2nd & 3th N.U.T.S. Workshops
Gulu University
Naples FEDERICO II University
3 – Imaging
(thin lenses and spherical mirrors)
2nd & 3th NUTS Workshop ( Jan 2010)
Mirrors: Importance of the Shape
Plane Mirror
Spherical Mirror
The same scene as seen
reflected by:
 plane mirror
spherical mirror 
3- Imaging
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2nd & 3th NUTS Workshop ( Jan 2010)
Spherical Mirrors
Spherical mirrors appear to be cut from a section of a sphere.
They can be concave or convex.
Each obeys the law of reflection
but each kind of mirror makes a different kind of image
Center of curvature
Axis of the mirror
or
Optical Axis
Center or vertex of the mirror
CONCAVE
3- Imaging
CONVEX
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2nd & 3th NUTS Workshop ( Jan 2010)
Beams from a Distant Light Source are Parallel
Here the beams from a distant light source
such as the sun or a star
By the time they arrive here (into a
camera or mirror) only the nearby
almost parallel beams enter
Concave mirror
Concave mirror
Whenever we speak of incoming
parallel beams you can always
Beams which arrive at the mirror
visualize beams from the sun or a star from a close source, like Alex’s
nose, are not almost parallel
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2nd & 3th NUTS Workshop ( Jan 2010)
Type #1 Useful Rays: Parallel to Opt. Axis
Rule #1: all rays parallel to the axis are
reflected such that to cross the focal point
F (type #1 rays)
R/2
C
F
Rule #2: incident rays coming towards
the centre of curvature are reflected back
on themselves (type #2 rays)
The focal point F is halfway between the
curvature centre C and the centre of the
mirror. The focal distance is positive for
concave and negative for convex mirror
given the sign convention for distance axis
R/2
F
(f=R/2 concave , f=-R/2 convex)
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C
2nd & 3th NUTS Workshop ( Jan 2010)
Type #3 Useful Rays: Headed for F
Rule #3: incident rays headed for F
(type #3 rays) are reflected so that they
are parallel to the axis (this rule is just
the reverse of Rule #1)
Rule #4: Any ray striking the centre of
the mirror (type #4 rays) reflects
symmetrically about the mirror axis
R/2
C
F
R/2
F
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C
2nd & 3th NUTS Workshop ( Jan 2010)
Spherical Mirror Equation
1
1 1


xO xI f
xI>0 →real image in
front of the mirror
hO
C
xI<0 →virtual image on the
back of the mirror
hI
f
F
xO
xI
Magnification
hI
xI
M

hO
xO
3- Imaging
if M<0 image inverted
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2nd & 3th NUTS Workshop ( Jan 2010)
Images from Concave Spherical Mirrors
Concave Spherical Mirrors Rules:
xI 
xO f
xO  f
 f 0 M
(a) Object beyond C, image is real, reduced, and inverted.
(b) Object between C and F, image is real, magnified, and
inverted.
(c) Object inside F, image is virtual, magnified, and upright.
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xI
xO
2nd & 3th NUTS Workshop ( Jan 2010)
Convex Spherical Mirrors
R
V
C
F
f=-R/2  f <0
1
1 1


xO xI f

1 1 1
 
0
xI f xP
M
3- Imaging
f=R/2
virtual image (in the back side of mirror)
xI
 M>0 (upright) and M<1 (reduced)
xO
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2nd & 3th NUTS Workshop ( Jan 2010)
Image Using Convex Mirror
xO f
xI 
xO  f
R
 f  0
2
The image is always:
- virtual (xI<0, on the mirror back)
- upright (xI<0, xO>0 , M>0)
- reduced in size (|xI|< |xO|, M<1)
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2nd & 3th NUTS Workshop ( Jan 2010)
Concave Mirror Exercise
A technician stands 3.85m (from
the vertex) in front of concave
mirror with focal length = 5.52 m.
a) The location of the tech’s image is?
b) Is tech’s image virtual?
c) What is its magnification ?
d) What is the curvature radius of
the mirror?
The technician moves to 15.00 m from the concave mirror. Answer again
the questions a,b and c.
xI
1
1 1
M




