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Good morning, we start with the fourth lecture
on the subject on physical metallurgy, and
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we continue with the first chapter that is
on atomic bond and crystal structure. And
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in the
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last class, we talked about point lattice,
which is used to represent crystal structure
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in any
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material, we looked at atomic arrangement
in a few simple crystal structures that we
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find
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in metals namely, face centered cubic, body
centered cubic and hexagonal close packed
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structure. We also looked at the packing density,
the relationship between the atomic
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diameter and the lattice parameter, and also
introduced the concept of Miller indices to
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represent planes and directions in a crystal.
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In the end; we talked about stereographic
projection which is an angle true projection,
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and
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this is very frequently used to solve problems
in crystal; crystallography. Now today, we
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shall look at particularly; stereographic
projection in great detail.
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And to recapitulate, look at this octahedral
site; we talked about in the last class; this
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is a
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body centered cubic structure. The main atoms,
they are located at the corner of the cube,
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and the interstitial sites are the edge center
and face center. We calculated the number
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of
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interstitial site in an unit cell, and here
you will clearly see that number of interstitial
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sites.
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How to calculate that; is given here each
of these edge center is shared by 4-unit cell,
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which
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are placed around this edge.
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So, therefore, contribution of edge centered
atom will be 12 times one-fourth, and then
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we
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also have face centered and this is shared
by two neighboring unit cells, the contribution
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of each towards the unit cell is half. So,
therefore, if you add these up, it comes out
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to be
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6. So that means, there are 6 interstitial
sites of octahedral type in a unit cell, and
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this
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incidentally happens to be three times the
number of atoms in a BCC lattice. Similarly,
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we
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looked at tetrahedral sites; tetrahedral sites
are made up of four triangular faces; now
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this
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each of these triangle is an isosceles triangle,
two edges are equal to half the diagonal say
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like this and this, and the other edge is
equal to the edge of the cube that is lattice
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parameter
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‘a’. So, this is ‘a’, this will be
root 3 ‘a’ over 2.
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And here also you can calculate the number
of interstitial site, which is given here,
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here all
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the four sites in a face, we know they are
they will belong to 2-unit cell; one this,
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one top
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of this; and there are 6 faces, so total number
of site is 4 times 6 that is 24, and each
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face
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being shared by 2-unit cell, the net contribution
comes out to be 4 into 6 times half.
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So, this comes out to be 12. So, therefore,
it is 6 times the number of atom in a unit
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cell.
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And last time, you also calculated the packing
density; so, this packing density, you know
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it is; the gaps are divided into octahedral
and tetrahedral sites. So, here you have more
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number of sites so obviously, this is going
to be smaller than octahedral sites. (Refer
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Slide
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Time: 05:01)
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And here is a method, which is illustrated
to calculate the gap or dimension of the
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interstitial site; octahedral site is much
easier to visualize. These are the 4-corner
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atom on
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the face; and this is the face center, which
is an octahedral site in a BCC structure,
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and you
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have eight isosceles triangles basically.
So, this is dimension ‘a’, and this dimension
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is half
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of the diagonal. So, you have eight faces
surrounding this interstitial site. And the
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atomic
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diameter we estimated last time, this is equal
to root 3 a by 2. So, once you know this,
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then
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it is possible to calculate this gap.
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Now this distance between this atom and this
atom is equal to ‘a’ and therefore, the
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gap
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along c axis will be; this distance ‘a’
minus this atomic diameter, because half of
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this atom
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and half of this atom, add together to one
atom. So, one atomic diameter, you subtract,
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you
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get the gap along c axis. But; if you calculate
the gap along the diagonal in the same way,
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the diagonal is; this is a, this is a, so
the diagonal is ‘a’ root two; so, if the
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gap in the face
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diagonal therefore, will be ‘a’ times
root 2 minus 3 by root 3 by 2 and obviously,
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this is
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larger than this.
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So, therefore, what it says; that this site
is asymmetric, if you try to put an atom,
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it will get:
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these two will get this displaced more than
these two. In the same way, it is possible
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to
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calculate the interstitial sites’ dimension
in case of tetrahedral site, which is shown
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here.
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It is possible to find out the coordinates
of each of the corner atoms basically, you
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have
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one corner atom here, one corner atom, and
this distance is ‘a’. Similarly, you have
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one
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face centered atom, another face center atom.
So, this coordinate if you say 0 0 0 this
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coordinate is half, half, half and this diagonal,
this is half the face diagonal, which is equal
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to ‘a’ root 3 over 2.
