Transcript Chapter 5: Geometrical Optics

Chapter 5 Geometrical optics
January 21,23 Lenses
5.1 Introductory remarks
Image: If a cone of rays emitted from a point source S arrives at a certain point P, then P
is called the image of S.
Diffraction-limited image:
The size of the image for a point source is not zero. The limited size of an optical system
causes the blur of the image point due to diffraction effect:
d l
f
D
Geometrical optics:
When l <<D, diffraction effects can be neglected, and light propagates on a
straight line in homogeneous media.
Physical optics:
When l ~ D or l>D, the wave nature of light must be considered.
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5.2 Lenses
Lens: A refracting device that causes each diverging wavelet from an object to converge
or diverge and to form the image of the object.
Lens terminology:
• Convex lens, converging lens, positive lens
• Concave lens, diverging lens, negative lens
•
•
•
•
•
S
Focal points
Real image: Rays converge to the image point
Virtual image: Rays diverge from the image point
Real object: Rays diverge from the object point
Virtual object: Rays converge to the object.
P
S
P
P
S
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5.2.1 Aspherical surfaces
Determining the shape of the surface of a lens: The optical path length (OPL) from the
source to the output wavefront should be a constant.
Example: Collimating a point source (image at infinity)
OPL  ni FA  nt AD  constant
y
 FA  nti AD  constant
A(x,y)
D
 x 2  y 2  nti (d  x)  constant
F
 x 2  y 2  nti x  c
 x  y  n x  2cnti x  c
2
2
2
ti
2
d
ni
2
x
nt
 (nti2  1) x 2  2cnti x  y 2 c 2  0
The surface is a hyperboloid when nti>1, and is an ellipsoid when nti<1.
Example: Imaging a point source. The surface is a Cartesian oval.
(n
2
ti
 1)( x 2  y 2 )  2nti2dx  nti2d 2  c 2

2
 4c 2 ( x 2  y 2 )
Aspherical lens can form perfect image, but is hard to manufacture.
Spherical lens cannot form perfect image (aberration), but is easy to manufacture.
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5.2.2 Refraction at spherical surfaces
Terminology: vertex, object distance so, image
distance si, optical axis.
lo
S
qi
A
R qt li
j
V
P
C
so
si
n
n1
2

so  R
lo 


sin(   q i ) sin j 
so  R
si  R 

n1 n2 1  n2 si n1so 

 n1 ( so  R) n2 ( si  R)




  

si  R
li
l
sin
q
l
sin
q

o
i
i
t


l
l
l
l
R
l
l
o
i
o
i
o 
 i

sin q t sin(   j ) 

n1 sin q i  n2 sin q t 
Gaussian (paraxial, first-order) optics: When j is small, cosj ≈1, sinj ≈j,
lo  ( so  R ) 2  R 2  2( so  R ) R cos j  so
li  ( si  R ) 2  R 2  2( si  R ) R cos j  si
 Paraxial imaging from a single spherical surface:
n1 n2 n2  n1
 
so si
R
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Paraxial imaging from a single spherical surface:
n1 n2 n2  n1
 
so si
R
Note:
1) This is the grandfather equation of many other equations in geometrical optics.
2) For a planar surface (fish in water): si  
3) Magnification: M T 
n2
so . (A bear needs to know this.)
n1
yi
ns
  1 i . (P5.6).
yo
n2 so
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Object (first) focal length: when si = ,
Fo
n1
f o  so 
R.
n2  n1
Image (second) focal length: when so = ,
f i  si 
fo
n2
R.
n2  n1
Fi
fi
Virtual image (si< 0) and virtual object (so< 0):
EVERYTHING HAS A SIGN!
Sign convention for lenses
(light comes from the left):
• so, fo
+ left of vertex
• si, fi
+ right of vertex
• xo
+ left of Fo
• xi
+ right of Fi
•R
+ curved toward left
• yo, yi + above axis
Fi C
V
si
C
Fo
V
so
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Read: Ch5: 1-2
Homework: Ch5: 1,5,6
Note: In P5.1 the expression should be (so+si-x)2.
Due: January 30
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January 26, 28 Thin lenses
5.2.3 Thin lenses
Thin lens: The lens thickness is negligible compared to object distance and image distance.
Thin lens equations:
nm
Forming an image with two spherical
surfaces:
P'
2nd surface
1st surface
S
P'
(R1, nm, nl)
(R2, nl, nm)
V1
S
P
R1
R2
C2
P
V2
nl
C1
so1
si1
d
so2
si2
nm nl nl  nm 
 
