### Video Transcript

In this video, our topic is the deviation that a ray of light experiences when it passes through a prism. Through some careful geometric analysis, were going to quantify the angle, weve symbolized it πΌ in this sketch, by which array passing through a prism is deviated. In that process, weβll see what the angle πΌ depends on. And well also see how to make it as small as possible.

To get started, say that we have a prism and that this prism has an angle at its apex. Weβll call it the apex angle, capital π΄. Along with this, lets say that this prism has an index of refraction that weβll call π. And well say further that this prism is surrounded by air which has an index of refraction of one. And weβll also say that π is greater than one. This is usually the case with prisms. This scenario means that if a ray of light is incident on one of the faces of the prism like this, rather than passing straight on through, this ray will actually be refracted or bent at this interface.

If we draw a line thats perpendicular to the face of the prism where the ray meets it, we know that rather than continuing in its current direction, the ray of light will instead be refracted in the direction of this normal line. But then when the ray traveling through the prism reaches this face, once more, it will be refracted. But this time, since its moving from a material of higher index of refraction to lower, it will be refracted farther away from the normal line, in a direction that might look like this. What we have then is this ray of light whose path is being deviated thanks to these two refractions.

In this lesson, we want to quantify just how much deviation occurs. We do that by identifying an angle in this diagram. The angle was measured between the path the ray was following before it reached the prism and the pathway the ray ends up traveling because of these two refractions. So, if this dashed line here is the path the ray was traveling and this solid line here is the path it actually travels after passing through the prism, the angle weβre interested in solving for is this one right here. On our diagram, thats this angle.

Just like we did on our opening sketch, weβll name this angle πΌ. And it turns out that we can solve for this angle in terms of the apex angle π΄ of our prism as well as a few of the angles involved as a ray moves through the prism. To begin solving for πΌ, lets think about both refractions of our ray of light which occur at the faces of the prism and label the angle of incidence as well as the angle of refraction for each one.

We start with a ray of light traveling through air and then being incident on this face of the prism. This happens at a certain angle called the angle of incidence. And we measure that angle with respect to the line drawn normal to the surface where the ray meets it. So lets call this angle π, and well add a subscript one to it. After this ray passes into the prism, its refracted. And that refraction angle is again measured with respect to the normal line to the surface where the ray meets it.

Lets label this angle π. And again, well give it a subscript one. The subscript one on both our angle of incidence and angle of refraction are to indicate that this is for the first refraction of a ray of light. We know, though, that theres a second one which occurs at this other face of the prism. For this second refraction, theres also an angle of incidence. That would be this angle right here. And we can label that π sub two. And in addition, theres a second angle of refraction. Again, thats measured from the normal line, this dashed blue line, to the final direction of the ray. And well label this angle π sub two.

Now, the point in labeling all these angles on this diagram is so that we can express the deviation angle πΌ in terms of some of them. To move ahead in doing this, lets look more closely at some of these paths within our prism. If we zoom in a bit, we can say that this dashed line is the path that our ray of light wouldve traveled through the prism if it wasnt refracted or bent at all. And then this dotted line here shows the line along which our ray ends up traveling after it passes through the prism. So weve said that the angle between these two lines, called πΌ, is the angular deviation that our beam experiences.

Well, if that angle is πΌ, then that must mean that this angle here is equal to 180 degrees minus πΌ. And this is actually an interior angle in a triangle we can draw. If we say that this line segment here is the path our ray of light actually travels as it passes through the prism, then our dashed line can be extended to meet up with that line. And its worth noting that the place where these two lines meet is at the face of the prism. And, as well, the point where our dotted line meets with this ray also occurs at an interface between the prism and its surroundings.

So now, we can start to see a triangle. Heres one side, hereβs a second side, and heres the third side of it. Were going to look carefully into the angles of this triangle because we already see that one of them involves the angle we ultimately want to solve for, πΌ. So then lets consider what the other two interior angles are. First, we can look into this angle right here. Based on the angles that weve already labeled above, the angles of incidence and the angles of refraction, once we have our dashed normal line drawn in, we can say that this angle right here, that original angle of refraction, is π sub one. And if we draw our ray incident on that first interface this way, then we can say that this angle, right here, is π sub one.

Now, with the angle of incidence π sub one defined on this side of our prismβair interface, we can see how this angle could be represented on the other side of the interface too. If we drew that angle in, it would be equal to this angular extent. So this angle also is π sub one. And the reason its helpful to know this is because knowing π sub one and π sub one, we can define this interior angle of our triangle. That angle was equal to π sub one minus π sub one. Knowing this, letβs now use a similar process to determine this interior angle.