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So, therefore, the distance between any of
these; any of these corner points with this
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interstitial site, it is possible to find
out from geometrical relationship. Say if
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so; this is the
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coordinate half, one fourth, 1/4 0 is the
coordinate of this site, and if you try to
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find out
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distance between this coordinate and this,
it will clearly come out to be ‘a’ root
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5 over 4.
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We also looked at Miller indices of crystal
plane, say these are the crystal axis a, b,
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c, and
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if you represent an atom in terms of this
fractional indices or intercepts, like this
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is the
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plane which intercepts ‘a’ axis at 1 over
h, b axis one over k, c axis at one over l,
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then
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indices of this plane, we call it, we defined
it as h k l, which is the reciprocal of these
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intercepts. And we represent this particular
plane as h k l, but in the crystal, because
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of
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symmetry, there will be several planes, which
may have, may be similar having h k l
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indices the atomic arrangement to be similar,
and this set of such planes are represented
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by this curl bracket. In the same way; it
is possible to represent the directions; a
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particularly
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this is a crystallographic direction which
we represent simply by these coordinates in
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terms
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of these crystallographic axes. So, you move;
if this coordinate is u v w that means, you
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move u distance along ‘a’ axis, v distance
along ‘b’ axis, and w distance along ‘c’
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axis.
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And here also similar directions if you want
to represent, we represent it in terms of
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this
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angular brackets. And always in a crystallography
type of system that we use all similar
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planes have similar indices. Now in case of
a hexagonal crystal, these are the crystal
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axis
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h k l; and here also you can apply this same
concept, but here last time it was shown that
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say this particular plane, this is a prism
plane.
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Now, this prism plane; its indices is 1 0
0; Miller indices, where as if we take this
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particular
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plane, this is also exactly similar plane,
this also should have same indices and this
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in fact,
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it is as 0 1 0. But if you look at this particular
plane, say this particular plane, which
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intersects h at 1, k at minus 1, so this is
actually 1 minus 1 0. So, this and this although
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they are same; indices are different. That
is why in this case we make a; there is a
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slight
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difference, we say that we use four indices,
this is called Miller-Bravais indices h k
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i l. ‘i’
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is actually redundant, and this i is equal
to minus h plus k. And a system of planes
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is
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represented like this.
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Same thing, you can extend to direction as
well. Now here it is little tricky; so, which
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is
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shown over here, because all these directions
say suppose if you followed that earlier
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convention, then this direction should have
been; this direction should have been 1 1
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0, but
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crystal-graphically this direction and this
direction they are same. So, here also, we
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use
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four indices system u v t w, the t is as such
redundant, this is equal to minus one third
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of
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U plus V. The capital U V W is the normal
Miller indices, and smaller u v w is the Miller
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Bravais system.
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So, with this set of conversion, it is possible
to show that all these equivalent directions
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will have identical or similar indices. Later
on, we will see that this type of representation
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allows us to do lot of matrix calculations
or vector calculation or this will help you
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to
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geometrically to represent crystal planes
and directions in a projection. Now then we
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looked at the concept of stereographic projection;
say normal engineering projection the
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distance; the relationship between the distances
are maintained. If you take a projection as
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an engineering projection of a 3D object,
the distance between two points even if it
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is
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scaled down, it maintains a definite relationship
even on the projection; whereas in case of
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a stereographic projection, it is no longer
a distance true projection, it is an angle
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true
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projection.
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And the concept of this projection is illustrated
here, you imagine the crystal to be very
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small, and it is placed in a three dimension;
at the center of a hemisphere like this. And
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the
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size of the crystal is so small in comparison
to the sphere that any plane that you draw
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here, even on the top surface or bottom surface,
they all will coincide, they will pass
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through the center of this sphere; and this
will intersect the reference sphere along
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these
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great circles. So, this is one plane, this
is another vertical plane, so we call, so
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therefore,
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the list of assumptions is here, the reference
sphere dimension is very large, crystal is
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very
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small. So, that all planes pass through the
origin.
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Now, what is a stereographic projection? We
try to illustrate this here, say suppose here
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we have placed this crystal; now imagine;
and here this is the reference sphere. Now
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draw
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a plane in the crystal; extend it to meet
the reference sphere, it meets the reference
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sphere
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along this great circle. So, radius of this
circle and the radius of the reference sphere,
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they
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are same. Now, draw a perpendicular to this;
and extend it to meet the reference sphere,
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it
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meets at point P.
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Now, let us try to project this point P on
to this projection plane, projection plane
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is this
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circle is the projection plane by placing
a light source here. Imagine that this pole
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is marked
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by a dark color say, black, this is a transparent
sphere, this is a black dot over here, and
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put
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a flash bulb over here. And try and find out
the line joining this and this; where does
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it
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intersect the projection plane? This is the
projection plane, and this is the line joining
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this
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and this particular point, and this is where
it intersects the projection plane, we call
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this the
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stereographic projection of point P.
