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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)
SVsO )object distance(
VPsi )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.
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.
Let’s return to Fermat’s Principal
io lnlnOPLwithOPLdx
d210
Using the Law of cosines:
cos)cos(
cos2
cos2
cos2
2/122
2/122
222
with
RsRRsRland
RsRRsRl
RsRRsRl
iii
ooo
ooo
2/122
2
2/1221
21
cos2
cos2
,
RsRRsRn
RsRRsRnOPL
lnlnOPLwithTherefore
ii
oo
io
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
0sin)(2cos22
1
sin2cos22
1)(
2/1222
2/1221
RsRRsRRsRn
RsRRsRRsRnd
OPLd
iii
ooo
We can express this result in terms of the original variables lo and li:
o
o
i
i
io
i
i
o
o
l
sn
l
sn
Rl
n
l
n
andl
RsRn
l
RsRn
1221
21
1
0sinsin
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 .
R
nn
s
n
s
nthen
io
1221
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 Snell’s law: n1sin 1= n2sin2 and used sin .
Again, subscripts o and i refer to object and image locations, respectively.
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
Rnn
nf
R
nn
f
n
R
nnn
s
nthen
oo
o
12
1121
1221
Suppose now that plane waves (parallel rays) are incident from a point emitting light from a point very far away (i.e., so = ).
ii) Plane waves refracted into spherical waves.
Rnn
nsf
R
nn
s
n
R
nn
s
nnswhen
iii
io
12
2122
1221
Diverging rays revealing a virtual image point using concave spherical surfaces.
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.
R < 0
fi < 0
si < 0
Signs of variables are important.
Rnn
nsf ii
12
2
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 < 0 here.
012
1121
1221
Rnn
nsf
R
nn
s
n
R
nn
s
n
s
nswhen
ooo
ioi
The combination of various surfaces of thin lenses will determine the signs of the corresponding spherical radii.
S
)a(
)b(
)c(
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.
We will use virtual image points to locate conjugate image points.
In the paraxial approximation:
R
nn
s
n
s
nfrom
R
nn
s
n
s
n
io
ml
i
l
o
m 1221
111
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
dssssssdss ioiiooio 12112212 ,,,
)A(
)B(Thus, at the 2nd surface:221 R
nn
s
n
ds
n lm
i
m
i
l
Add Equations (A) & (B)
112121
11
ii
lml
i
m
o
m
sds
dn
RRnn
s
n
s
n
Let d 0 (this is the thin lens approximation) and nm 1:
)(11
111
21
cRR
nss l
io
and is known as the thin-lens equation, or the Lens maker’s formula ,
in which so1 = so and si2 = si, V1 V2, and d 0. Also note that
oos
iis
fsfsio
lim,lim
For a thin lens (c) fi = fo = f and
fss
RRn
f
io
l
111
111
1
21
Convex f > 0
Concave f < 0
Also, known as the Gaussian lens formula
Location of focal lengths for converging and diverging lenses
1m
llm n
nn 1
m
llm n
nn
If a lens is immersed in a medium
with
21
111
1
RRn
f lm
m
llm n
nn
f2f
2ffObject
Real image
Convex thin lens
Simplest example showing symmetry in which so = 2f si = 2f
Concave, f < 0, image is upright and virtual, |si| < |f|
f
fObject
Virtual image
2
3
1siso
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.
For case (b) below
1,1
111111
fswheref
fs
sfs
sfsfss
oo
oi
oiio
0, ii sfs
Tracing a few key rays through a positive and negative lens
yo
S2
S1
Consider the Newtonian form of the lens equations.
From the geometry of similar triangles:
fxfxfssfy
y
s
s
ioioi
o
i
o
111111,
22
1
fxxfxx
ff
fxx
fxf
fxfxf
ffx
ioio
io
o
io
o
Newtonian Form:
xo > 0 if the object is to the left of Fo .
xi > 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.
Define Transverse (or Lateral) Magnification:
f
x
x
f
xfx
xxff
fx
fxf
fx
fx
s
s
y
yM
i
o
oo
oo
o
o
o
i
o
i
o
iT
1//2
2f f
f 2f
Image forming behavior of a thin positive lens.
MT > 0 Erect image and MT < 0 Inverted image. All real images for a thin lens will be inverted.
f2f
2ffSimplest example 2f-2f conjugate imaging gives
1f
f
x
fM
oT
Define Longitudinal Magnification, ML
022
22
To
Lo
io
iL M
x
fM
x
fxand
dx
dxM
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.
The number-2 ray entering the lens parallel to the central axis limits the image height.
The transverse magnification (MT) is different from the longitudinal magnification (ML).
Image orientation for a thin lens:
)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.
Two thin lenses separated by a distance smaller than either focal length.
Note that d < 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.
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.
Note that d > si1, so that the object for Lens 2 (L2) is real.
11
112
11
1122
21
21
22
222
222
2
2
12
11
111
111
,111
)(0
)(0
,111
fsfs
fd
fsfsf
df
fsd
fsd
fs
fss
sfs
reals
virtualssds
fs
fss
sfs
o
o
o
o
i
i
o
oi
oi
o
o
io
o
oi
oi
For the compound lens system, so1 is the object distance and si2 is the image distance.
The total transverse magnification (MT) is given by
1111
21
2
2
1
121 fsfsd
sf
s
s
s
sMMM
oo
i
o
i
o
iTTT
For this two lens system, let’s 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
21
211
211112
2
11111
ffd
fdfsffl
fdfsfS i
i
soiso
From the previous slide, we calculated si2. Therefore, if so1 we get,
21
12
12
12
12
12
1222
111
,0
fff
fff
fffflbfldfor
ffd
fdf
ffd
ffdfsbfl
ef
ef
i
fef = “effective focal length”
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.
1 2 3……… NThen, in the thin lens approximation:
Nef fffff
1...
1111
321
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.
Example A Example B
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.
111
111
fss io
si1 = -30 cm at (O’) which is virtual and erect
Then so2 = |si1| + d = 30 cm + 6 cm = 36 cm
222
111
fss io
si2 = i’ = +26 cm at I’ Thus, the image is real and inverted.
The magnification is given by
17.236
26
10
30
2
2
1
121
o
i
o
iTTT s
s
s
sMMM
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.
111
111
fss io
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
222
111
fss io
si2 = i’ = +25 cm; The magnification is given by
57.314
25
18
36
2
2
1
121
o
i
o
iTTT s
s
s
sMMM
Thus, if yo1 = 1 cm this gives yi2 = -3.57 cm
Image is real and Inverted