WEBVTT
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i make formation by lenses by m k srivastava
department of physics indian institute of
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technology roorkee uttarkhand
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see in the first lecture geometrical optics
you have seen that both the laws of reflection
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and diffraction follow from fermats principle
of least time now in this lecture we shall
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review diffraction and image formation by
thin lenses
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a lens is called ten if the thickness is small
compared to for example radii of curvature
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of the two surfaces focal lengths and imagine
object distances we shall also consider coaxial
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and systems and so called thick lenses
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now the line joining the two centres of curvature
of the two surfaces of the lens is known as
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the axis of the lens which is perpendicular
to the two faces at the points of intersection
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the first principal focus is defined as the
point on the axis such that any ray coming
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from it or proceeding towards it travels parallel
to the axis after reflection the second principal
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focus is an axial point such that any incident
ray traveling parallel to the axis will after
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reflection proceed towards are appear to come
from it
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so this picture shows these features primary
may be called as first secondary may be called
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a second in the primary from f the old diverging
rays they become parallel after reflection
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or if the lens is a concave lens all these
rays which was about to converge to f after
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refraction again become parallel to the axis
for the second focal point the rays which
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are parallel to begin with after reflection
they converge to a point called us second
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difficult point or second focal point second
focus and similarly for the rays which after
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refraction appear to come from the focal point
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now the distance between the center of a lens
and either of its focal points is its focal
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length for a lens with the same medium on
both sides the two focal lengths are equal
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now a plane perpendicular to the axis and
passing through a focal point is called a
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focal plane parallel incident rays making
an angle theta with the axis or brought to
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focus at a point q prime in line with the
chief ray which is defined the chief ray is
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defined as the ray which passes undeviated
to the center of the lens the object and image
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are called conjugate points
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and the planes through these perpendicular
to the axis are called conjugate planes the
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plane mq and the plane m prime q prime are
conjugate planes m prime is the image corresponding
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to the object point m q prime used image corresponding
to the object point q
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now before we proceed further we should fix
the sign convention which shall be used during
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these lectures for finding out where the image
will be formed for a given position of the
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object this is the usual coordinate geometric
convention it is as follows the point of contact
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of the center of the lens with the axis is
taken as the origin of the coordinate system
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remember we are talking about thin lenses
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now the rays are always incident from the
left on the refracting surface the incident
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from the left going towards right all distances
to the right of the origin or positive and
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distances to the left of the origin or negative
usual coordinate geometry arrangement
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the angle that the ray makes with the axis
is positive if the axis has to be rotated
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in the anti-clockwise direction to coincide
with the ray and all distances major upwards
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from the axis are positive and all distances
measured downward direction or negative i
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mean the same convention should be pretty
familiar now with this convention the first
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focal length f1 is negative for a converging
lens because this point comes from the left
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of the lens
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and positive for a diverging lens which is
a double concave lens and a positive the case
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for the second focal length f2
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let us consider this an axial image formation
by a thin lens but axial means they are very
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close to the axis they are not making a large
angle with the axis the thin lens formula
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is n3 by v - n1 upon u = n2 - n1 upon r1 +
n3 - n2 upon r2 you see n1 n2 n3 are the refractive
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indices n1 is for the medium on the left of
the lens and n2 is for the material of the
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lens and n3 is for the medium on the right
of the lens
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now the first in the second focal length f1
and f2 can be obtained from this expression
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by putting v equal to infinity and u becomes