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And this plane also intersects the projection
plane at these two points; and later on, we
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will
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see that trace of this line, this great circle
will also be projected on the projection plane
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as
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a great circle. Now the problem comes that
if we take a projection like this, we can
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project
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all the point in the part of the hemisphere,
which is away from the source of light, what
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happens to the points or to the poles, which
appear in the hemisphere, which contains the
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light source? Suppose, this example is taken
here, how do you represent the point Q. What
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we do? We draw a line join this point Q with
the center. Extend it to meet the other half
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of
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the reference sphere. So, this is the diametrically
opposite point of this; and we say that
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this is small q. And project this on to the
projection plane, and represent it by an open
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circle, not a filled in circle, so, which
signifies this is a point in this hemisphere.
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Now, we look at the same thing, say, so far,
I was trying to explain you with respect to
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a
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3-dimensional object. Now what happens if
we do the same thing on a 2D or represent
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in
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a two dimension, this is the projection plane.
Now here, this is the plane, which was drawn,
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this is the perpendicular and this is the
projection here. So, this is still now here
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also this
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is a 3-dimensional representation, we convert
it into 2-dimension, then it will look like
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this. This is point east, this is point west;
so, this is getting projected and east and
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west,
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these are the two points. This point P, it
comes here; that means, this pole is represented
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by its projection in a 2D like this. And this
great circle, this great circle is represented
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here
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as a plane; and we will see later the distance;
the angular, this is an angle true projection
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and P is normal to this plane; so, this angular
relationship must be maintained. So here,
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we
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will see that the distance; angular distance
between this and this great circle this will
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be 90
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degree.
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Now, let us see how do you represent angles
between planes? This is illustrated here;
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and
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00:20:54,350 --> 00:20:59,789
here also we use the reference sphere so that
means, in this drawing, I am trying to still
232
00:20:59,789 --> 00:21:09,539
represent it in terms of a 3D diagram. So,
take this as a plane; draw a normal to the
233
00:21:09,539 --> 00:21:10,539
plane,
234
00:21:10,539 --> 00:21:17,919
which intersects the reference sphere at this
point. Draw another plane; this is the other
235
00:21:17,919 --> 00:21:26,309
plane, draw the perpendicular to this plane
passing through the origin. So, this is the
236
00:21:26,309 --> 00:21:27,309
other
237
00:21:27,309 --> 00:21:42,429
plane; and these two planes, they intersect
along this line, which is shown here. Now
238
00:21:42,429 --> 00:21:43,429
to
239
00:21:43,429 --> 00:21:48,799
measure this angle, how do you measure this
angle? You have to visualize a plane passing
240
00:21:48,799 --> 00:21:57,269
through these two lines, which is shown here.
And then on that plane, therefore, you can
241
00:21:57,269 --> 00:21:59,710
put a protector and measure the angle.
242
00:21:59,710 --> 00:22:09,100
So, this is the basic principle of measuring
angle in a, on a 3-dimensional reference here.
243
00:22:09,100 --> 00:22:28,820
Now, poles in a reference sphere; now in this
reference sphere, we draw certain reference
244
00:22:28,820 --> 00:22:36,769
lines or planes. So now, this particular plane,
this is the great circle, we also use a set
245
00:22:36,769 --> 00:22:37,769
of
246
00:22:37,769 --> 00:22:43,990
small circles, which are drawn like this;
so, they are called latitudes and these are
247
00:22:43,990 --> 00:22:44,990
great
248
00:22:44,990 --> 00:22:59,379
circles. And we can draw similar great circles
at regular interval; these are all great circles.
249
00:22:59,379 --> 00:23:05,899
Now we represent any angle or any angular
position of any pole on this sphere, in terms
250
00:23:05,899 --> 00:23:14,690
of these angular coordinates. Suppose we want
to represent this point here, we will
251
00:23:14,690 --> 00:23:22,190
represent it by two angles; say; one angle,
say, joining this pole; rather this perpendicular
252
00:23:22,190 --> 00:23:31,919
line, that angle it subtends with this reference
line. So, this angle is phi, which is measured
253
00:23:31,919 --> 00:23:43,950
along this great circle; another angle is
theta, say, we take, say this is the line
254
00:23:43,950 --> 00:23:44,950
here, this
255
00:23:44,950 --> 00:23:52,860
great circle it meets here; and we say this
angle is theta. So, by these two coordinates,
256
00:23:52,860 --> 00:23:53,860
theta
257
00:23:53,860 --> 00:23:59,789
and phi, we represent a point on the globe.
