Consider Refraction at Spherical Surfaces:

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Consider Refraction at Spherical Surfaces
Starting point for the development of lens
equations Vast majority of quality lenses that
are used today have segments containing spherical
shapes. The aim is to use refraction at surfaces
to simultaneously image a large number of object
points which may emit at different wavelengths.
Point V (Vertex)
(object distance)
(image distance)
?i - Angle of incidence ?t - Angle of
refraction ?r - Angle of reflection
The ray SA emitted from point S will strike the
surface at A, refract towards the normal,
resulting in the ray AP in the second medium (n2)
and strike the point P.
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All rays emerging from point S and striking the
surface at the same angle ?i will be refracted
and converge at the same point P.
Lets return to Fermats Principal
Using the Law of cosines
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Note Si, So, R are all positive variables here.
Now, we can let d(OPL)/d? 0 to determine the
path of least time. Then the derivative becomes
We can express this result in terms of the
original variables lo and li
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However, if the point A on the surface changes,
then the new ray will not intercept the optical
axis at point P. Assume new small vaules of the
radial angle ? so that cos ? ? 1, lo ? so, and li
? si.
Again, subscripts o and i refer to object and
image locations, respectively.
This is known as the first-order theory, and
involves a paraxial approximation. The field of
Gaussian Optics utilizes this approach. Note
that we could have also started with Snells law
n1sin? 1 n2sin?2 and used sin ? ? ? .
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Using spherical (convex) surfaces for imaging and
focusing
i) Spherical waves from the object focus
refracted into plane waves. Suppose that a point
at fo is imaged at a point very far away (i.e.,
si ?). so ? fo object focal length
Object focus
ii) Plane waves refracted into spherical waves.
Suppose now that plane waves (parallel rays) are
incident from a point emitting light from a point
very far away (i.e., so ?).
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Diverging rays revealing a virtual image point
using concave spherical surfaces.
R lt 0 fi lt 0 si lt 0
Signs of variables are important.
Virtual image point
Parallel rays impinging on a concave surface. The
refracted rays diverge and appear to emanate from
the virtual focal point Fi. The image is
therefore virtual since rays are diverging from
it.
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A virtual object point resulting from converging
rays. Rays converging from the left strike the
concave surface and are refracted such that they
are parallel to the optical axis. An object is
virtual when the rays converge toward it. so lt 0
here.
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The combination of various surfaces of thin
lenses will determine the signs of the
corresponding spherical radii.
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As the object distance so is gradually reduced,
the conjugate image point P gradually changes
from real to virtual. The point P indicates the
position of the virtual image point that would be
observed if we were standing in the glass medium
looking towards S.
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We will use virtual image points to locate
conjugate image points.
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In the paraxial approximation
(A)
The 2nd surface sees rays coming towards it
from the P (virtual image point) which becomes
the 2nd object point for the 2nd
surface. Therefore
(B)
Thus, at the 2nd surface
Add Equations (A) (B) ?
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Let d ? 0 (this is the thin lens approximation)
and nm ? 1
and is known as the thin-lens equation, or the
Lens makers formula, in which so1 so and si2
si, V1 ? V2, and d ? 0. Also note that
For a thin lens (c) ? fi fo f and
Convex ? f gt 0 Concave ? f lt 0
Also, known as the Gaussian lens formula
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Location of focal lengths for converging and
diverging lenses
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If a lens is immersed in a medium with
f
Object
2f
Real image
f
2f
Convex thin lens
Simplest example showing symmetry in which so
2f ? si 2f
Concave, f lt 0, image is upright and virtual,
si lt f
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Note that a ray passing through the center is
drawn as a straight line.
Ideal behavior of 2 sets of parallel rays all
sets of parallel rays are focused on one focal
plane.
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For case (b) below
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Tracing a few key rays through a positive and
negative lens
Consider the Newtonian form of the lens equations.
S2
yo
S1
From the geometry of similar triangles
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Newtonian Form xo gt 0 if the object is to the
left of Fo. xi gt 0 if the image is to the right
of Fi.
The result is that the object and image must be
on the opposite sides of their respective focal
points.
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Define Transverse (or Lateral) Magnification
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Image forming behavior of a thin positive lens.
f 2f
2f f
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MT gt 0 ? Erect image and MT lt 0 ? Inverted image.
All real images for a thin lens will be inverted.
Simplest example 2f-2f conjugate imaging gives
Define Longitudinal Magnification, ML
This implies that a positive dxo corresponds to a
negative dxi and vice versa. In other words, a
finger pointing toward the lens is imaged
pointing away from it as shown on the next slide.
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The number-2 ray entering the lens parallel to
the central axis limits the image height.
Image orientation for a thin lens
The transverse magnification (MT) is different
from the longitudinal magnification (ML).
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(a) The effect of placing a second lens L2 within
the focal length of a positive lens L1. (b) when
L2 is positive, its presence adds convergence to
the bundle of rays. (c) When L2 is negative, it
adds divergence to the bundle of rays.
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Two thin lenses separated by a distance smaller
than either focal length.
Note that d lt si1, so that the object for Lens 2
(L2) is virtual.
Note the additional convergence caused by L2 so
that the final image is closer to the object. The
addition of ray 4 enables the final image to be
located graphically.
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Note that d gt si1, so that the object for Lens 2
(L2) is real.
Fig. 5.30 Two thin lenses separated by a
distance greater than the sum of their focal
lengths. Because the intermediate image is real,
you could start with point Pi and treat it as if
it were a real object point for L2. Therefore, a
ray from Pi through Fo2 would arrive at P1.
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For the compound lens system, so1 is the object
distance and si2 is the image distance.
The total transverse magnification (MT) is given
by
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For this two lens system, lets determine the
front focal length (ffl) f1 and the back focal
length (bfl) f2. Let si2 ? ? then this gives so2
? f2. so2 d si1 f2 ? si1 d f2 but
From the previous slide, we calculated si2.
Therefore, if so1 ? ? we get,
fef effective focal length
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Suppose that we have in general a system of N
lenses whose thicknesses are small and each lens
is placed in contact with its neighbor.
Then, in the thin lens approximation
Fig. 5.31 A positive and negative thin lens
combination for a system having a large spacing
between the lenses. Parallel rays impinging on
the first lens enable the position of the bfl.
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Example B
Example A
Example A Two identical converging (convex)
lenses have f1 f2 15 cm and separated by d
6 cm. so1 10 cm. Find the position and
magnification of the final image.
si1 -30 cm at (O) which is virtual and erect
Then so2 si1 d 30 cm 6 cm 36 cm
si2 i 26 cm at I
Thus, the image is real and inverted.
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The magnification is given by
Thus, an object of height yo1 1 cm has an image
height of yi2 -2.17cm
Example B f1 12 cm, f2 -32 cm, d 22
cm An object is placed 18 cm to the left of the
first lens (so1 18 cm). Find the location and
magnification of the final image.
si1 36 cm in back of the second lens, and thus
creates a virtual object for the second lens. so2
-36 cm 22 cm -14 cm
Image is real and Inverted
si2 i 25 cm The magnification is given
by
Thus, if yo1 1 cm this gives yi2 -3.57 cm