equal to f1 or u equal to minus infinity and
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v = f2 respectively and this gives 1 upon
f1 = - 1 upon n1 n 2 - n1 upon r 1 + n3 - n2
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upon r2 and for the second one 1 upon f2 = 1
upon n3 n2 - n1 upon r1 + n3 - n2 upon r2
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you see this capital r1 and r2 they are the
ray of radii of curvature of the two curved
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surfaces of the lens
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for a thin lens placed in a medium such that
the refractive indices on both sides of the
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lens are the same that is n3 = n1 the values
of f1 and f2 can readily be obtained by this
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relation 1 upon f2 naturally in this case
it is equal to -1 upon f1 i mean they are
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equal in magnitude there is a sign difference
because one focal point is on the left that
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is on the right so this is equal to n – 1
n is the refractive index of the material
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of the lens relative to the medium outside
multiplied by the factor 1 upon r1 - 1 upon
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r2
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now if this factor 1 upon r1 - 1 upon r2 is
a positive quantity then the focal length
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f which is equal to f2 which is the second
focal length is a positive quantity and the
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lens acts as a converging lens this is a situation
for a by convex lens similarly if that is
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a negative quantity then the lengths acts
the diverging lens
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now the relation between u v and f is 1 upon
v - 1 upon u = 1 upon f the solution is very
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very popular when most commonly used for finding
out the distance of the image when the object
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is that distance u and f in the focal length
of lens now once we know f1 and f2 and therefore
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the positions of the first and the second
principle for side the paraxial image can
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be graphically constructed from the following
simple rules
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a ray passing through the first principal
focus will up the refraction image parallel
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to the axis that is characteristic of the
focal point in a parallel to the axis will
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after refraction either pass through or happier
to come from depending on the sign of f2 depending
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on whether the lens is a converging lens or
diverging lens coming from the second principal
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focus a ray passing through the center of
the lens called the chief ray will pass through
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undeviated
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consider these figures you see from the point
a considered that the ray two which is parallel
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to the axis after refraction pass it through
f2 which is the focal point then a ray from
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a going through the focal point f1 after refraction
becomes parallel to the axis they meet at
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the point me which is the image of the point
a if you consider the chief ray now this should
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also reach the point b apb
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similarly for the case of the a concave lens
parallel ray starting from a after refraction
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appears to come from f2 a ray because going
towards f1 after refraction becomes parallel
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at the point l becomes parallel to the axis
and these two meet at the point b which is
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the image of the point a
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the above graphical procedure is called the
parallel ray method sometimes another graphical
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method called the oblique ray method that
is preferable the basic principles are the
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same parallel rays incident on the lens are
always brought to a focus at the focal plane
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and the rays of the center being the only
one undeviated that is the chief ray
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let us consider the picture here you consider
the rays starting from the point m if we actually
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have real diverging from m we can find the
direction of any one of them after it passes
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through the lens by making it intersect the
parallel line are r prime through a let us
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consider this ray mt this is ray now we wish
to find out the reflected ray corresponding
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to this so consider a parallel chief ray and
consider the focal plane through f prime
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then one can construct the rate tx because
the two parallel rays they must meet in the
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focal plane the ray rr prime is not an actual
ray in this case and stated as such only the
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means of locating the point x and once we
have located the point x we extend this tx
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to m prime and m prime is the image of the
object at the point m
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newtons formula you see this is another relation
which is used to relate the object and image
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distances with the focal length of the lens
the difference here is the distances are not
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measured from the lens you see let x1 be the
distance of the object from the first principal
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focus not from the lens f1 will be positive
if the object point is on the right of f1
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that is as per sign convention and let x2