So, this is very much similar to geographical
258
00:23:59,789 --> 00:24:09,240
globe, where we have a set of lines say, longitudes;
they are great circles and the latitudes
259
00:24:09,240 --> 00:24:13,970
are the small circles.
260
00:24:13,970 --> 00:24:20,859
If we take a projection of that kind of a
ruled globe, we get a net called Wulff net
261
00:24:20,859 --> 00:24:21,859
and this
262
00:24:21,859 --> 00:24:30,529
helps us to measure angles on the stereographic
projections. Now here these great circles,
263
00:24:30,529 --> 00:24:37,820
they are the longitudes, and these are the
projections of the small circles called latitudes.
264
00:24:37,820 --> 00:24:44,869
Now in a Wulff net, these lines are drawn
at a regular angular interval and usually
265
00:24:44,869 --> 00:24:45,869
for most
266
00:24:45,869 --> 00:24:53,479
calculation we use Wulff net, where these
lines are drawn at an interval of 2 degrees.
267
00:24:53,479 --> 00:25:01,139
Now, I will try to show you an illustration.
So, suppose this is the 2-D projection
268
00:25:01,139 --> 00:25:08,731
stereographic projection, now here are two
poles, two crystallographic directions; how
269
00:25:08,731 --> 00:25:09,731
do
270
00:25:09,731 --> 00:25:20,369
you measure this angle using Wulff net? Now,
place it over the Wulff net, now here to
271
00:25:20,369 --> 00:25:29,149
measure an angle both these poles must lie
on one plane, and here it is not. And here
272
00:25:29,149 --> 00:25:30,149
one
273
00:25:30,149 --> 00:25:34,749
plane in a Wulff net is represented by this
longitude. So, they are not lying on any
274
00:25:34,749 --> 00:25:35,749
longitude.
275
00:25:35,749 --> 00:25:40,890
So, therefore, you cannot; it is difficult
to measure with this position, what we do?
276
00:25:40,890 --> 00:25:41,899
We try
277
00:25:41,899 --> 00:25:53,520
and rotate this and by rotating, we try to
bring it in such a position that these two
278
00:25:53,520 --> 00:25:54,609
points,
279
00:25:54,609 --> 00:26:03,399
they lie on one longitude; as shown over here.
And this longitude is graduated at regular
280
00:26:03,399 --> 00:26:09,749
intervals; look at the coordinates that latitude
over here, latitude over here, and read the
281
00:26:09,749 --> 00:26:21,299
difference. So, therefore this angular graduation,
read along this longitude; it’s the angle
282
00:26:21,299 --> 00:26:24,669
between the two poles.
283
00:26:24,669 --> 00:26:34,159
Now, we very often use standard projection
of certain crystals, we will mostly concentrate
284
00:26:34,159 --> 00:26:41,509
on cubic crystal and let as try and understand,
how do we construct standard projection of
285
00:26:41,509 --> 00:26:44,349
a cubic crystal.
286
00:26:44,349 --> 00:26:55,179
This is illustrated here, imagine, say this
is a reference sphere, place a crystal at
287
00:26:55,179 --> 00:26:56,179
the center;
288
00:26:56,179 --> 00:27:01,029
the crystal is very small with respect to
the reference sphere; these are the crystallographic
289
00:27:01,029 --> 00:27:09,789
axes, and we will try to construct 001 standard
projection. Now, if you extend these cube
290
00:27:09,789 --> 00:27:20,909
planes, they will intersect this plane here,
here, and the other one will intersect this
291
00:27:20,909 --> 00:27:22,049
plane;
292
00:27:22,049 --> 00:27:30,289
great circle at this point. So, these are
the three poles. Now, let us try to construct
293
00:27:30,289 --> 00:27:31,289
the
294
00:27:31,289 --> 00:27:36,840
projection of this particular plane in the
crystal. What is the indices of this plane?
295
00:27:36,840 --> 00:27:37,840
This
296
00:27:37,840 --> 00:27:48,539
plane is joining two opposite edges, so that
means and it is; it contains the face diagonal.
297
00:27:48,539 --> 00:28:00,359
So, therefore the indices of this plane is
(011), we extend this plane. So, this is the
298
00:28:00,359 --> 00:28:01,359
plane.
299
00:28:01,359 --> 00:28:10,779
Then we construct the normal to this and extend
it to intersect the reference sphere. It
300
00:28:10,779 --> 00:28:18,529
intersects this here, and also it intersects
this. It is halfway between this pole and
301
00:28:18,529 --> 00:28:19,529
this pole
302
00:28:19,529 --> 00:28:22,730
on this great circle, this is 011.
