WEBVTT
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This is College Physics
Answers with Shaun Dychko.
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Two glass slides are separated
at one end by a piece of hair
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that is 0.1 millimeters in diameter.
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And so the distance here is 0.1 millimeters.
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Then the pieces of glass
are touching at this end.
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So this forms a triangular wedge of
air between the pieces of glass.
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Now light incident here will interfere.
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It'll one Ray will bounce off of
this interface between glass and air
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and then go upwards.
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And then the another Ray will bounce
off the bottom piece of glass
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and then go back up.
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And this drawing is not quite accurate
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because it says that the light
is incident perpendicularly.
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So all these rays are in fact overlapping
but it would be too difficult to draw that
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and explain where the different
rays are coming from.
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And so that's why it's showing an angle.
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But keep in mind that it's
meant to be overlapping
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so rays one and two will interfere.
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And in this case because
we're looking for dark bands
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the interference is going to be destructive.
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And then at some position later further
down here some distance *delta x*
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you'll have the next dark
band occurring when
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there is destructive interference again
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as a result of this increased
path length difference.
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So let's figure out expressions for
the phase shifts of Rays one and two
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and then and then we'll see how that
changes with horizontal position.
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Now the phase shift of Ray one is zero
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because this reflection
here occurs at an interface
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where you're beginning in a medium with high
index refraction and some kind of glass
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and then going to a medium of low
index refraction which is air.
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And so there is no phase shift there.
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And then for Ray two however
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there's a reflection off of this interface
between the air in the glass
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and this does have a phase shift
automatically of half of a wavelength
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when going from a low index
refraction to a high index refraction
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and then additionally
there's a phase shift
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due to this additional path
length travelled by ray two.
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And so we calculate the number of
wavelengths that path length corresponds to
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and then multiply by the wavelength.
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And this works well these are
actually going to be the same
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because normally I have a subscript *n* here
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to say you know the additional path
divided by the wavelength in that medium.
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But in this case the medium is air.
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So these cancel and we have a two there
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because the ray is going down
once over that thickness
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and then up again over the
thickness it's doing a round trip
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traveling that thickness twice the
thickness of that air gap there.
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And this works out to *Pi* over two plus
two times the thickness of the air gap.
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So now we have phase shifts
for each of the two rays.
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Now the total phase shift
when you add them together
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has to be some integer plus
a half times the wavelength
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in order to have destructive interference
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and this total phase shift is going
to be *Delta Phi 1* plus *Delta Phi 2*
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equals *m* plus half times wavelength.
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And now we can substitute these
expressions for *Delta Phi*
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that we found before in
blue zero four every one
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and then *lambda* over two plus *2t* for ray
two and we're going to solve this for *t* .
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And so I multiply both sides by two here
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and then distribute the
*lambda* into the brackets
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and end up with *Lambda* plus 40 equals
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I guess I just multiplied the
two into the brackets there.
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So that's *2m* plus one times *lambda*
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and you end up with *2m lambda* plus *lambda*
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and then subtract a lambda from both sides
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and you have 40 equals *2m* *lambda*
and then t* is *m* lambda over two.*
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So this is an expression for
the thickness of the air gap
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needed to have destructive interference.
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And there will be many such thicknesses
that cause destructive interference
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because we can choose any value
any integer value for *m*.
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So we're going to have here's the use
with the glass pieces are touching
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and that's some thickness *t1* will
have destructive interference.
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And that's at a horizontal position *x1*
from this point where they're touching
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and then again will have destructive
interference that happening again at *t2* .
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So this will be corresponding
to *m* equals one
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and this corresponds to *m* equals
two these two thicknesses.
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We would plug in the number
one or the number two
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into this expression for thickness
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and this will happen at a position *x2* .
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And the question is asking us
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what is the difference between
these horizontal positions.
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How far apart are the
dark bands horizontally.
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What is delta *x* in other words.
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So these are similar triangles and
because the angles are all the same.
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I mean they share this common angle
here and this is a 90 degree angles.
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That means this angles the same
two and because they're similar
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that means that the ratio of
corresponding sides is the same.
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So when you take *x2* divided by *t2* that's
gonna be the same as *x1* divided by *t1* .
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So let's just call it *x* over *d* .
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And d being, this is,
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this here because we know what
the ratio for this triangle is
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and the ratios where all
the triangles inside here
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are all going to be the same
because they're all similar.
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And I mean similar in a mathematical
technical sense of the word
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meaning that all the time all the
angles in the Triangle are the same
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and therefore ratios of
corresponding sides are the same.
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And we know the opposite and the
adjacent for this triangle here
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it's 0.1 millimetres high
and 7.5 centimeters long.
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So we can say that *x2* is *t2* times *x*
over *d* and *x1* is *t1* times *x* over *d*
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and we'll substitute for each of those
in this expression for *delta x*.
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And that's what I've done here and
the *x* over *d* can be factored out
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and it's gonna be *t2*
minus* t1* times *x* over *d*
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and *Delta x* then substituting for *t2*
and *t1* using this expression here
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substituting the numbers two and one for *n* .
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We have two times lambda over two minus
one times lambda over two times *x* over *d*
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and then this works out to Lambda over two
and then we can substitute in numbers.
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So that's 589 nanometres
is the wavelength we told
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divided by two times seven
and a half centimeters
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written as times ten to
the minus two meters
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divided by 0.1 millimetres written as
times ten to the minus three meters
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gives us this many nanometres
which is 0.221 mm.
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So this is the horizontal
separation between dark bands.
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Now what you would expect
in this picture is
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to see lines like this spaced
apart 0.221 millimeters
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and in reality though you don't see that
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because that's showing you that there are
imperfections in these pieces of glass
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it means that they're not perfectly
straight as shown in the drawing
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and so that's kind of interesting
and this is one of the ways
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that they actually engineer really nearly
perfect mirrors for telescopes and so on
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is by looking at interference
patterns like this.