Question Video: Calculating the Angle of Deviation of a Ray of Light Passing through a Triangular Prism Physics

The diagram shows the path of a light ray passing through a triangular prism with a refractive index of 1.5 that is surrounded by air. The angle Φ₁ = 42° and the apex angle of the prism 𝐴 = 58°. Find the angle 𝛼 to the nearest degree.

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Video Transcript

The diagram shows the path of a light ray passing through a triangular prism with a refractive index of 1.5 that is surrounded by air. The angle Φ one equals 42 degrees and the apex angle of the prism 𝐴 equals 58 degrees. Find the angle 𝛼 to the nearest degree.

In our diagram, we see this angle 𝛼 indicated here. Physically, 𝛼 indicates the difference between the direction of the light ray originally and its final direction of travel after being refracted through the prism. Before we can calculate 𝛼, we’ll want to generate an equation for it in terms of our other given information. Our diagram also indicates the apex angle 𝐴 of the prism, the initial angle of incidence Φ one, and initial angle of refraction 𝜃 one, as well as the final angle of incidence Φ two and the final angle of refraction 𝜃 two of our ray.

We’re told that Φ one equals 42 degrees and the apex angle 𝐴 equals 58 degrees. We’re also told that the refractive index of our prism is 1.5 and that the prism is surrounded by air, which has an index of refraction of 1.0. Knowing all this, let’s clear some space on screen and work on writing the angle 𝛼 in terms of the other values in the scenario.

If we begin by focusing on this highlighted orange triangle, we can draw an expanded view of the triangle and start labeling its interior angles. Because this angle right here is 𝛼, that must mean that this angle is 180 degrees minus 𝛼. That’s because these two angles added together must equal 180 degrees. Knowing further that this angle here is Φ one and this angle here is 𝜃 one, we can identify this interior angle of our triangle as Φ one minus 𝜃 one. Finally, knowing that this angle here is 𝜃 two and this angle here is Φ two, we can conclude that this interior angle of our triangle is 𝜃 two minus Φ two.

Let’s now recall that for any three-sided object, if we take the interior angles of that triangle and we add them all together, we always get a value of 180 degrees. Applying that rule to the interior angles of this triangle, we can write out this equation. Here, Φ one minus 𝜃 one is our one angle, 180 degrees minus 𝛼 is another angle, and 𝜃 two minus Φ two is the third. Notice that 180 degrees appears on both sides of this equation. This means that if we subtract 180 degrees from both sides, this number will cancel out entirely.

As a next step for our resulting equation, if we add the angle 𝛼 to both sides, then 𝛼 minus 𝛼 on the left equals zero, giving us this expression, which, if we switch the right and left sides, tells us that the angle 𝛼 is equal to 𝜃 two minus Φ two plus Φ one minus 𝜃 one. Of all these values on the right-hand side of this equation, we know just one of them, Φ one. We don’t yet know enough then to solve for the angle 𝛼.

Let’s continue working in that direction by using the fact that we do know the apex angle 𝐴. Clearing away our expanded orange triangle, let’s now think about this pink quadrilateral in our prism. One of the interior angles of this four-sided shape is the apex angle 𝐴. Another interior angle here is actually a right angle. This is because this angle is defined by the surface of the prism and a line perpendicular to it. The same thing is true at the opposite corner of the quadrilateral. The last interior angle of this four-sided shape is this unlabeled angle here.

Whenever we have any sort of shape with four sides, if we add together the four interior angles, we get a result of 360 degrees. This is different to the result for a triangle. We can say then that 𝐴, the apex angle of this quadrilateral, plus 90 degrees, this other interior angle, plus 90 degrees for this interior angle added to our fourth unknown interior angle that we’ll leave blank for now is all equal to 360 degrees.

Now, 90 degrees plus 90 degrees is 180 degrees. Subtracting this angle from both sides of the equation, 180 degrees minus 180 degrees on the left equals zero. And on the right, we have a result of 180 degrees. Lastly, if we subtract the apex angle 𝐴 from both sides, canceling that angle on the left, we find that we have now solved for our unknown angle in the quadrilateral. It’s 180 degrees minus 𝐴.

With this in mind, let’s turn our attention to this triangle highlighted in green. An expanded view of this triangle looks like this, where the interior angles are 𝜃 one, Φ two, and 180 degrees minus 𝐴. Let’s now recall the rule for triangles that we used earlier. If we add together all three of its interior angles, then they add up to 180 degrees. Subtracting 180 degrees from both sides of the equation cancels this angle out entirely. And then, with the resulting equation, if we add the angle 𝐴 to both sides so that that angle cancels on the left, we arrive at an expression for the apex angle 𝐴 in terms of 𝜃 one and Φ two.