303
00:28:22,730 --> 00:28:32,979
Now, project it, we look at it from this particular
point, just opposite diametrically opposite
304
00:28:32,979 --> 00:28:42,570
to 001, to get a (001) standard projection
we must look at 001 from its diametrically
305
00:28:42,570 --> 00:28:50,029
opposite point. And try to project it on this
projection plane, and this is what is done
306
00:28:50,029 --> 00:28:51,029
over
307
00:28:51,029 --> 00:28:57,619
here. Then it projects at this particular
point which is halfway, angular distance wise
308
00:28:57,619 --> 00:28:58,619
half
309
00:28:58,619 --> 00:29:07,649
way between this point and this point.
310
00:29:07,649 --> 00:29:15,259
Now to draw this type of projection we often
use the angular relationships between
311
00:29:15,259 --> 00:29:24,809
different crystallographic planes. For example,
one plane has an indices h1 k1 l1 - this is
312
00:29:24,809 --> 00:29:32,320
the Miller indices, and this is another plane,
has h2 k2 l2. Geometrically it can be shown
313
00:29:32,320 --> 00:29:44,759
that angle between the two is given by this
relationship; cos phi can be determined, if
314
00:29:44,759 --> 00:29:45,759
the
315
00:29:45,759 --> 00:29:54,129
indices of the planes are known. And it is
also possible to show that in cubic crystal
316
00:29:54,129 --> 00:29:55,129
where
317
00:29:55,129 --> 00:30:03,619
these axes; they are orthogonal. So, any crystal
system where the axes are orthogonal, if a
318
00:30:03,619 --> 00:30:13,799
plane is like this, if a plane is like this;
and its indices is h k l and normal to this
319
00:30:13,799 --> 00:30:14,799
plane is
320
00:30:14,799 --> 00:30:27,620
drawn here, normal to this plane is u v w,
the angle between this plane h k l and its
321
00:30:27,620 --> 00:30:28,620
normal
322
00:30:28,620 --> 00:30:38,090
u v w is 90 degrees. So, therefore, so in
place of h2 k2 l2, if you substitute u v w
323
00:30:38,090 --> 00:30:39,090
you get
324
00:30:39,090 --> 00:30:45,159
this relationship, if ‘hu’ plus ‘kv’
plus ‘lw’ equal to 0.
325
00:30:45,159 --> 00:30:54,879
So, this relationship will be valid; for example,
the plane 111 in a cubic crystal and the
326
00:30:54,879 --> 00:31:02,450
pole; or perpendicular direction which is
represented that also by 111, the angle between
327
00:31:02,450 --> 00:31:11,950
these two will be 90 degree, and this relationship
is very frequently used to construct a
328
00:31:11,950 --> 00:31:17,289
standard projection.
329
00:31:17,289 --> 00:31:24,789
Let us try to construct the standard projection
of a cubic crystal on the projection plane
330
00:31:24,789 --> 00:31:39,629
001. Now, obviously projection plane 001;
so, center of this is 001 and on; and this
331
00:31:39,629 --> 00:31:40,629
is the
332
00:31:40,629 --> 00:31:51,940
pole 001 and this is plane 001. And the pole
100, will be located here, and the pole 010
333
00:31:51,940 --> 00:32:08,360
will be located here. And it is best illustrated
by; let us try to do this exercise on paper.
334
00:32:08,360 --> 00:32:19,919
Now here this is pole 001 which is represented
by this point, this is 100 which is
335
00:32:19,919 --> 00:32:28,139
represented by this point, this is 010 is
represented by this point. Now, the opposite
336
00:32:28,139 --> 00:32:29,139
of this
337
00:32:29,139 --> 00:32:40,710
the point diametrically opposite of this is
this. So, indices of this will be minus 100
338
00:32:40,710 --> 00:32:41,710
we call
339
00:32:41,710 --> 00:32:55,809
it bar 100; like this it will be 0 bar 1 0,
and we have shown a little while ago on the
340
00:32:55,809 --> 00:32:56,809
reference
341
00:32:56,809 --> 00:33:09,979
sphere how to locate the pole 011. So, 011
will be located between these 2 poles on this
342
00:33:09,979 --> 00:33:27,169
equator, this is 011. Now, here if we recollect
the relationship between two planes of
343
00:33:27,169 --> 00:33:47,690
indices h1 k1 l1 and h2 k2 l2, we try and
find out the angle between 011 and angle 001.
344
00:33:47,690 --> 00:34:01,340
So, this is equal to is given by cos phi equal
to the product of 0 times 0, so, this is 0.