Once again, we have our normal line. And we saw earlier that this angle here, what we call the second angle of incidence, is represented by π sub two and that if our ray finally bends like this as it leaves the prism, then we saw that this angle here is the second angle of refraction. Recall we name that π sub two. If we consider this angle π sub two on this side of our interface, it helps us identify that same angle on the other side. π sub two is equal to this angle right here. And this tells us that the interior angle of our triangle that we want to solve for can be expressed as π sub two minus π sub two.

So then this triangle weβre focusing on has these three interior angles and were taking such care with them so that we can solve for this angle πΌ. Thats our goal. At this point, we can recall a fact about the interior angles of all triangles. Given a triangle with interior angles π΄, π΅, and πΆ, its always the case that the sum of those angles is 180 degrees. So that means if we add up all three of the angles weve determined, their sum must equal 180 degrees. And now notice that 180 degrees appears on both sides of this equation. So if we subtract 180 degrees from both sides, then this term will cancel out. And well end up with this equation here.

Now, if we add the angle πΌ to both sides of this equation, then we wind up with this expression. And we see that we now have an equation for the angular deviation of our ray πΌ in terms of the angles of incidence of our beam and its angles of refraction. Weve come a long way, but theres one more step we want to take in finding an expression for πΌ. We mentioned that weβd like to express this angle in terms of the apex angle π΄ of our prism. There is a way to do that, and it involves, once again, looking carefully at the geometry of our situation. Lets clear away our zoomed-in view of our triangle, and now were going to look at a different shape.

Lets look more closely at this quadrilateral here, highlighted in orange. Now, if we take a zoomed-in view of this four-sided shape, we know the apex angle π΄ is at the top. And we also know that this angle here, as well as this angle here, are right angles; theyβre 90 degrees. We know that because these two sides of our quadrilateral represent faces of the prism, while these two dashed lines represent normal lines to those faces. So then, by definition, these interior angles must be 90 degrees. Now, just as we recall the rule for the sum of the interior angles of a triangle, we can do the same thing with a quadrilateral.

Say that we have a four-sided shape and the interior angles of that shape are π, π, π, and π. If we add up these angles, their sum is 360 degrees. This is true for all quadrilaterals. And so its also true for our quadrilateral over here. Now, if the sum of all four of these angles is 360 degrees and this angle here and this angle here are both 90 degrees, then that means we can say that the apex angle, π΄, if we add that angle to this one here in our quadrilateral, then their sum must be equal to 180 degrees. Thats because all four together need to add up to 360. And these two already account for 180 of those 360.

So to help ourselves out, lets give this angle a name. Weve already had an angle πΌ. So why dont we call this one π½? From what weve learned so far, we can say that the apex angle π΄ plus π½ is 180 degrees. But remember that were working toward a goal of expressing our angle of deviation πΌ in terms of our apex angle π΄. So really, wed like to express π΄ in terms of some of the angles of incidence and angles of refraction by which πΌ is already expressed. In other words, what we really like to do is express π½ in terms of our already-defined angles of incidence, π sub one and π sub two, and/or the angles of refraction, π sub one and π sub two. So lets see how we can do that.

If we return to our quadrilateral, we know that these two angles right here are actually joined by the ray of light as it passes through the prism. That ray looks like this in our diagram above. This shows us that this angle π½ is not only a part of the quadrilateral weve been talking of, but its also an interior angle to this triangle right here, defined by this side, this side, and this side. Perhaps using this fact, we will be able to express π½ in terms of other variables. Now, the nice thing about this triangle with the sides highlighted in pink is that weve already labeled the other two interior angles in it, besides π½.

If we look up at our full-scale sketch, we can see that this interior angle here is the first angle of refraction of our ray. We call that π sub one. And then this angle right here is the second angle of incidence π sub two. So then we have names for all three of the interior angles. And if we refer once more to this rule for all triangles, then we can say that π sub one plus π sub two plus π½ equals 180 degrees. Recalling that we want an expression for this angle π½, lets subtract the angles π sub one and π sub two from both sides of this equation which leaves us with this result.

This is a very good thing because, just like we wanted to do, weve now expressed this angle π½ in terms of, in this case, the original angle of refraction and the second angle of incidence. So now, lets take this whole expression for π½ and substitute it into our equation that reads π΄ plus π½ equals 180 degrees. And having done that, notice that once again we have 180 degrees appearing on both sides of this equation, which means that term can cancel out from both sides. And now, and were getting close, we can add π sub one and π sub two to both sides of the equation.

