the index of refraction for each of our three substances. We notice that these values are all different from one another. This explains why when the rays of light reach the boundaries between these substances, they refract. This refraction occurs in a way described by Snell’s law. This law says that if we have an interface between two materials of different index of refraction, then if a ray of light is incident on the interface at an angle of incidence 𝜃 sub 𝑖, it will be refracted at an angle 𝜃 sub 𝑟 that is consistent with this equation: the index of refraction in the material the ray originally traveled through multiplied by the sine of the angle of incidence 𝜃 sub 𝑖 is equal to the index of refraction in the material the ray traveled into multiplied by the sine of the angle of refraction 𝜃 sub 𝑟.

Note that these angles 𝜃 sub 𝑖 and 𝜃 sub 𝑟 are measured with respect to a line that is normal or perpendicular to the interface. By applying Snell’s law to our situation, we can solve for the difference between the angles 𝜃 one and 𝜃 two. Let’s start by considering 𝜃 two here. We’ve seen that this angle is associated with the solid black ray. If we go to the point where this ray of light interacts with substance two and draw a normal line in pink, we can say that this angle here between that normal line and the ray is the alternate interior angle of this angle here, 𝜃 two. In other words, this angle in pink is 𝜃 two. And notice that because it’s defined with respect to a line that is perpendicular to the interface between our substances, it’s also the angle of incidence of our solid black ray as it reaches the interface between substances one and two.

Applying Snell’s law to the interaction of a solid black ray at this interface then, we can say that 𝜃 sub 𝑖 is equal to 𝜃 two. We can also say that 𝑛 sub 𝑖, the index of refraction in the material that ray is originally in, is 1.50. So then, we have 1.50 times the sin of 𝜃 two being equal to the index of refraction of substance two, that’s 1.33, times the sine of the angle of refraction of this ray. As we’ve seen, when this ray of light is refracted, it travels along the interface between these two substances. In other words, relative to a line at normal to this interface, it has an angle of refraction of 90 degrees. But then if we think about the sine of the angle 90 degrees, we know that the sine of 90 degrees is exactly one. Therefore, we can leave out this part of our expression.

So, dividing both sides of his equation by 1.50, causing that factor to cancel on the left, we have the sin of 𝜃 two being equal to 1.33 divided by 1.50. And if we then take the inverse sine of both sides of the equation, the inverse sine of the sin of 𝜃 sub two is equal just to 𝜃 sub two. Now that we have an expression for 𝜃 two, let’s store it off to the side and work on coming up with a similar expression for 𝜃 one.

Just as we found to be the case for our first ray, if we consider the second ray, the one indicated by the dotted black line, when it reaches the boundary between substances one and substance two, the angle of incidence of that ray, indicated here, is the alternate interior angle to 𝜃 one. Therefore, its angle of incidence at this boundary is 𝜃 one. If we were to take an up close view of the dotted black ray as it reaches this boundary, we would see that while we know the angle of incidence, 𝜃 one, and the index of refraction of the material the ray originally travels in, 1.50, as well as the index of refraction of the material the ray refracts in two, that’s 1.33, we don’t know this angle right here, the angle of refraction. For now, we’ll just call this angle 𝜃 sub 𝑟.

This leaves us then with an incomplete Snell’s law equation. We want to solve for 𝜃 one. But not knowing 𝜃 sub 𝑟, we’re not quite able to do that. Continuing on with this dotted black ray, let’s consider what happens when it reaches this boundary between substances two and three. Once again, looking at a zoomed-in view of this interaction, we know that a ray comes in at some angle of incidence and then refracts at an angle of 90 degrees so that the ray travels along the interface. For this interaction, the index of refraction of the material the ray is originally in is 1.33.

And then if we consider the angle of incidence of this ray, notice that if we follow the ray on our original diagram, that this angle of incidence will be the same as the angle of refraction when the ray crossed from substance one into substance two, that is, the angle of incidence will be what we’ve called 𝜃 sub 𝑟, the angle of refraction from the previous interaction. 1.33 times the sin of 𝜃 sub 𝑟 is equal to the index of refraction of the material the ray is refracted by 1.00 times the sine of the angle of refraction, which is 90 degrees. And as we saw earlier, the sine of 90 degrees is one.

Our equation then simplifies to this expression. And recall that this expression applies to the interaction of a ray of light between substances two and three. If we divide both sides of this equation by 1.33, cancelling out that factor on the left, we get an expression for the sin of our angle of refraction 𝜃 sub 𝑟. And now look at this; we can substitute this fraction 1.00 divided by 1.33 in for the sin of 𝜃 sub 𝑟 in our equation above. And when we do this, note that we have 1.33 in the numerator and denominator of the right side of this expression, so they cancel out. So taking this simplified equation, dividing both sides by 1.50, canceling that factor out on the left, and then with the resulting expression, taking the inverse sine of both sides, the left side of this equation will simplify to 𝜃 one. We find that 𝜃 one equals the inverse sin of 1.00 divided by 1.50.

Now that we have expressions for 𝜃 one and 𝜃 two, we can move ahead with solving for the difference between them. Clearing some space at the bottom of our screen, we can write that 𝜃 two the larger of the two angles minus 𝜃 one is equal to the inverse sin of 1.33 divided by 1.50 minus the inverse sin of 1.00 divided by 1.50. Calculating this expression to the nearest degree gives us an answer of 21 degrees. This is the difference in degrees between the angles 𝜃 one and 𝜃 two.