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Topic 5 - Image Formation
Image formation
• Based on 2 simple laws
– Law of reflection
– Snell’s Law
• The two laws can be applied by simply tracing the
geometrical paths of the light rays
• Two types of images
• Real images – the rays from the object after passing
through the optical element really do pass through the
image point. E.g. images formed on photographic film,
on a projection screen and in the eye
• Virtual images – the rays from the object after passing
through the optical element do not pass through the
image point, they only appear to come from the image,
cannot be projected onto a screen, only view through
the optical element
Reflection - Plane Mirror
• Nomenclature
P = object
P’ = image
s = object distance from the mirror
s’ = image distance from the mirror
Where is the image located?
the image point P’ is located
exactly opposite the object point P,
as far behind the mirror as the
object point is in front of the
Textbook – University Physics
mirror
Young and Freedman
Figure 35-4
Page 1087
Extended Object
• Object of finite size
y = object height
y’ = image height
Lateral magnification m
y′
=
y
Upright image m>0
Inverted image m<0
Plane mirror, m = 1
But the image is reversed
Textbook – University Physics
Young and Freedman
Figure 35-5 and 35-6
Page 1087 and 1088
Finding Images
• Lab experiment – how to find images in practice
• Method of No Parallax
Parallax occurs when you move your head from side to
side, things near move faster than those further away
but if they coincide they appear to move together –
then there is no parallax
Reflection at a Concave Mirror
• Application: astronomical telescopes
use large concave mirrors to
photograph celestial objects
• R = radius of curvature
• C = centre of curvature
• V = vertex of the mirror (i.e. centre
of the mirror surface)
• CV = the optic axis
http://surendranath.tripod.com
Textbook –
University Physics
Young and Freedman
Figure 35-9
Page 1089
From geometry and using the paraxial
approximation can obtain
1 1 2
+ =
s s' R
Young and Freedman
Section 35-3
Spherical Aberration
• Hubble Telescope, 1990
• Image “smeared out”
Textbook – University Physics
Young and Freedman
Figure 35-10
Page 1090
Concave Mirror 2
• When the object point P is very far
from the mirror and the incoming rays
are parallel, then the image is formed
at the focal point
• The distance from the vertex to the
focal point is the focal length, f
1 1 2
+ =
s s' R
Case (a)
s = ∞, s ' = f
1
=0
s
1 2
=
f R
R
f =
2
1 1 1
+ =
s s' f
Sign Rules
1 1 1
+ =
s s' f
1) When the 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) When the image is on the same side of the reflecting
or refracting surface as the outgoing light the image
distance s’ is positive (real), otherwise it is negative
(virtual)
3) When the centre of curvature C is on the same side as
the outgoing light the radius of curvature is positive
otherwise it is negative
so for the concave mirror R>0, f>0
Concave Mirror 3
Orientation of the extended image?
From the geometry, similar triangles give the relationship
m=
y'
s'
=−
y
s
Young and Freedman
Page 1092
Minus because the object and image are on opposite sides
of the optic axis
• Can also look at plane mirror with the equation
1 1 2
+ =
s s' R
1
R = ∞, = 0
R
s = − s'
As expected from the sign rules
And this also gives m=1
Question
An object is placed 2.0. cm in front of a concave sherical
mirror whose radius of curvature is 8.0 cm. Locate the
position of the image and its size.
Graphical Approach – Ray Diagrams
• Principal Rays
1) A ray drawn parallel to the optic axis, after reflection
passes through the focal point F of a concave mirror
2) A ray through the focal point F is reflected parallel to
the optic axis
3) A ray through the centre of curvature C intersects
the surface normally and is reflected back along the
original path
4) A ray to the vertex V is reflected forming equal
angles with the optic axis
If the principal rays do not converge at a real image
point, extend them back to locate a virtual image point
as for convex mirror case
Concave Mirror – ray diagrams
a) s > R
The 4 rays converge after
reflection at point Q’
Image is real, inverted and
smaller than the object,
s’ > 0, 0 > m > -1
b) s < f
Rays appear to diverge from Q’,
Virtual image, upright and larger
than object
s’ < 0, m > 1
As in earlier numerical example
Concave Mirror
•
What will be the effect on the image if half of the
mirror surface is covered with a non-reflecting
coating?
a)
b)
c)
d)
No effect
Can’t see any image
Can only see half the object at the image
The image is of the whole object but less bright
Convex Mirror - Summary
• See text book
• Focus is virtual
f < 0, R < 0
• Image is always virtual, lies on the opposite side of the mirror to
the outgoing light, s’ < 0
• Image is upright and reduced in size, 0 < m < 1
Example: car side door mirrors – “objects are closer than they
appear”
• The same formulae can be used with the same sign convention
Question
A flea is located 3.0 cm from a convex spherical mirror of
radius 10 cm. Where is the image of the flea?
