Question Video: Critical Angle for Total Internal Reflection | Nagwa Question Video: Critical Angle for Total Internal Reflection | Nagwa

# Question Video: Critical Angle for Total Internal Reflection Physics • Second Year of Secondary School

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Which of the following formulas correctly shows the relation between the critical angle for total internal reflection π_π for a light ray, the refractive index π_π of the substance the light is propagating in, and the refractive index π_π of the substance when the light is reflected from its surface? [A] sin π_π = π_π/π_π [B] sin π_π = π_π/π_π [C] sin π_π = sin(π_π/π_π) [D] π_π = π_π/π_π [E] π_π = sin(π_π/π_π)

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### Video Transcript

Which of the following formulas correctly shows the relation between the critical angle for total internal reflection π sub π for a light ray, the refractive index π sub π of the substance the light is propagating in, and the refractive index π sub π of the substance when the light is reflected from its surface? (a) sin of π sub π is equal to π sub π divided by π sub π. (b) sin of π sub π is equal to π sub π divided by π sub π. (c) sin of π sub π is equal to the sin of π sub π over π sub π. (d) π sub π is equal to π sub π over π sub π. (e) π sub π is equal to sin of π sub π over π sub π.

Okay, in all these answer options, we see possible relations between these three variables. π sub π, the critical angle of incidence; π sub π, the index of refraction of the substance a ray of light starts in; and π sub π, the refractive index of the substance that the ray of light reflects off of.

We can begin by making a sketch of whatβs taking place here physically. Weβre told that we have a surface. And letβs say that this is our surface. And itβs there because on one side of it, say above it, the index of refraction of that material is π sub π. While below this boundary, the index of refraction of that material is called π sub π.

Our problem statement tells us that π sub π is the refractive index of the substance that a ray of light is initially propagating in. So letβs draw that ray of light like this. And then hereβs what we know. If the angle of incidence of this incoming ray of light is π sub π, the critical angle, then in that case when this ray reaches the surface, the boundary between these two materials, it will travel along that surface. In other words, it will neither refract into this material nor reflect back into this one. Rather, it moves right along the boundary between the two. If we consider the boundary that a ray of light moves along and the line normal to that boundary, then we can see that this ray is at 90 degrees to that normal line.

Now, even though this ray of light is in some sense not refracting, in this limiting case, we can nonetheless say that the angle of refraction is equal to 90 degrees. Indeed, this is the condition for the angle of incidence to be the critical angle π sub π. So thatβs whatβs taking place in this situation. We have these two refractive indices and a light ray approaching the boundary between these materials at the critical angle.

Knowing all this though, we still need to choose from among these five candidates for the correct mathematical relationship between these three variables. Hereβs how we can get started doing that. We can recall an optical law called Snellβs law. This law says that if we have a ray of light here incident on a boundary between two optically different materials. Then the angle of incidence π sub π of that ray and the angle of refraction π sub π, along with the indices of refraction of the two materials, are related according to this equation. π sub π times the sine of the angle of incidence is equal to π sub π times the sine of the angle of refraction.

Now, this law is generally true, but itβs also helpful to us in the specific case when the angle of incidence is equal to the critical angle π sub π. When that happens, when π sub π is equal to π sub π, then we can see from our diagram here that π sub π is equal to 90 degrees. So we have these two substitutions we can make into the variables in Snellβs law. If we write a version of Snellβs law using these values, then looking at the right-hand side, we see we can do a bit of simplification. The sin of 90 degrees is equal to one, which means we can write the equation this way.

And next, if we divide both sides of this equation by π sub π, the refractive index of the substance the light is propagating in, then π sub π cancels on the left-hand side. And that leaves us with this expression. The sine of the critical angle π sub π is equal to π sub π divided by π sub π. We can now take this expression and compare it against our five answer options. And we see right away that itβs a match for option (a). So by using Snellβs law and then applying the specific conditions implied by a critical angle of incidence, we found that the sin of π sub π is equal to π sub π divided by π sub π.

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