345
00:34:01,340 --> 00:34:02,340
Product
346
00:34:02,340 --> 00:34:10,950
this; times this that is 0; plus this times
this that is one and then it is root over
347
00:34:10,950 --> 00:34:13,379
h1 square
348
00:34:13,379 --> 00:34:22,410
plus k1 square plus l1 square. So, which is
root two and in this case particularly it
349
00:34:22,410 --> 00:34:23,410
is root
350
00:34:23,410 --> 00:34:37,260
one. So, therefore this equal to one over
root two, so that means, phi is 45. So, likewise
351
00:34:37,260 --> 00:34:38,260
it
352
00:34:38,260 --> 00:34:47,750
is possible to index all these, this is also
the angle 45 and its indices is 110. Opposite
353
00:34:47,750 --> 00:34:48,750
of
354
00:34:48,750 --> 00:35:00,600
this is 1 bar 1 bar 0, this point is 1 bar
00.
355
00:35:00,600 --> 00:35:08,890
Now, it is quite simple, if we represent;
recollect this zone relationship, that is
356
00:35:08,890 --> 00:35:12,860
hu plus kv
357
00:35:12,860 --> 00:35:27,290
plus lw equal to 0, in that case it is possible
to show that if you add this and this you
358
00:35:27,290 --> 00:35:28,290
get
359
00:35:28,290 --> 00:35:35,800
110. So, that also will lie; that indices
will lie on this particular plane; it may
360
00:35:35,800 --> 00:35:36,800
not give you
361
00:35:36,800 --> 00:35:43,110
the exact where; but somewhere in between
these two. And then we can find out the angle
362
00:35:43,110 --> 00:35:50,700
between this, and this which also can be shown
in the same way to be 45, so this is here
363
00:35:50,700 --> 00:36:03,570
like this. Similarly, what is the index of
this? It’s 1 1 bar 0 which is I think this
364
00:36:03,570 --> 00:36:04,570
is one bar;
365
00:36:04,570 --> 00:36:12,230
there is a mistake; it is 1 bar 1 0. So, it
is diametrically opposite to this is 1 bar
366
00:36:12,230 --> 00:36:13,310
1 0, this is
367
00:36:13,310 --> 00:36:24,620
1 1 bar 0. Now, what is the index of this;
how do we find out, now here if we add this
368
00:36:24,620 --> 00:36:25,620
and
369
00:36:25,620 --> 00:36:34,990
this what do we get? We get one bar 1 1, so
that means, 1 bar 1 1 lies on this great circle.
370
00:36:34,990 --> 00:36:44,750
Similarly, if we add this and this we came
get 1 bar 1 1, so that means, this lies on
371
00:36:44,750 --> 00:36:45,750
this
372
00:36:45,750 --> 00:36:52,340
great circle. So, therefore, the point of
intersection is 1 bar 1 1. In the same way
373
00:36:52,340 --> 00:36:53,340
you can
374
00:36:53,340 --> 00:37:01,130
show that this also you get by adding this
and this you get 1 1 1, similarly adding this
375
00:37:01,130 --> 00:37:02,130
and
376
00:37:02,130 --> 00:37:11,330
this also you get 1 1 1, so index of this
is 1 1 1. And then what is this? This is 1
377
00:37:11,330 --> 00:37:12,860
bar 1 bar
378
00:37:12,860 --> 00:37:41,120
1, this is 0 1 bar 1, this is 1 1 bar 1. So,
we have indexed all these.
379
00:37:41,120 --> 00:37:50,520
So, now it is also possible to index the other
directions as well. So, for example, you take
380
00:37:50,520 --> 00:37:57,020
somewhere in between the two, in between the
two you will get a pole, so if we add this
381
00:37:57,020 --> 00:38:05,990
and this, you get 2 1 0 - 2 1 0 will be halfway
somewhere between the two, you can find
382
00:38:05,990 --> 00:38:13,440
out the angle using this type of that angular
relationship that cos phi. So, it will subtend
383
00:38:13,440 --> 00:38:20,820
some particular angle with this, another angle
with this. Similarly, opposite of that will
384
00:38:20,820 --> 00:38:21,820
be
385
00:38:21,820 --> 00:38:36,560
2 bar 1 bar 0, you join this and this. So,
basically you join this line you get this,
386
00:38:36,560 --> 00:38:37,560
and then
387
00:38:37,560 --> 00:38:42,800
you try and find out what are the indices
of this. So, this is how it is possible to
388
00:38:42,800 --> 00:38:43,800
construct
389
00:38:43,800 --> 00:38:47,490
a standard 001 projection.
