Chapter 31: Images and Optical
Instruments
Reflection at a plane surface
 Image
formation
The reflected rays entering eyes look
as though they had come from image P’.
virtual
P
image
P’
Light rays radiate from a point object
at P in all directions.
Reflection and refraction at a plane surface
 Image
formation
Reflection and refraction at a plane surface
 Image
formation
i (or s’) is the image distance
s is the object distance:
|s| =|i|
s’
Sign Rules:
(1) Sign rule for the object distance:
When object is on the same side of the reflecting
or refracting surface as the incoming light, the object
distance s is positive. Otherwise it is negative.
(2) Sign rule for the image distance:
When image is on the same side of the reflecting or
refracting surface as the outgoing light, the image
distance i ( or s’) is positive. Otherwise it is negative.
(3) Sign rule for the radius of curvature of a spherical
surface:
When the center of curvature C is on the same side
as the outgoing light, the radius of the curvature is
positive. Otherwise it is negative.
Reflection at a plane surface
 Image
formation
image is erect
image is virtual
h
Multiple image due to multiple
Reflection by two mirrors
h’
m = h’/h=1
lateral magnification
Reflection at a plane surface
 Image
formation
When a flat mirror is rotated, how
much is the image rotated?
Reflection at a spherical mirror
 Concave
and convex mirror
Reflection at a spherical mirror
 Focal points at concave and convex mirror
Focal point or focus: Point F at which rays from a source point are
brought together (focused) to form an image.
Focal length: Distance f from mirror where focus occurs.
f=R/2 where R is the radius of a spherical mirror.
Reflection at a spherical mirror
 Focal points at a concave mirror
   
   
    2
tan   h /( s  d )
tan  h /( s'd )
h
object
d
image
If
s  , s'  R / 2
s’
tan  h /( R  d )
  h/s
  h / s' if d  s, s'
  h/R
1 1 2 1
  
s s' R f
Reflection at a spherical mirror
 Image
of an extended object at a concave mirror
real image
Principle rays: Light rays that can be traced (more easily) from the source
to the image:
1. Parallel to optical axis
2. Passing through the focal point
3. Passing through the center of curvature
4. Passing through the center of the mirror surface or lens
Reflection at a spherical mirror
 Magnification
of image at a concave mirror
h
h’
h'
s'
f
m  
h
s
f s
When s,s’ >0 , m<0 inverted
s/s’<0, m>0 upright or
erect
Reflection at a spherical mirror
 Example
with a concave mirror
real image
real image
real image
virtual image
Reflection at a spherical mirror
 Example
with a concave mirror
Reflection at a spherical mirror
 Image
at a convex mirror
1 1 2 1
  
s s' R f
s
s’
f
f
s'
f
m 
s f s
R
s positive
s’ negative (virtual image)
R negative
f negative
Reflection at a spherical mirror
 Magnification
of image at a convex mirror
height at s height at s'

s
s'
For a convex mirror f < 0
height at s s' 1 1 1
m
 ,  
height at s' s s s' f
s’

s'
f
m 
s f s
m > 1 magnified m < 1 minimized
m > 0 image upright
m < 0 image inverted
Refraction at a spherical surface
 Refraction
at a convex spherical surface
1
For small angles
12
sin     n11  n22
BF (1   2 )  AB  f  AB /(1   2 )
AB  R1  R1 /(1   2 )  Rn2 /(n2  n1 )
1  1
f (
n2
)R
n2  n1
Refraction at a spherical surface
 Refraction
at a concave spherical surface
For a concave surface, we can use the same formula
n2
f (
)R
n2  n1
But in this case R < 0 and f < 0. Therefore the image is virtual.
Refraction at a spherical surface

Relation between source and image distance
at a convex spherical surface
s’
1      2     n1 (    )  n2 (    )
AB  R AB  s  s' 
Snell’s law
AB AB
AB AB
n n
n n

)  n2 (

) 1  2  2 1
R
s
R
s'
s s'
R
For a convex (concave) surface, R >(<) 0.
 n1 (
Refraction at a spherical surface

Example of a convex surface
|s’|
Refraction at a spherical surface

Example of a concave surface
|s’|
Refraction at a spherical surface

Example of a concave surface
Refraction at a spherical surface

Example of a concave surface
Convex Lens

Sign rules for convex and concave lens:
Sign Rules:
(1) Sign rule for the object distance:
When object is on the same side of the reflecting
or refracting surface as the incoming light, the object
distance s is positive. Otherwise it is negative.
(2) Sign rule for the image distance:
When image is on the same side of the reflecting or
refracting surface as the outgoing light, the image
distance i (or s’) is positive (real image). Otherwise it is negative
(virtual image).
(3) Sign rule for the radius of curvature of a spherical
surface:
When the center of curvature C is on the same side
as the outgoing light, the radius of the curvature is
positive. Otherwise it is negative.
Convex Lens

