Determine the maximum angle θfor which the light rays incident on the end of the pipe in Figure

Determine the maximum angle θfor which the light rays...

Determine the maximum angle $\theta$ for which the light rays incident on the end of the pipe in Figure $\mathrm{P} 35.36$ are subject to total internal reflection along the walls of the pipe. Assume the pipe has an index of refraction of 1.36 and the outside medium is air. Your answer defines the size of the cone of acceptance for the light pipe.

Key Concepts

Total Internal Reflection

Total internal reflection is the phenomenon where light traveling within a medium of higher refractive index is completely reflected at the boundary with a medium of lower refractive index, provided the light strikes the interface at an angle greater than a specific value known as the critical angle. This is the fundamental principle that enables light to be confined within optical fibers or light pipes.

Critical Angle

The critical angle is the minimum angle of incidence relative to the normal at which total internal reflection occurs when light travels from a denser medium to a rarer medium. It is calculated using the relation sin(?_c) = n2/n1, where n1 is the refractive index of the denser medium and n2 that of the rarer medium.

Refractive Index

The refractive index is a measure of how much light slows down when passing through a material compared to its speed in a vacuum. It determines the bending of light as it enters a new medium and plays a central role in defining both the critical angle and the conditions for total internal reflection.

Acceptance Cone

The acceptance cone refers to the range of incident angles at the entrance of a light pipe or optical fiber that will result in light propagating via total internal reflection. This cone is defined by a maximum acceptance angle, beyond which light will not be confined within the guide. The acceptance angle is directly linked to the numerical aperture of the system, which quantifies the light-gathering ability.

Transcript

00:01 So we're being asked to find what is the maximum angle theta for which light rays incident on the pipe are subject to total internal reflection.

00:17 So let's draw this pipe, right? and it's divided along the center here.

00:25 And we got some light rays incident on this end of the pipe at an angle theta.

00:32 So these, this light will be reflected and hit the wall of the pipe, right? so we're going to call this theta 1.

00:43 We're going to call this theta 2 as it is conventional.

00:47 And we're going to note that if we were to split this in half, we would not split this in half, but if we take into account the normal line to this surface, we are going to get a third angle right here.

01:03 We're going to call that phi.

01:05 And since this forms a triangle, we're going to note, this is a 90 degree triangle, that 90 minus theta 2 will give us, sorry, 90 minus phi will give us theta 2.

01:23 So, we need to take into account that the n of this cone, not this cone, but this cylinder, is 1 .36.

01:33 So now we know that.

01:37 And we are being asked about total internal reflection.

01:40 So there's nothing coming out of here.

01:41 So when that is the case, we have the equation sine theta c is equal to n2 over n1.

01:49 We have n1 right here because that is the index of what's inside the pipe...

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