Optical fibers are constructed with a cylindrical core...

Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica $n_2 = 1.4502$ for the cladding and silica doped with germanium $n_1 = 1.4652$ for the core. (a) What is the critical angle $\theta_{crit}$ for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence $theta_i$ at the flat end of the cable for which light is incident on the core-cladding interface at angle $\theta_{crit}$ ($\textbf{Fig. P33.46}$). Show that sin $\theta_i$ =$ \sqrt {n^2_1 - n^2_2}$ . (c) What is the value of $\theta_i$ for $n_1$ = 1.465 and $n_2$ = 1.450?

Transcript

0:00 This question.

00:01 It is about how light travels in, uh, the optical fiber.

00:07 So first, um, in an optical fiber, there's a call.

00:12 And outside that there's a cladding.

00:14 So we want to find a critical angle for total internal reflection to take place at the core cladding in the face.

00:22 Then we need to find, uh, uh, expression for incident angle sign of the incident angle in terms of the credit, effective in diseases of the core and cladding.

00:37 Okay.

00:37 And then lastly, calculate that critical angle if you are given the reflective in this is okay, so, um, this question is about total internal reflection.

00:50 Mm.

00:51 It's now it's low.

00:52 Okay, so you want to find a critical angle in the car cladding interface, so we'll be using a sales law.

01:06 Okay, so now so i say that and one yes, i did.

01:10 One equals two and two scientists to eight.

01:14 Where one is the call.

01:17 Uh, two is the cutting pain.

01:21 And then, uh, the other one is our critical angle data to is, uh, 90 degrees.

01:29 Okay, we are also given that, uh, that one is 1.465 and two is 1.45 p.

01:41 So, uh, substitute everything.

01:44 We have 1.465 sigh data, see, basically to 1.45 sign 90 degrees.

01:53 Okay.

01:54 So here you can calculate our critical angle.

01:58 Yes, i didn't see is 1.45 do i buy 1.465 so to see, yes, 81.8 degrees.

02:08 Okay.

02:12 Okay, then in part b.

02:14 Um, i want to show that the science theater i which is the incident angle, uh, on this part of this diagram, ok, is equal to square and one square minus and two square.

02:30 Okay, so, uh, that's just redraw the diagram.

02:50 Okay, so we have the light ray entering, uh, pick of five.

02:54 Oh, you know, something like this.

02:59 Okay, the eye...

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