xO
x
x
f
O
I
1
1
1


 0.0786  x I  12.72m (back to the mirror  virtual image)
x I 5.52 3.85
M=xI/xO=+12.72/3.85=+3.30,upright
3- Imaging
R =2f = 11.04 m [ 8.74 m; real; -0.58]
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2nd & 3th NUTS Workshop ( Jan 2010)
More on Spherical Mirror Image
Where is Alex's image when he is between the
mirror centre and it's focal point?
Let's find the image of Alex’s nose
Here is a ray of type 1 from his nose reflecting off
the mirror
Here is a ray of type 3 from his nose reflecting off
the mirror
The image of the nose is at the intersection of
reflected rays of type 1 and type 3. Why?
Axis
Can we use a ray of type 2?
Here is Alex's image
Is it (a) real or (b) virtual? -Magnified or reduced?
From which points can your eye see his nose?
a) From all points (A, B, C)?
b) From A and C, but not B?
c) From B and C, but not A?
d) From A only?
C
e) From C only
3- Imaging
Mirror
surface
Center
Focal
point
A
B
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2nd & 3th NUTS Workshop ( Jan 2010)
Cylindrical Mirrors
Mirrors shaped like a cylinder can make you look fat or skinny!
By a concave mirror, your image is
vertically expanded like a bathroom
magnifying mirror but in one dimension
only. Your image looks skinny
Virtual image on other
side of mirror is
compressed vertically
3- Imaging
By a convex mirror, your image is
compressed vertically like rear view mirror
but in one dimension only.
Your image looks fat
Virtual image on other
side of mirror is
stretched vertically
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2nd & 3th NUTS Workshop ( Jan 2010)
Thin Lenses
A thin lens consists of a piece of glass or plastic, ground so that each
of its two refracting surfaces is a segment of either a sphere or a plane.
A lens is thin when the radii of curvature (of the spheres) are much
bigger than its thickness.
Converging lenses are
Diverging lenses are thickest
thickest in the middle and have
positive focal lengths
at the edges and have negative
focal lengths
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2nd & 3th NUTS Workshop ( Jan 2010)
Converging Lens and Its Focal Length, f
The parallel beams
pass through the lens
and converge at the
focal point.
The parallel rays can come from the left
or right of the lens. A thin lens has two
focal points equidistance from the lens,
corresponding to parallel rays from the
left and from the right.
A thin lens is one in which the distance between the surface of
the lens and the center of the lens is negligible compared to f.
3- Imaging
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2nd & 3th NUTS Workshop ( Jan 2010)
Diverging Lens and Its Focal Length, f
The parallel beams
diverge after passing
through the diverging
lens.
The focal point is the point
where the beams appear to have
originated.
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2nd & 3th NUTS Workshop ( Jan 2010)
Imaging by a Lens
f
xO
xI
Three principal beams (rays) can help you to locate the image:
1. A beam parallel to the lens’ axis on the front side passes through the
focal point on the back side.
2. A beam passing through the lens’ center does not get refracted.
3. A beam passing through the focal point on the front side emerges
parallel to the lens’ axis on the back side.
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2nd & 3th NUTS Workshop ( Jan 2010)
Thin Lenses Equation
Optical axis
xO
xI
1
1 1
 
xO xI f
f > 0 converging lens
3- Imaging
f < 0 diverging lens
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2nd & 3th NUTS Workshop ( Jan 2010)
Thin Lenses Equation: Sign Convention
XI > 0
xO
xI
real image, in the
opposite “space” of the
object with respect to
the lens
XI < 0
xP
3- Imaging
xI
f
virtual image, in the
same “space” of the
object with respect to
the lens
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2nd & 3th NUTS Workshop ( Jan 2010)
Lens magnification
Optical axis
xO
xI
The two shaded triangles, blue and gold, are right-angled and
similar. Therefore, for the lens magnification, M, we have:
h'
x
M  I
h
xO
3- Imaging
M>0 image upright
M<0 image inverted
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2nd & 3th NUTS Workshop ( Jan 2010)
Ray Diagram for Converging Lens #1
If the object is further away from the lens than the front focal
point, xO > f  the image is real, inverted, and located
behind the lens
If the object is located at more than 2f, xO > 2f  the image is
reduced. If the object is located at f < xO < 2f the image is
magnified.
xO f
1
1 1
   xI 
xO xI f
xO  f
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2nd & 3th NUTS Workshop ( Jan 2010)
Ray Diagram for Converging Lens #2
If the object is closer to the lens than the front focal point,
xO<f  the image is virtual, upright, magnified, and in front
of the lens
xO f
xI 
xO  f
h'
xI
M 
h
xO
The image can only be viewed from behind the lens!
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2nd & 3th NUTS Workshop ( Jan 2010)
Ray Diagram for Diverging Lens
The situation is pretty much the same no matter where the object is located.
xO
xI
f
The image is virtual and located in the object space.
The image is upright and is always reduced in size compared with the
object.
xO f
h'
xI
xI 
M 
xO  f
h
xO
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2nd & 3th NUTS Workshop ( Jan 2010)
Lens and Apertures
I
O
I
O
What happens to
the image if we put
in an aperture?
The image becames
less bright !
3- Imaging
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2nd & 3th NUTS Workshop ( Jan 2010)
Masking Half a Lens
I
O
I
O
3- Imaging
What happens to
the image if we
mask the top half
of the lens ?
The image becames
less bright !
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2nd & 3th NUTS Workshop ( Jan 2010)
Given the Object and its Image, Find the Lens!
I
O
I
O
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2nd & 3th NUTS Workshop ( Jan 2010)
Given the Object and its Image, Find the Lens! #2
O
I
O
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I
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2nd & 3th NUTS Workshop ( Jan 2010)
Image Formation with a Lens #1
A light bulb is +56 cm from a converging lens. The image appears on a
screen +31cm on the other side of the lens.
(a) What is the focal length of the lens?
31  56
1
1
1
1
1 1


