WEBVTT

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This is College Physics Answers
with Shaun Dychko.

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We have an optical fiber
made out of crown glass

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and it's surrounded by air.

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So that means <i>n one</i> is 1.00

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and the index of refraction
of the fiber is <i>1.52</i>.

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And we're going to show that no matter
what the angle of incidence is here,

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<i>theta one</i>,

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there will always be
total internal reflection

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of any ray inside this fiber.

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So, we're going to figure out

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what is the minimum <i>theta
three</i> that's possible.

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And then, we'll also calculate
the critical angle

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for the crown glass-air interface

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and show that this <i>theta three</i>

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is always going to be greater than
the critical angle.

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And since that's the case,

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there will always be total
internal reflection.

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So, let's find the smallest <i>theta
three</i> that's possible though.

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Because the smaller this
angle of incidence here,

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the better chance it has of going
through the interface.

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So, there is an angle small enough

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such that this ray will go through,

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but we'll show that <i>theta three</i>
will never get that small.

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Okay.

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So, <i>theta three</i> plus <i>theta two,</i>

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this angle here,

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add up to 90 because they're
part of this triangle here,

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which is a right triangle.

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And, you know, the total angles
inside a triangle is 180.

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And then this ones 90 though,

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so that means 90's left over for <i>theta
two</i> and <i>theta three</i>.

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In which case, <i>theta three</i> is 90
minus <i>theta two</i>

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when you subtract <i>theta
two</i> from both sides.

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Now we want to get the minimum
possible <i>theta three</i>,

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and that will occur

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when you have the maximum
possible <i>theta two</i>.

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So, let's figure out what the maximum
possible <i>theta two</i> is.

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Now, we're going to turn our attention
to this interface here,

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and we'll use Snell's law

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and show what the maximum
<i>theta two</i> is

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based on the maximum <i>theta one</i>.

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And we can control <i>theta one</i>;

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that's when it's maximum,
it's gonna be 90 degrees.

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I mean, strictly speaking, at 90 degrees,

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it's not going to enter the fiber at all,

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but, you know, that's the border between
when it does and doesn't.

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So, we're gonna take <i>theta one</i>
to be 90 degrees...

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and... which we do down here.

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And we're gonna use Snell's law to figure out

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what <i>theta two</i> will be at its maximum.

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So, Snell's law says

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the index of refraction of the first medium,

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which is air,

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times sine of the angle of incidence
<i>theta one</i>,

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equals <i>n two sine theta two</i>.

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And we'll solve for <i>theta two</i> by
dividing both sides by <i>n two</i>.

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And then, take the inverse
sine of both sides.

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So <i>theta two</i> is the inverse sine of
<i>n one sin theta one over n two</i>.

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Now, <i>theta two</i> maximum is gonna be

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the inverse sine of <i>n one times
sine theta one</i> maximum.

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And that's where I talked about having
<i>theta one</i> being 90 degrees.

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And then we plug in numbers here.

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So it's the inverse sine of
index of refraction of air,

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times the maximum possible angle
of incidence from the air,

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which is 90 degrees,

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divided by the index of refraction
of the crown glass,

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which is 1.52, giving a maximum
<i>theta two</i> 41.1395 degrees.

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So that means <i>theta three</i>
minimum is 90 minus that

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<i>theta two</i> maximum,
which is 48.86 degrees.

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So now, we're going to see how this
compares with the critical angle.

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Now, we need to consider
this interface here now

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and we have the index of refraction
of the medium we're starting in,

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which is crown glass, times
sine of the critical angle,

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equals the index of refraction
that we're going to,

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which is air,

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times sine of the angle of refraction
at the critical angle.

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So when we have whatever the
critical angle is here,

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the angle of refraction is going to be 90.

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So we can solve for <i>theta c</i> then.

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It's the inverse sine of the second medium,

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which is air,

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which has an index of refraction of one,

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times sine of 90, divided by 1.52,
which is 41.1395 degrees.

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And we can see that this maximum...

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or this minimum, sorry, possible
<i>theta three</i>

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is always going to be greater
than this number.

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So, because <i>theta three</i> will always
be more than the critical angle,

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there will always be total
internal reflection.