Download Lecture 3.3.

Lecture 3.3.
Image formation
Ray tracing
Calculation
Lenses
Convex
Concave
Mirrors
Convex
Concave
Optical instruments
Image formation
Laws of refraction and reflection can be used
to explain how lenses and mirrors operate
Spherical Lenses
Convex
(converging) lens
Parallel rays from the Sun passing through
a convex lens
F
F: focal point
f: focal length
f
Sunlight concentrated by magnifying glass
may burn hole in paper placed at focal point F
Image formation
F
f
F is the focal point and f is the focal length
Refractive power (or Strength) of the lens
in diopters (D) =
1
f
( f is in metres)
Example
A farsighted person requires an eyeglass of
strength 2.5 diopters. What is the focal length
of the eyeglass lens?
(a) 4 cm, (b) 40 cm or (c) 400cm
D
=2.5m-1
f = 1 =1/2.5m = 0.4 m = 40cm.
D
Image Formation (ray tracing)
Object at a distance greater than the focal
length from the lens
1
h
3
2
F
h′
F
s
s′
Ray 1 entering lens parallel to optic axis will
exit and pass through the focal point
Ray 2 passing through the focal point will exit
the lens and travel parallel to optic axis
Ray 3 will undergo small deviation
(displacement (not shown) (thin lens)
Real, inverted image formed
Real image (may be projected and displayed
on a screen)
Rays are reversible
Image Formation (ray tracing)
Object at a distance less than the focal
length from the lens
h′
h
F
s
s′
Image
• virtual
• upright
•magnified.
Simple magnifier
F
Concave (diverging) lens
Optic
axis
F
f
Rays entering lens parallel to axis appear to
originate at focal point
Dashed lines indicate the direction from which
the rays appear to come
Concave (diverging) lens
Object outside the focal length
h
h′
s′
F
Optic
axis
s
Virtual image
always produced by a concave lens
Cannot be viewed on screen since rays are
diverging on the right of the lens
However can be viewed with the eye since the
eye converges the rays onto the retina.
Image Formation---calculation
Ray diagrams are useful in sketching the
relationship between object and image
Relationship may also be calculated
A
E
h
F
B
C
O
F
h′
D
s
s′
Triangles AOB and
DOC are similar
Triangles EFO and
DFC are similar
h
h'
= −
s
s'
h
h'
= −
f
s '− f
h'
s'
= −
h
s
s ' s '− f
=
s
f
Equating the expressions
h'
s '− f
= −
h
f
1 1 1
+ =
s' s f
Image ---calculation
1 1 1
Thin lens formula
+ =
s' s f
h
F
B
h′
F
s
s′
Object distance s
positive if object is in front of lens
negative if object is behind lens
Image distance s′
positive if image is formed behind the lens (real)
negative if is formed in front of the lens (virtual)
Focal length f
•positive -- convex lens
•negative --concave lens
Magnification
h
F
B
h′
F
s
s′
Magnification is defined as
s'
M= −
s
or
M Negative :
 inverted image
M Positive :
 upright image
h'
M=
h
Simple magnifier
h′
h
F
s
F
s′
Object placed inside focal length of converging
lens;
Image viewed.
•Virtual,
•Magnified
•Upright
Example
An object 0.5 cm in height is placed 8 cm from a
convex lens of focal length 10 cm. Determine the
position, magnification, orientation and height
of the image.
1 1 1
1 1 1
+ =
s' s f
1 1 1
=
−
s ' 10 8
f
−
s
10 x 8
s′ =
= - 40cm.
8-10
s'
−40cm
M=
− =
−
=
5
s
8cm
h'
M=
h
=
s'
M +ve image upright
h′ = M x h = 5 x 0.5cm = 2.5cm
h′
h
F
s
s′
F
Combining Lenses
Effective focal length (feff) of combination of a
number of thin lenses close together
1
1 1
= + + ......
f eff
f1 f 2
Effective refractive power (or strength) (Seff) of
combination of thin lenses close together
Seff = S1 + S 2 + .....
Determine the combined strength of a thin convex lens
and a thin concave lens placed close together if their
respective focal lengths are 10cm and -20cm.
Strength S, in diopters (D) =
1
1 1
= Seff=
+
f eff
f1 f 2
1
f
( f is in metres)
1
1
S=
−
eff
.1m .2m
Seff = 10 − 5 = 5diopters
Example
An object is placed 45 cm from a lens of focal
length -25 cm. Determine the position,
magnification, and orientation of the image.