be the distance of the image from the second
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principal focus
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so object distance to be measured from first
principle focus image distance to be measured
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from the second principal focus
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considered this picture i mean to obtain a
relationship in this case let us first fix
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up the distances you see they consider the
distance for example x2 which is the distance
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of image from the focal point f2 usually your
object distance u which is the distance of
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the object from the center of the lens the
distance of the image i from p is again from
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the center to obtain a relationship in this
case we shall make use of the similar triangles
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consider the triangle for example aop and
the triangle bip this they will give information
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some sort of a relationship between the various
quantities involved here
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so as i said considering the similar triangles
aof1 and plf1 you have a relationship like
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this -y prime upon y these are the heights
of the object points is equal to - f1 upon
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-x1 the minus and plus signs are s per sign
convention similarly from another set of similar
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triangles the ratio -y prime to y = x2 to
f2 the above equation gives the relationship
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f1 f2 = x1 into x2 this is known as the newtons
formula
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the medium is same on the two sides has is
the situation in most cases the formula basically
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now is x1 x2 = - f square
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let us consider virtual image real images
are the ones which can be made visible on
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a screen they are characterized by the fact
the rays of light are actually brought to
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a focus in the plane of the image a virtual
image on the other hand cannot be formed on
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a screen the rays from a given point on the
object do not actually come together at the
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corresponding point on the image instead they
must be projected to find this point
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these images cannot be taken on a screen virtual
images are produced with converging lenses
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when the object is placed between the focal
point and the lens and with diverging lenses
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this is the situation and the object is in
any position diverging lenses always lead
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to virtual images
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is this is a very typical example here the
object is very close to the lens between the
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lens and the focal point if you consider a
parallel ray starting from the point q then
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naturally after reflection it passes through
the f prime and it should then you consider
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the chief ray from starting from q goes through
undeviated and if we extend them backwards
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they meet at the point q prime which is the
image of the point q
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so that what it says here rays emanating from
the object point q are reflected by the lens
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but are not sufficiently deviated to come
to a real focus to the observer’s eye at
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these rays appear to be coming from a point
q prime on far side of the lens the distance
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of q prime from the lens is more than the
distance of q this point represents a virtual
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image because the rays do not actually pass
through q prime they only happier to come
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from there
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the ray qt parallel to the axis is reflected
through f prime while there really qa to the
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center of the lens which is the chief ray
is undeviated
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now these two rays when extended backwards
as i said earlier we have seen in the figure
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the intersect at q prime the third ray picture
qs traveling outward as though it came from
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f actually misses the lens in this case but
it will ends for larger the day would be refracted
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parallel to the axis as it should when projected
backward it also intersects the other projections
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at q prime
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cardinal points of an optical system in the
case of at a thick lens l in a combination
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of two or more lenses separated by a finite
distance it will be extremely tedious to consider
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refraction at each surface to overcome this
difficulty a set of points called cardinal
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points have been suggested to deal with any
number of coaxial refracting systems the system
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is treated as one unit without bothering about
its actual details
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the cardinal points are pairs of focal points
principal points and nodal points we have
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already discussed focal points first principal
focus and the second principal focus the planes
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passing through the principal foci perpendicular
to the axis are called focal planes the main
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property of the focal planes is that the rays
is starting from a point in the focal plane
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in object space they correspond to a set of