Thin Lenses - Converging
• Converging lens – when a beam of
rays parallel to the optic axis pass
through the lens they converge to a
point F2 and form a real image at
that point (a)
Rays passing through F1 emerge
from the lens as a beam of parallel
rays (b)
The focal length is a positive
quantity and the lens is also called a
positive lens
Thin lenses have 2 foci each the same
distance f from the centre of the
lens even when the two sides have
different curvatures
Thin Lenses - Diverging
• Diverging lens – a beam of parallel
rays incident on this lens diverges
after refraction, the rays appear
to come from the virtual image at
F2 (a)
Incident rays converging toward
F1 emerge from the lens parallel
to the optic axis (b)
The focal length is a negative
quantity and the lens is also called
a negative lens
NB positions of F1 and F2 have
swapped compared with the
converging lens
Thin Lens Equation
• From the geometry can derive
the same equations as for the
1 1 1
spherical mirrors and use the same
+ =
s s' f
sign conventions
y'
s'
m= =−
y
s
For a converging lens see:
s > f: the image is real and inverted, s’ > 0, on rhs of lens
s < f: image is virtual, located on the same side of the
lens as the object, it is upright and larger than the
object
Ray Diagrams for Lenses
•
Three principal rays for lenses
1) A ray parallel to the optic axis emerges from the lens in a
direction that passes through F2 for a converging lens, or
appears to come from this point for a diverging lens
2) A ray through the centre of the lens is not deviated
appreciably so this ray emerges from the lens following the
same line
3) A ray through (or proceeding toward) the F1 emerges
parallel to the axis
When the image is real, the position of the image point is
determined by the intersection of any 2 rays
When the image is virtual, we extend the diverging outgoing
rays backward to their intersection point to find the image
point
Many cases examined for the converging lens in text book
Converging Lens - ray diagram example
• NB the 3 principal rays
1) A ray || to the axis emerges from the lens in a direction
that passes through F2
2) Ray through the centre is undeviated
3) Ray through F1 emerges || to the axis
Real inverted smaller image
Diverging Lens – ray diagram example
• NB the 3 principal rays
1) A ray || to the axis emerges from the lens in a direction
that passes through F2
2) Ray through the centre is undeviated
3) Ray through F1 emerges || to the axis
Virtual upright smaller image
Lens Combination 1
• Image from the 1st lens is the object for the 2nd lens
The Simple Magnifier
• Simple imaging tool – magnifying
glass
• Angle subtended by the object
at the eye = angular size
• To examine something small in
detail - increase the angular
size (fill retina with image of
the object)
• Eye cannot focus sharply on
objects closer to the eye than
the near point
20-30 years old: 10 – 14 cm
Standard used is 25 cm
θ =
h
h
=
near point
25 cm
Magnifying Glass
Textbook – University Physics
Young and Freedman
Figure 35-10
Page 1126
A converging
lens can be used
to create an
image that is
larger and
further from
the eye than
the object
Virtual image most comfortable to view at infinity
The angular size of the image can be substantially larger than
the angular size of the object at the near point without a lens
A MAGNIFYING GLASS
y
To find M:
Angular Magnification
θ'
M=
θ
θ =
25 cm
y
θ '=
f
θ'
M =
=
θ
y
f
y
25 cm
=
25 cm
f
Microscope
s1 '
m1 = − ; s1 = f1
s1
s1 '
m1 = −
f1
25 cm
M2 =
f2
M = m1M 2 =
•
•
•
•
•
25 s1 '
f1 f 2
Lens combination – objective and eyepiece
Objective – converging lens - forms real enlarged image
Eyepiece – converging lens – magnifier of the real image
Overall magnification M = m1 M2
Objective: small f1, increases size of image I, maximize overall
magnification
• Eyepiece: small f2
• Manufacturers specify m1 and M2 rather than f1,2 or s1’
Telescope
θ′
M=
θ
ab y ′
θ=
=
f1
f1
cd y ′
=
θ′ =
f2
f2
f1
M =−
f2
http://www.walter-fendt.de/ph11e/refractor.htm
Textbook – University Physics
Young and Freedman
Figure 36-15
Page 1129