Lens-makers (thin lens) formula
surface 2
surface 1
s’
Image due to surface 1:
s’1 becomes source s2 for
1 n n 1
1 n 1 1
 ' 
 ' 

s1 s1
R1
s1 nR1 ns1
n 1 1 n
1
1 1 n
 ' 
 ' ' 
surface 2: s2 s2 R2
 s1 s2
R2
n( 
s1 = s and s’2 = s’:
R1>0
R2<0
n 1 1
1 1 n

) ' 
nR1 ns1
s2
R2
1 1
1
1
1
  (n  1)(  ) 
s s'
R1 R2
f
Parallel rays (s=inf.)
w.r.t. the axis converge
at the focal point
Convex Lens

Magnification
s’
m  II ' / SS'
SS' P  II ' P
m
s'
f

s f s
s'
m
s
same as for mirrors
Convex Lens

Object between the focal point and lens
A virtual image
Convex Lens

Object position, image position, and magnification
real inverted image
m<1
real inverted image
m >1
virtual erect image
m >1

Types of lens
Lens
Lens

Two lens systems
Lens

Two lens systems (cont’d)
Lens

Two lens systems (cont’d)
Lens

Two lens systems (cont’d)
Eyes

Anatomy of eye
Eyes

Near- and far-sightedness and corrective lenses
farsightedness
nearsightedness
Angular size
h
d
h
s 
d
i
m 
s
In general the minimum distance d=dmin~25 cm at
which an eye can see image of an object comfortably
and clearly.
Magnifying glass
s’
hi
i
s
h
virtual image
1 1 1
 
s' f s
the eye is most relaxed
 s'
m


when s  f  s'   but
s
the minimum distance at
which an eye can see image
of an object comfortably and clearly.
hi m h h
h
tan i   i 

 ( )s f 
| s' | | s' |
s
f

h/ f
d
M  m  i 
 min , d min  25 cm for human eye.
 s h / d min
f
Microscope
2i
L  i  f1 , f 2
f1 , f 2 small
magnifier
h1 | m | h0  Lh0 / f1 ( | m | i / s1  L / s1  L / f1 )
Object is placed
near F1 (s1~f1).
Image by lens1
is close to the
focal point of
lens2 at F2.
image ang. size
 2i  h1 / f 2  ( Lh0 ) /( f1 f 2 ) ( i  hi / i  h / f for a magnifier)
M  m   2i /  object  [(Lh0 ) /( f1 f 2 )] /[h0 / d min ]  ( Ld min ) /( f1 f 2 )
Refracting telescope
Image by lens1 is at its focal point which is
the focal point of lens 2
image distance
after lens1
h1  m h0  s' h0 / s  s' s  f1 s ( s  ) image height by lens1 at its focal point
magnifier
 i  h1 / f 2   s f1 / f 2
m   i /  object   i /  s  f1 / f 2
angular size of image by lens2; eye
is close to eyepiece
Reflecting telescope
m  i / object  f1 / f 2
Aberration
sphere
paraboloid
Chromatic aberration
Gravitational lens
Problem 1
Exercises
What is the size of the smallest vertical plane mirror in which a woman
of height h can see her full-length?
Solution
x
The minimum length of mirror for
a woman to see her full height h
Is h/2 as shown in the figure right.
x/2
(h-x)/2
h-x
Exercises
Problem 2 (focal length of a zoom lens)
ray bundle
r0
f2=-|f2|
f1
Q
r0
x
f1
I’
r’0
d
s2
d (variable)< f1
|f2|>f1-d
s’2
f
'
(a) Show that the radius of the ray bundle decreases to r0  r0 ( f1  d ) / f1
x  r0  r0'  r0'  r0  x  r0  r0 (d / f1 )  r0 ( f1  d ) / f1
Exercises
Problem 2 (focal length of a zoom lens)
ray bundle
r0
f2=-|f2|
f1
Q
f1
I’
r’0
r0
x
d
s2
s’2
d (variable)< f1
f
(b) Show that the final image I’ is formed a distance
to the right of the diverging lens.
s2  d  f1 
s2'  f 2 ( f1  d ) /( f 2  f1  d )
f  f1  d
f ( f  d)
1
1 1
1 d  f1  f 2
 '   ' 
 2
 s2'  2 1
d  f1 s2 f 2
s2
f 2 (d  f1 )
f 2 ( f1  d )
f 2  f1  d
Exercises
Problem 2 (focal length of a zoom lens)
ray bundle
r0
f2=-|f2|
f1
Q
r0
x
f1
I’
r’0
d
s2
s’2
d (variable)< f1
f
(c) If the rays that emerge from the diverging lens and reach the final
image point are extended backward to the left of the diverging lens,
they will eventually expand to the original radius r0 at some point Q.
The distance from the final image I’ to the point Q is the effective focal
length of the lens combination. Find the effective focal length.
f 2 ( f1  d )
f1 f 2
r0' r0
r0 '
f1


f

s


f

2
s2'
f
r0'
f1  d f 2  f1  d
f 2  f1  d