conjugate parallel rays in the image space
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we have seen that similarly they set a parallel
rays in the object space they correspond to
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a set of rays intersecting at a point the
focal plane in the image space
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now the principal points and the principal
piece let us see what they signify there are
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two principal planes and two principal points
the principal plane in the object space the
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locus of the points of intersection of the
emergent trace in the major space parallel
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to the axis and their conjugate incident rays
in the object space the second principal plane
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in the image space is the locus of the points
of intersection of the incident rays in the
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object space parallel to the axis and there
conjugate emergent rays in the image space
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we can see it from here it is clear from the
figure then the two incident rays are directed
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towards h1 and after refraction seem to come
from h2 therefore h2 is the image of h1 thus
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h1 and h2 are the conjugate points and the
planes h1 p1 and h2 p2 are a pair of conjugate
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planes further h1 p1 = h2 p2 so the lateral
magnification of the plane says +1 unit positive
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little magnification and that characterizes
the principal points and the principal planes
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the range is starting from any point on the
axis and cutting the principal plane had given
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heights from the axis will have their conjugate
emergent rays starting from the points in
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the second principal plane and the same respective
height that is the main thing that the same
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respective heights from the access all these
emergent rays converge to the image point
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on the axis
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let is consider the nodal points these points
are a pair of conjugate points on the axis
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having unit positive angular magnification
principal points were having unit positive
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little magnification nodal points have a unit
positive angular magnification this simply
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means that a ray of light directed towards
one of these points after refraction through
25:12.930 --> 25:21.530
the optical system appears to proceed from
the second point in a parallel direction
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lets consider let h1 p1 and h2 p2 the first
and the second principal planes of an optical
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system and let f1 af1 and bf2 be the first
and second focal planes respectively consider
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a point af1 the first focal plane from a consider
a ray ah1 parallel to the axis the conjugate
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ray will proceed from h2 a point in the second
principal plane such that h1 p1 = h2 p2 and
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will pass through the second focus f2
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take another ray at1 parallel to the emergent
ray h2 f2 and striking the first principal
26:20.420 --> 26:32.150
plane at t1 it will emerge from t2 point on
the second principal plane such that t2 p2
26:32.150 --> 26:46.180
= t1 p1 and we will proceed parallel to the
ray h2 f2 as the two rays originate at a a
26:46.180 --> 26:55.190
point on the first focal plane then the points
of intersection of the incident ray at1 and
26:55.190 --> 27:07.220
conjugate emergent ray t2 are with the axis
give the position of the two nodal points
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n1 and n2
27:10.730 --> 27:18.410
it is clear that the two points n1 and n2
are a pair of conjugate points an incident
27:18.410 --> 27:29.320
ray a n1 is parallel to the conjugate emergent
ray t2r the planes passing through the nodal
27:29.320 --> 27:37.440
points and perpendicular to the axis are called
nodal planes
27:37.440 --> 27:45.280
now the distance between the nodal points
n1 and n2 is equal to the distance between
27:45.280 --> 27:55.830
the principal points p1 and p2 further p1
n1 = p2 n2 which is really equal to f1 + f2
27:55.830 --> 28:05.660
now if the medium on both sides of the system
is optically similar f1 will will be equal
28:05.660 --> 28:19.350
to -f2 and this means p1 n1 = p2 n2 = 0 it
means that the nodal points coincide with
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the principal points so in this situation
if you have the same medium on both sides
28:26.740 --> 28:36.690
this pair of points have equal lateral magnification
and wholesome equal angular magnification
28:36.690 --> 28:45.090
let us consider the question and system its
equivalent focal length and it is cardinal
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points we consider now here to thin lenses
l1 and l2 or focal lengths f1 and f2 respectively
28:55.030 --> 29:03.250
a point object o is placed at a distance u
from the first lens and if final image is
29:03.250 --> 29:11.480
formed at i a distance of v from the second
lens you see after the reflections of the
29:11.480 --> 29:20.270
first lens an intermediate image can be thought
of at the point i prime which then serves
29:20.270 --> 29:28.900
as a virtual object for the second lens and
the final image is formed at the point r the
29:28.900 --> 29:37.010
d is the distance between the lenses use the
object distance from the first lens v prime
29:37.010 --> 29:41.770
is the distance of that intermediate image
which is formed
29:41.770 --> 29:49.160
so all the distances are given here the first
image due to the first lens is formed by primers
29:49.160 --> 29:57.070
i pointed out earlier at a distance of v prime
from it so we have the standard relation 1
29:57.070 --> 30:07.490
upon v prime - 1 upon u = 1 upon f1 so this
can be written like this 1 upon v prime = u
30:07.490 --> 30:18.811
+ f1 divided by uf1 now image i primve that
an object for the second lens the object distance
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for the second lens is v prime - d and the
final image is formed at i so here we have
30:27.660 --> 30:33.580
the relation 1 upon v v the major distance
from the second lens
30:33.580 --> 30:43.830