so1 si1
R1 



nl d
nl nm nm  nl   nm  nm  (nl  nm ) 1  1  
 R R  (s  d )s
For the 2nd surface :


s
s
o
1
i
2
2 
i1
i1
 1
so 2 si 2
R2 

so 2 | si1 |  d   si1  d 
For the 1st surface :
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1 1 
nm nm
nl d

 (nl  nm )   
so1 si 2
 R1 R2  ( si1  d ) si1
If the lens is thin enough, d → 0.
Assuming nm=1, we have the thin lens equation:
1 1 
1 1
  (nl  1)  
so si
 R1 R2 
lim si  lim so  f 

so 
si 
Lens maker’s equation:
1
1
1 
 (nl  1)  
f
 R1 R2 
Gaussian lens formula:
1 1 1
 
s o si
f
Remember them together
with the sign convention.
Question: what if the lens is in water?
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Optical center:
All rays whose emerging directions are parallel to their incident directions pass through
one special common point inside the lens. This point is called the optical center of the lens.
Proof:
incident// emergent  θi  θt  θt 'θi '
 sin θi 
 sin θt ' 
 θi  arcsin 

θ
'

arcsin
 t


 n 
 n 
C O BC 2 R2
 θt '  θi  AC1// BC 2  2 

OC1 AC1 R1
qi
C2 A q t
Oq i '
R2
C1
B
qt '
R1
 O does not depend on A and B.
Conversely, rays passing through O refract parallelly.
C2O R2 BC 2
C2O OC1 rule of sine
Proof :




 θi '  θt  incident// emergent
OC1 R1 AC1
BC 2 AC1
 For a thin lens, rays passing through the optical center are straight rays.
Corollary: For a thin lens, with respect to the optical center, the angle subtended by
the image equals the angle subtended by the object.
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Focal plane: A plane that contains the focal point and is perpendicular to the optical axis.
In paraxial optics, a lens focuses any bundle of parallel rays entering in a narrow cone
onto a point on the focal plane.
Proof: 1st surface, 2nd surface
C’
C
Focal
plane
Fi
Focal
plane
S
C
P
Image
plane
Image plane:
In paraxial optics, the image formed by a lens of a small planar object normal to the
optical axis will also be a small plane normal to that axis.
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Read: Ch5: 2
Homework: Ch5: 7,10,11,15
Due: February 6
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January 30 Ray diagrams
Finding an image using ray diagrams:
Three key rays in locating an image point:
1) Ray through the optical center: a straight
line.
2) Ray parallel to the optical axis: emerging
passing through the focal point.
3) Ray passing through the focal point:
emerging parallel to the optical axis.
S'
yo
Fo
S
2
A
1
O
Fi
P
3
yi
B
P'
f
xo
xi
f
so
si
Newtonian lens equation:
xo  so  f
xi  si  f
Meanings of the signs:
yo
f
x
  o  xo xi  f 2
| yi | xi
f
+
Real object
Real image
Converging
Erect object
Erect image
Erect image
Transverse magnification: M T 
Longitudinal magnification:
dxi
f2
ML 
  2   M T2
dxo
xo
yi
s
 i
yo
so
• so
• si
•f
• yo
• yi
• MT
–
Virtual object
Virtual image
Diverging lens
Inverted object
Inverted image
Inverted image
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Thin lens combinations
4
I. Locating the final image of L1+L2 using
F
Fi2
O1
O2 i1
ray diagrams:
Fo1 Fo2
1) Constructing the image formed by L1 as
if there was no L2.
2) Using the image by L1 as an object (may
d<f1, d<f2
so2
d
be virtual), locating the final image. The
si1
ray through O2 (Ray 4, may be
backward) is needed.
1
1
1 
II. Analytical calculation
 

s
s
f
of the image position:
o1
i1
1 
1
1
1 
1
1
1
si2 is a function of (so1, f1, f2, d)