390
00:38:47,490 --> 00:38:57,360
Now, if you want to convert it to any other
projection, like, let us say, we want (011)
391
00:38:57,360 --> 00:39:06,320
projection. Then, how will you do it? You
can, this can be done by rotating this projection;
392
00:39:06,320 --> 00:39:12,640
and rotation of this projection is possible
with the help of Wulff net, and I will leave
393
00:39:12,640 --> 00:39:13,640
it to
394
00:39:13,640 --> 00:39:24,290
you as an exercise.
395
00:39:24,290 --> 00:39:33,400
Now often you may have to represent Miller
indices of a pole in a cubic crystal or a
396
00:39:33,400 --> 00:39:44,090
standard projection. Say, suppose this is
a standard projection. How do you locate a
397
00:39:44,090 --> 00:39:54,480
particular pole which subtends a set of specific
angles? Say suppose, we consider this plane
398
00:39:54,480 --> 00:40:07,690
here, this plane has an indices hkl; the normal
to this plane is this. Now we want to
399
00:40:07,690 --> 00:40:11,270
represent this pole over here.
400
00:40:11,270 --> 00:40:21,520
Now look at, or measure the angle, this angle
is rho, this angle is sigma, this angle is
401
00:40:21,520 --> 00:40:24,540
tau,
402
00:40:24,540 --> 00:40:34,790
and now particularly look at this angle tau,
this direction hkl say, the pole h k l subtends
403
00:40:34,790 --> 00:40:44,830
angle tau with respect to this axis c. So
that is, this axis c is 0 0 1; 0 0 1 is located
404
00:40:44,830 --> 00:40:46,830
here. So,
405
00:40:46,830 --> 00:40:54,370
along the great circle or along the equator
here, you can measure this angle tau to locate
406
00:40:54,370 --> 00:41:08,550
this point, and then you locate this angle
sigma along this great circle and rho along
407
00:41:08,550 --> 00:41:09,550
this
408
00:41:09,550 --> 00:41:14,470
great circle, and this can be done by vice
versa. So, suppose here is a pole, you can
409
00:41:14,470 --> 00:41:15,470
find
410
00:41:15,470 --> 00:41:22,700
out these three angles and you can convert
it. Take cos rho; I make it cos rho cos sigma
411
00:41:22,700 --> 00:41:28,050
cos tau will be in proportion to h k l.
412
00:41:28,050 --> 00:41:38,420
So, basically if you can; magnitude of cos
of any angle is a fraction. So, if you can
413
00:41:38,420 --> 00:41:39,420
with a
414
00:41:39,420 --> 00:41:50,600
suitable multiplier make it in integer, then
h k l will be the indices of the plane. So,
415
00:41:50,600 --> 00:41:51,600
with
416
00:41:51,600 --> 00:42:02,950
we finish this chapter on atomic bond and
crystal structure some of these exercises
417
00:42:02,950 --> 00:42:03,950
that I
418
00:42:03,950 --> 00:42:16,620
have given, it will be worthwhile for you
to pursue that. I think the concept of stereographic
419
00:42:16,620 --> 00:42:24,670
projection is a little abstract, and you have
to go through some exercises say suppose over
420
00:42:24,670 --> 00:42:39,760
here, we try to find out say we had just drawn
a standard 001 projection which was shown
421
00:42:39,760 --> 00:42:41,540
here.
422
00:42:41,540 --> 00:42:59,050
Now, using this say suppose we said that this
is 001, this is 100, this is 010, this is
423
00:42:59,050 --> 00:43:00,210
110,
424
00:43:00,210 --> 00:43:21,220
this is 1 bar 1 bar 0, this is 0 1 bar 0,
this is 001 bar, and this is 011 bar. So,
425
00:43:21,220 --> 00:43:24,970
this is 1 1 bar
426
00:43:24,970 --> 00:43:47,350
0, this is bar 1 bar 1 0, this is 1 1 bar
1, this is 111. Now, let us try to find out,
427
00:43:47,350 --> 00:43:48,350
say suppose
428
00:43:48,350 --> 00:44:04,890
the plane 1 1 1. So, which of these directions,
say try to find out the directions, type of
429
00:44:04,890 --> 00:44:05,890
110
430
00:44:05,890 --> 00:44:21,150
directions, which lie on plane 111. Now here
this is the pole; pole of 111. On this try
431
00:44:21,150 --> 00:44:22,170
to
432
00:44:22,170 --> 00:44:30,380
construct the plane 111; plane 111 will be
represented by a great circle, and that will
433
00:44:30,380 --> 00:44:31,380
be at
434
00:44:31,380 --> 00:44:33,150
90 degree from this.