1 1 1
+ =
s' s f
1
S’
1 1 1
=
−
s' f s
1
1
= -25 cm- 45 cm
M=-
S’
S
S’ = -16.1 cm
16.1 cm
M=45 cm
= 0.36
M +ve image upright
h
h′
s′
F
s
Optic
axis
Mirrors
Flat Mirror
Concave Mirror
Convex Mirror
Curved mirrors analogous to lenses
Ray tracing and thin lens equation also valid.
real and virtual images are also formed
Flat Mirror
image
object
d
d
Object and image distance equal
Upright, virtual image
Spherical Mirrors
Hollow sphere
Spherical mirror
C
R
Principal or optic axis
Spherical mirror is a section hollow sphere
Radius of curvature R = 2f
Curved (Spherical) Mirrors
Concave mirror
(converging)
F
f
Positive focal length
Convex mirror
(diverging)
Negative focal length
f
F
Thin lens formula may be used to determine
object and image distances and focal lengths etc
For all single lenses and mirrors:
Real image : inverted (h′ negative), positive
image distance s′ (RHS of lens, LHS of mirror)
Virtual image: upright (h′ positive), negative
image distance s′ (LHS of lens, RHS of mirror)
Mirrors
Concave shaving/makeup mirrors
C
F
Object placed at distance < f from mirror
Image is virtual, upright and enlarged.
Question: if object is placed at the focal point,
where is image located?
Application: searchlight
Mirrors
Example
An object is positioned 5 cm in front of a
concave mirror of focal length 10 cm.
Determine the location of its image and its
characteristics.
s = 5 cm
1 1 1
+ =
s' s f
1 1 1
=
−
s' f s
f = 10 cm
1
1
1
=
−
s ' 10cm 5cm
1
1
2
1
=−
=
−
s ' 10cm 10cm
10cm
Characteristics.
•Image virtual
•Located behind mirror
S’ = -10cm
s'
M= −
s
−10cm
M= −
5cm
M =2
Optical instruments
Microscopes, telescopes, cameras etc
System may have many optical elements
(example, lenses and mirrors)
Thin lens formula or ray tracing may be
used to analyse behaviour of such systems
Simple compound microscope
two convex lenses
Objective
lens
eyepiece
Fo
Object (height h)
Fe
h′
Fe
Fo
Final image
h′′
image formed by objective lens is
inside focal length of eyepiece lens.
Optical instruments
Simple refracting telescope
Objective lens
f0
fe
eyepiece
Fo,Fe
Virtual image at infinity,
Magnified and inverted
f0
M= −
fe
Objective lens forms image (real, inverted) at
focal point F0 which is also the focal point Fe of the
eyepiece; virtual image is then formed at infinity.
Optical instruments
Simple reflecting telescope
Flat mirror
Objective,
concave mirror
Eyepiece
lens
Example
Effective focal length of the objective in the
Hubble telescope is 57.8 m. What focal length
eyepiece is required to give a magnification of
-8.0 x 103.
f
M= − 0
fe
fe = −
f0
M
57.8m
−3
fe =
7.23
10
m
−
=
×
3
8.0 ×10
Optical instruments
Medical loupes
Galilean Design
objective
Operating
site
eyepiece
Typical
working distance
28-38 cm
Typically Magnification
Optical design allows
m ≈ 2.5 → 4.5
observer focus at infinity
thereby relieving eyestrain
Optical instruments
Optical fibre: Total Internal Reflection
Endoscope
Endoscope for medical investigations—
inserted through small incision or orifice to
inspect and facilitate operation on interior parts
of the body
flexible shaft includes:
light source to illuminate area,
image channel to view area under investigation,
air or water conduit to clear debris,
instrument conduit
eyepiece
Typical endoscope
Instrument entry
Flexible shaft
Transmits light, air , water
Optical instruments
Camera
aperture
CCD array
Lens translated to change image distance to
adjust for different object distances.
Real, inverted image formed on CCD array
Example
Microscope as shown in the previous figure
has the following characteristics
objective lens focal length = 0.7 cm
eyepiece lens focal length = 4.5 cm
separation between lenses = 21 cm.
Object size = 0.01 cm.
object to objective lens distance = 0.73 cm.
Calculate total magnification produced and
size of the final image.
image distance s′ due to objective lens
1 1 1
+ =
s' s f
1 1 1
=
−
s' f s
1
1
1
=
−
s ' 0.7 0.73
s′ = 17 cm
Magnification M0 = -s′/s = -17/0.73 = -23.3
Magnification negative because image is inverted
M0 = h′/h
size of the image = h′ =M0h
= (-23.3)(0.01)cm = -0.233 cm.
Image produced by objective is situated at a
distance (21-17) cm = 4 cm from the eyepiece.
1 1 1
+ =
s' s f
1 1 1
=
−
s' f s
1
1 1
=
−
s ' 4.5 4
s′ = -36 cm
s′ negative: image is on same side of eyepiece
as object
Magnification Me = -s′/s = -(-36/4) = 9
Size of the final image
= h′ ′ =Meh′ = (9)( -0.233) = -2.1cm.
Total magnification MT = (-2.1cm)/(0.01 cm)
= -210
Total magnification MT = M0 x Me
= (-23.3) x 9 = - 210