Notice that in our equation for 𝛼, both 𝜃 one and Φ two appear. In fact, we could group the quantities on the right-hand side of this expression so that they read 𝜃 two plus Φ one minus the quantity 𝜃 one plus Φ two. We’ve just seen that the apex angle 𝐴 is equal to 𝜃 one plus Φ two. So if we make this substitution, we arrive at an expression for 𝛼 in terms of the variables 𝜃 two, Φ one, and 𝐴. This is certainly progress because, as we’ve seen, we’re given Φ one and 𝐴.

To solve for 𝛼, the last value we’ll need to calculate is 𝜃 two. Let’s store our equation for 𝛼 off to the side for now, clear some space on screen, and consider how we can solve for this second angle of refraction 𝜃 two. To help us do this, we can use a law of optics known as Snell’s law. This law tells us that when a ray of light is incident on a boundary between two optically different materials, then the index of refraction of the material the ray is originally in, 𝑛 sub 𝑖, multiplied by the sin of the angle of incidence, 𝜃 sub 𝑖, is equal to the index of refraction of the material the ray crosses into, 𝑛 sub 𝑟, multiplied by the sin of the angle of refraction, 𝜃 sub 𝑟.

Notice that in our diagram of the prism 𝜃 sub two, the angle we want to solve for, is an angle of refraction of this light ray. If we were to focus on this interaction of the light ray, where it leaves the prism and enters the air, we could write that 1.5, the index of refraction of the prism, times the sin of Φ sub two, the incident angle for this case, is equal by Snell’s law to the index of refraction of air, 1.0, times the sin of 𝜃 sub two, the angle of refraction of this ray. Knowing that it’s 𝜃 sub two that we want to solve for in this equation, we can’t quite do it because we don’t know Φ sub two.

But returning to our green triangle, we could solve for Φ sub two if we knew both the angle 𝜃 sub one and this angle here. And in fact, we do know the second angle. We solved for it earlier; it’s 180 degrees minus 𝐴. Therefore, if we could solve for 𝜃 sub one, we could add that angle to this angle here and use that sum to solve for Φ sub two.

The only way to solve for 𝜃 one though is actually to use another application of Snell’s law. Let’s now shift our focus to this interaction of the light ray as it enters the prism from the air. In this case, we have 1.0, the index of refraction of the air that the ray is originally in, multiplied by the sin of the original angle of incidence, Φ one. Snell’s law says this is equal to the index of refraction of the prism, 1.5, times the sin of the angle of refraction, 𝜃 one. Here, it’s 𝜃 one we want to solve for. And in this case, we do actually know this angle Φ one. That angle, we’re told, is 42 degrees.

So let’s do this. Let’s divide both sides of this equation by 1.5, canceling that factor on the right. Then, with the remaining expression, we can isolate this angle 𝜃 one by taking the inverse sine of both sides of the equation. On the right, the inverse sine of the sine of an angle is simply equal to that angle itself. That is, we now have an expression where 𝜃 one is the subject. 𝜃 one equals the inverse sin of 1.0 times the sin of Φ one divided by 1.5. And now, we can use the fact that the angle Φ one is 42 degrees.

Making this substitution, when we calculate 𝜃 one, we get an approximate result of 26.49 degrees. Storing this result off to the side and clearing a bit of space, let’s now turn our attention back to this green triangle and the interior angles of the triangle. Those interior angles, as we’ve seen, are 𝜃 one, 180 degrees minus 𝐴, and Φ two. According to our rule for triangles, when we add these three angles together, we must get 180 degrees. If we subtract 180 degrees from both sides of the equation, then that angle measure cancels out completely. And we get this expression, where note that we know 𝜃 one and we also know the angle 𝐴.

If we add 𝐴 and subtract 𝜃 one from both sides of this equation, then on the left-hand side, those two terms cancel out. And we find that Φ two equals 𝐴 minus 𝜃 one. 𝐴 is 58 degrees, and 𝜃 one is about 26.49 degrees, giving us an angle of 31.51 degrees for Φ two. We can substitute this angle in for Φ two in our original Snell’s law equation. Notice that now we know everything in this equation except for the angle 𝜃 two. That’s the angle we want to solve for because if we know it, then we’ll know the last unknown value in our equation for 𝛼, which is what we really want to know.

Looking at the right-hand side of this expression, notice that we’re multiplying the sin of 𝜃 two by 1.0, which doesn’t change that value at all, so we can leave it out. From there, if we take the inverse sine of both sides of the equation, then the inverse sin of the sin of 𝜃 two equals just 𝜃 two by itself. 𝜃 two then equals the inverse sin of 1.5 times the sin of 31.51 degrees. Calculating this angle, we get 51.63 degrees.

Returning to our equation for the angle 𝛼, we now know 𝜃 two, Φ one, and the angle 𝐴. Substituting in these values, 𝛼 equals 51.63 degrees plus 42 degrees minus 58 degrees. This equals 35.63 degrees, or, rounded to the nearest degree, 36 degrees. This is the angle of deviation of the ray of light to the nearest degree due to the fact that it was refracted.

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