  

f xO xI
f 56 31 56  31
f  20 cm
(b) What is the magnification factor of the image?
x
31
M   I    0.55
xO
56
Is the image inverted?
3- Imaging
YES!
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2nd & 3th NUTS Workshop ( Jan 2010)
Image Formation with a Lens #2
A lens has a focal length of +35 cm.
(a) Find the type and height of the image when a 2.2 cm high
object is placed at f +10 cm from the lens.
1
1
1
1 1 1

 
 
f xO xI
xI f xO
35  45
f  xO
f ( f  10)

 157.5 cm
xI 

10
xO  f
10
M
xI
157.5

 3.5
xO
45
h I  M  h O  3.5  2.2  7.7 cm
(b) Describe the image you got
The image is real, behind the lens, and inverted.
3- Imaging
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2nd & 3th NUTS Workshop ( Jan 2010)
Image Formation with a Lens #3
A lens has a focal length of +35 cm.
(a) Find the position and height of the image when a 2.2 cm high
object is placed at f -10 cm from the lens.
1
1
1
1 1 1

 
 
f xO xI
xI f xO
xI 
f  xO
35  25
f ( f  10)

 87.5 cm

10
xO  f
- 10
xI
87.5
M

 3.5
xO
25
h I  M  h O  3.5  2.2  7.7 cm
(b) Describe the image you got
The image is virtual, in front of the lens, and upright.
3- Imaging
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2nd & 3th NUTS Workshop ( Jan 2010)
Translucent Screen at the Location of Real Image
1. To see the Alex image your eyes
have to catch beams refracted by
the lens (you have to stay behind the
lens)
2. A translucent screen is put at the
location of the image
3. Beams from each image point are
now scattered by the screen in all
directions
4. Now we can see Alex image from
the front, the side and the back of
the screen
3- Imaging
Beams scattered off translucent
screen particles at each point
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2nd & 3th NUTS Workshop ( Jan 2010)
Meaning of Negative Numbers in the Lens Equation
focal length f
means the lens is diverging.
Negative
• Otherwise it is converging.
Negative
image distance xI
from image to lens means the
image is at the left of the
lens, on same side as object
and virtual (beams coming
from it never really meet)
• Otherwise image is real
1
1 1
 
xO xI f
3- Imaging
Negative
object distance xO
means the object seen by the
lens is on the “wrong” side of
the lens (at the right of lens)
Negative
magnification M
means the image is upside
down (inverted) with respect
to the object.
M
xI
xO
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2nd & 3th NUTS Workshop ( Jan 2010)
The “Power” of a Lens
 Definition of Power, P, in
The power of a converging lens
terms of f is
is always positive because f is
1
P (in diopters) 
a positive number for a
f (meters)
converging lens
• The converging lens always
 Meaning of P
bends light beams towards the
• P is a measure of the beams
bending power of the lens
• Large P means the lens bends
beams more than if P was
small
• The eyeglass or contact lens
prescription is usually given in
diopters (P)
3- Imaging
axis behind the lens
The
power of a diverging
(concave) lens is always
negative because f is negative
for a diverging lens
• The diverging lens always bends
light beams away from the axis
behind the lens
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2nd & 3th NUTS Workshop ( Jan 2010)
Power for a Systems of Two Thin Lenses
For two lenses, of power P1 and P2, in contact to each
other, the total power of the lenses system is:
Ptotal = P1+P2
or in terms of the two lenses focal f1 and f2:
1
f total
3- Imaging
1 1
 
f1 f 2
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2nd & 3th NUTS Workshop ( Jan 2010)
Home Work #1
A candle is +36 cm from a concave mirror with a focal length +15 cm.
(a) Where is the image?
(b) What is the magnification?
(c) Is the image real or virtual, upright or inverted?
Home Work #2
The image of an object in a +27 cm focal length concave mirror is
upright and magnified by a factor of 3. Where is the object?
Home Work #3
A magnifying lens has a focal length of +30 cm. How far from the
page should you hold the lens in order to see the print enlarged 3X?
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