435
00:44:33,150 --> 00:44:42,600
So, along this diameter using Wulff net try
to locate a point which is 90degree, and try
436
00:44:42,600 --> 00:44:43,600
to
437
00:44:43,600 --> 00:44:58,820
construct a great circle which passes through
this point. And then you try and find out
438
00:44:58,820 --> 00:44:59,820
what
439
00:44:59,820 --> 00:45:10,540
is the indices of this. Same thing can be
done with respect to this 1 bar 1 bar 1 pole.
440
00:45:10,540 --> 00:45:11,540
So,
441
00:45:11,540 --> 00:45:28,760
corresponding to that, there will be a plane
like this, what is the indices of this? Say
442
00:45:28,760 --> 00:45:54,850
construct a plane, construct, try to find
out indices of each of these points of intersections.
443
00:45:54,850 --> 00:46:08,950
So, with this we end this topic on atomic
bond and crystal structure, and from time
444
00:46:08,950 --> 00:46:10,160
to time
445
00:46:10,160 --> 00:46:20,890
we will in subsequent lectures, will be referring
to these parts in detail, and if you go
446
00:46:20,890 --> 00:46:29,740
through this example this will help you to
understand subsequent lectures. To sum up
447
00:46:29,740 --> 00:46:30,740
we
448
00:46:30,740 --> 00:46:37,790
covered that atomic structures how that atomic
structures and atomic bond affect the
449
00:46:37,790 --> 00:46:43,710
properties of engineering materials. So, what
is the difference between metallic bond and
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00:46:43,710 --> 00:46:50,010
covalent bond, why metallic bond exhibits
good conductivity. Now, in metals what type
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00:46:50,010 --> 00:46:56,790
of crystals are there. There are three types
of crystal primarily face centered cubic,
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00:46:56,790 --> 00:46:57,790
body
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00:46:57,790 --> 00:47:00,520
centered cubic and hexagonal close packed.
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00:47:00,520 --> 00:47:06,580
We looked at some of the characteristics of
these crystal structures, we learnt how to
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00:47:06,580 --> 00:47:15,180
represent different planes and directions
in a crystal using concept of Miller indices.
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00:47:15,180 --> 00:47:16,180
Miller
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00:47:16,180 --> 00:47:22,860
indices in case of hexagonal close packed
structure has little problem. If we follow
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00:47:22,860 --> 00:47:23,860
the
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00:47:23,860 --> 00:47:30,320
conventional Miller indices, we often do not
come across similar indices for similar types
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00:47:30,320 --> 00:47:37,330
of plane. So, therefore, here we had to introduce
the concept of Miller Bravais indices
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00:47:37,330 --> 00:47:43,980
using four indices, but you don’t need four
indices to represent 3dimensional structures.
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00:47:43,980 --> 00:47:53,980
So, one of these is redundant. So, it follows
the definite relationship with the other two,
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00:47:53,980 --> 00:48:05,560
and then we looked at standard projections,
we looked at 001 standard projections, but
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00:48:05,560 --> 00:48:06,560
it
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00:48:06,560 --> 00:48:13,570
will be worthwhile to construct few other
standard projections using Wulff net or using
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00:48:13,570 --> 00:48:17,290
the angular relationship which I had given
you.
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00:48:17,290 --> 00:48:24,410
And if you try and do, or find out what will
be standard projection, how will the standard
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00:48:24,410 --> 00:48:32,830
projection 011 look like, or a standard projection
111 look like. And these standard
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00:48:32,830 --> 00:48:37,540
projections, looking at the standard projections
it is also possible to determine the
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00:48:37,540 --> 00:48:45,420
symmetry elements. So, may be if you look
at you will find out say in this particular
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00:48:45,420 --> 00:48:47,390
case,
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00:48:47,390 --> 00:48:56,800
say this particular cube axis, you know if
you rotate you have one point here, one point
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00:48:56,800 --> 00:49:02,820
here. So that means, if you give way ninety
- degree rotation it comes to occupy similar
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00:49:02,820 --> 00:49:03,820
projection.
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00:49:03,820 --> 00:49:15,490
So, this axis is actually a 4fold symmetry,
similarly try to find out what type of symmetry
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00:49:15,490 --> 00:49:27,420
element is represented by 110. Similarly;
what type of symmetry elements is represented
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00:49:27,420 --> 00:49:41,100
by 111? And try to be convinced using standard
projection that this exhibits, 111 exhibits
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00:49:41,100 --> 00:49:51,070
a 3fold symmetry, whereas this exhibits a
2fold symmetry. With this we end this chapter.
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00:49:51,070 --> 00:49:58,220
And we will in next chapter, we will look
at some of the tools and techniques which
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we
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00:49:59,220 --> 00:50:06,800
use in physical metallurgy to find out structure
property correlations, thank you.