we are writing this relation for the second
lens so 1 upon v - 1 upon v prime - d = 1
30:43.830 --> 30:51.830
upon f2 f2 is the focal length of the second
lens or this can be written as 1 upon v prime
30:51.830 --> 31:04.460
- d = f2 – v divided by f2v now from theset
relations we eliminate v prime and we get
31:04.460 --> 31:16.610
a relationship like this uv multiplied by
d - f1 - f2 + u multiplied by f1 f2 - d f2
31:16.610 --> 31:27.890
+ v multiplied by df1 - f1 f2 - d f1 f2 = 0
31:27.890 --> 31:34.020
now we treat this problem in an alternative
way if alpha represents the distance of the
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first lens from the first principal plane
and beta represents the distance of the second
31:41.780 --> 31:50.410
lens from the second principal plane then
the reduce object distances and the reduced
31:50.410 --> 31:58.560
image distances are u capital u is given by
u - alpha remember you were the distance of
31:58.560 --> 32:07.040
the object from the first lens else y is the
distance of the first principal point from
32:07.040 --> 32:08.640
the first lens
32:08.640 --> 32:15.840
so capital u is the object distance from the
first principal point similarly v capital
32:15.840 --> 32:26.360
v small v - beta is the distance of the image
from the second principal point if f is the
32:26.360 --> 32:31.810
focal length of this of the combination of
the equivalent lens we have the standard relation
32:31.810 --> 32:40.620
like this 1 upon capital v minus 1 upon capital
u = 1 upon f we put the values of capital
32:40.620 --> 32:51.940
v and capital u so write 1 upon v - beta - 1
upon u - alpha = 1 upon half
32:51.940 --> 33:02.280
on rearranging this again we have a equation
having terms like uv u multiplied by - beta
33:02.280 --> 33:15.390
– f v multiplied by - alpha + f and alpha
beta - beta + alpha =0 we can compare this
33:15.390 --> 33:20.080
equation with the only a similar equation
33:20.080 --> 33:27.690
and that gives us these relationship 1 upon
f comes out to be equal to 1 upon f1 +s 1
33:27.690 --> 33:37.480
upon f2 - d upon f1 f2 so this is the focal
length of the combination and alpha beta alpha
33:37.480 --> 33:54.411
is df upon f2 df 1 upon have 1 + f2 - d and
beta is - df upon f1 so -df2 –f1 f2 – d
33:54.411 --> 34:01.200
so now the idea is the two lens are not to
be treated separately anymore we have just
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one optical system object image distances
are to be measured from the principal points
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and we have the focal length equivalent focal
length of the combination for the power of
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the combination naturally we have this relationship
the total power is equal to p1 + p2 - d times
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d the distance between those two thin lenses
so d times p1 p2
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now we are in a position to consider the thickness
consider lens of thickness t and made from
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a material of refractive index mu and placed
in sir radii of curvature r1 and r2 consider
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look at the thickness between the points p
and pq that is t a luminous point object o
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is situated on the axis at a distance u from
the first reflecting surface and forms an
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image at i prime at a distance v prime from
p this image will serve as a virtual object
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the second surface
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so for the reflection from the first surface
one can write mu upon v prime - 1 upon u which
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is equal to mu - 1 upon r1 that is tendered
relation for reflection from a spherical surface
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the image i prime formed by the first surface
acts as the object for the second surface
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and he finally made it formed a i so here
you have the relationship 1 upon mu divided
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by v - 1 upon v prime -t whose p is coming
because it is a thick lens is equal to 1 upon
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mu – 1divided by r2 r2 the radius of curvature
of the second surface
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now v prime is to be eliminated between these
two equations to obtain an equation in uv
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r1 and r2
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and you now if alpha is the distance of the
first principal point from p and beta in the
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distance of the second principal point from
q then taking the capital v = v - beta capital
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u = u -l we have the lens equation as before
1 upon v - beta - 1 upon u - alpha = 1 upon
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m again substituting those relations as before
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and comparing this equation with the earlier
equation in u v i have alpha beta simpler
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as we did for the combination of two thin
lenses we get these results so 1 upon half
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= mu - 1 r1 - 1 upon r1 - 1 upon r2 + mu - 1
into t divided by mu r1 r2 that is the and
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defining the focal length remember if t =0
if it is the thin lens then this term will
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not be there you will have a simple relationship
mu - 1 into 1 upon r1- 1 upon r2
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for the thick lens this is the additional
term which is coming up similarly for the
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distances of the principal points alpha and
beta if t is 0 you do not have to bother about
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them but the center of the lens is the only
point but now for the thick lens alpha is
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given by this relationship theta is given
by the next one similar to those for the thin
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lens
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now as t goes to 0 as i pointed out the lens
thin the above reduce to the usual relation
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for the focal length of a thin lens okay with
this we have come to the end of this lecture
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now in the next two lectures we shall take
up the defects in the image formation that
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is the lens aberrations thank you