And when we do that, we find this wonderful result. The apex angle of our prism, the angle that we could say defines the prism, is equal to the first angle of refraction plus the second angle of incidence of our ray. And notice that in our equation for πΌ, both π sub one and π sub two appear. In fact, we can group them together. If we bring the angle π sub two over here with π sub one and the angle π sub one over here with π sub two, then we end up with this expression on the left-hand side of this equation.

And notice that this term right here, π sub one plus π sub two, is equal to the apex angle π΄. So we can replace whats in parentheses with this term π΄. And when weve done that, we finally have an expression for πΌ in terms of this apex angle. The great thing about this result is that its generally true that this angle of deviation is equal to the initial angle of incidence plus the second angle of refraction minus the apex angle of the prism the ray passes through.

Now, for a given prism, π΄ is a fixed value. And assuming that the index of refraction π of the prism is also fixed, then that means once we know π as well as the original angle of incidence of our ray, we could determine the second angle of refraction. All this to say, in this equation for πΌ, there are really only two variables, the angle of incidence π sub one and the angle πΌ. So we could think about varying this angle of incidence of our ray and seeing how it affects how much the beam is deviated. We could imagine plotting one of these variables against the other. We could vary the angle of incidence of our ray and see how it affects πΌ.

When we do this, a curve looking something like this typically results. Notice that there is an angle of incidence, we can call it π sub zero, which minimizes the angular deviation experienced by the beam. So we call that angular deviation πΌ sub zero. Now, if we were to draw a ray on our diagram up here, coming in at an angle of incidence, π sub zero, that ray could look like this. A hallmark of a ray deviated the least amount is that as it passes through the prism, it moves parallel to the bottom face of the prism. In addition to this, the original angle of incidence of the ray here and the final angle of refraction of the ray here are equal.

Now, whereas before we call these angles π sub one and π sub two, respectively, now, in recognition of the fact that they correspond to the minimum angular deviation of our beam, well label them π sub zero and π sub zero. And as we said, theyre equal in this case. This means that we can write a special case of this formula here for the angular deviation of our beam. We can write that the minimum angular deviation is equal to the original angle of incidence π sub zero plus the final angle of refraction π sub zero minus the apex angle π΄. But now, remembering that these two angles, π sub zero and π sub zero, are equal, we could also write them as two times π sub zero.

And now, given this expression, we can rearrange to solve for π sub zero. Its equal to πΌ sub zero plus π΄ all divided by two. Along with these two relationships weve generated so far, we can also write a relationship for the index of refraction π of our prism. We can do this by relying on Snells law. This law applies whenever a ray of light is crossing over from a material of one index of refraction to a material of another. In that case, given an angle of incidence, π sub i, and an angle of refraction, π sub r, Snells law says that π one times the sin of the angle of incidence equals π two times the sin of the angle of refraction.

Now, in the case of our prism, which has an index of refraction π and is surrounded by a material air with index of refraction one, we can apply Snells law by writing one times the sin of π sub zero equals π times the sin of π, where π is both the first angle of refraction and also the second angle of incidence. If we used a similar quadrilateral technique from earlier, we would find that two times this angle π is equal to π΄. Or, in other words, π is equal to π΄ over two, which we can substitute in here for π.

In a similar way, we can replace π sub zero with this expression here. And then well divide the resulting equation both sides by the sin of π΄ over two. And then we get this nice expression for the prisms index of refraction entirely in terms of the minimum angle of deviation and the apex angle of the prism. So this is another useful relationship. And now lets consider one last one of these. Working up at our sketch, what if the apex angle π΄ were very small, that is, if our prism looked like this? When we have a small angle, like π΄ is small here, then its generally the case that the sin of that small angle is approximately equal to the angle itself.

And we can apply this approximation to our equation here for π. We can say that for a small aspect angle π΄ β β in other words, for a thin prism β β we can say that π is equal to πΌ sub zero plus π΄ over two divided by π΄ over two. But then, in this overall fraction, the factors of one-half both can cancel out. And so we find that for a thin prism, the index of refraction of that prism is equal to the minimum angle of deviation πΌ sub zero plus the apex angle of the prism π΄ all divided by that apex angle.β

Lets now summarize this lesson by writing out these four equations. In this lesson, we first derived an equation for the angle of deviation of a ray of light passing through a prism. When that deviation angle reaches its minimum, we were also able to develop an expression for the angle of incidence corresponding to that minimum deviation. We also developed a relationship for the index of refraction of our prism. And lastly, we saw that when the apex angle π΄ is small β in other words, our prism is thin β that index of refraction of the prism equals the minimum angle of deviation plus the apex angle all divided by that apex angle.