 
 
so 2 si 2 f 2  si 2 f 2 d  f1so1
so1  f1

so 2  si1  d 
Total transverse magnification:
 s  s 
f1si 2
M T  M T 1M T 2    i1   i 2  
 so1  so 2  d ( so1  f1 )  so1 f1
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Back focal length (b.f.l.): Distance from the last surface to the 2nd focal point of the system.
Front focal length (f.f.l.): Distance from the first surface to the 1st focal point of the system.
1
1

b.f.l. si 2
1
1

f.f.l. so1

so1  

si 2  
1
1

f 2 so 2
1 1

f1 si1

so 1  

si 2  
1
1

f 2 d  si1
1
1

f1 d  so 2

1
1

f 2 d  f1

1
1

f1 d  f 2
so 1  
si 2  
Special cases:
1) d = f1+f2: Both f.f.l. and b.f.l. are infinity. Plane wave in, plane wave out (telescope).
1 1 1
1
1
 


2) d → 0: effective focal length f:
f
f1 f 2
f.f.l. b.f.l.
3) N lenses in contact:
1 1 1
1
  
.
f
f1 f 2
fN
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Read: Ch5: 2
Homework: Ch5: 20,25,26,32,33
Due: February 6
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February 2 Mirrors and prisms
5.4 Mirrors
5.4.1 Planar mirrors
1) |so|=|si|.
2) Sign convention for mirrors: so and si are positive when they lie to the left of the
vertex.
3) Image inversion (left hand  right hand).
5.4.3 Spherical mirrors
The paraxial region (y<<R):
y
( R  x )2  y 2  R 2  x  R  R 2  y 2
y2
y4

 3 
2 R 8R
 y 2  2 Rx  4 fx
x
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The mirror formula:
SC CP

SA PA
(paraxial)

A
1 1
2
so  R  si  R
 


so si
R
so
si
qi q
S
f
F
C
P
R 1 1 1
fo  fi   ,
 
2 so si f
V
f
si
R
so
Transverse magnification:
1
yi
si


1
2
1
R
MT 

      
.
yo
so
so  R so 
2 so  R
Ray diagrams of mirrors:
Four key rays in finding an image point:
1) Ray through the center of curvature.
2) Ray parallel to the optical axis.
3) Ray through the focal point.
4) Ray pointing to the vertex.
2
4
S
1
F
3
V
C
P
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5.5 Prisms
Functions of prisms:
1) Dispersion devices.
2) Changing the direction of a light beam.
3) Changing the orientation of an image.


qi1
qt1 qi2
qt2
5.5.1 Dispersion prisms
Apex angle, angular deviation
  (θi1-θt1 )  (θt 2 -θi 2 )
    θi1  θt 2  
  θt 1  θ i 2

θt 2  arcsin( n sin θi 2 )  arcsin n sin(   θt1 )
 arcsin[ n(sin  cos θt1  cos  sin θt1 )]
 arcsin[sin  n 2  sin 2 θi1  cos  sin θi1 ]
  θi1  arcsin[sin  n 2  sin 2 θi1  cos  sin θi1 ]  
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Minimum deviation:
  θi1  θt 2  
qi1  qt 2
dqt 2
d
 1
 0 
dqi1
dqi1
qt1  qi 2
The minimum deviation ray traverses the prism symmetrically.
At minimum deviation,
  θt 1  θi 2 


m 

θ



t1
sin
θt 1  θ i 2
2
sin θi1


2

n 

sin θt1
 m  θi1  θt 2   
m  
sin
  θi1 
2
θi1  θt 2
2 

This is an accurate method for measuring the refractive indexes of substances.
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Read: Ch5: 3-5
Homework: Ch5: 54,60,61,64,65,67,68
Due: February 13
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