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Transcript
The study of light based
on the assumption that
light travels in straight
lines and is concerned
with the laws controlling
the reflection and
refraction of rays of light.
CHAPTER 22:
Geometrical optics
(4 Hours)
1
UNIT 22 : GEOMETRICAL
OPTICS
22.1 Reflection at a spherical surface
22.2 Refraction at a plane and
spherical surfaces
22.3 Thin lenses
2
22.1 Reflection at a spherical surface
At the end of this topic, students should be able to:
( 1 H)
a) State laws of reflection.
b) Sketch and use ray diagrams to determine the
characteristics of image formed by spherical mirrors.
c) Use
1 1 1 2
f

u

v

r
For real object only
 Use sign convention for focal length:
+ f for concave mirror and – f for convex mirror.
 Sketch ray diagrams with minimum two rays.
 r = 2f only applies to spherical mirror.
3
The Law of reflection
The law of reflection states that
• The incident ray, the reflected ray and
normal, all lie in the same plane
• The angle of incidence i is equal to the
angle of reflection r.
4
22.1 Reflection at a spherical surface
Terms and Definitions
• A spherical mirror is a reflecting surface
with spherical geometry.
• Two types :
i) convex, if the reflection takes place on
the outer surface of the spherical shape.
ii) concave, if the reflecting surface is on
the inner surface of the sphere.
5
22.1 Reflection at
a spherical
surface
Terms and
Definitions
Imaginary spherical
A
C
A
P
F
P
F
B
B
f
r
A concave mirror
C
f
r
A convex mirror
C ~ centre of curvature of the surface mirror.
P ~ centre of the surface mirror (vertex or pole).
6
Line CP ~ principal or optical axis.
22.1 Reflection at
a spherical
surface
Terms and
Definitions
Imaginary spherical
A
C
A
P
F
P
F
B
B
f
r
C
f
r
AB ~ aperture of the mirror.
F ~ focal point of the mirror.
f ~ focal length (FP, distance between focal
point and the centre of the mirror).
7
r ~ radius of curvature of the mirror.
Focal point and focal length, f
• Consider the ray diagram for a concave and convex mirrors as
shown in Figures 1.8a and 1.8b.
Incident
Incident
rays
rays
C
F
f
P
P
f
C
F
Figure 1.8a
Figure 1.8b
• Point F represents the focal point or focus of the mirrors.
• Distance f represents the focal length of the mirrors.
• The parallel incident rays represent the object infinitely far
away from the spherical mirror e.g. the sun.
8
Focal point or focus, F
• For concave mirror – is defined as a point where the incident
parallel rays converge after reflection on the mirror.
– Its focal point is real (principal).
• For convex mirror – is defined as a point where the incident
parallel rays seem to diverge from a point behind the mirror
after reflection.
– Its focal point is virtual.
Focal length, f
• is defined as the distance between the focal point (focus) F
and pole P of the spherical mirror.
• The paraxial rays is defined as the rays that are near to and
almost parallel to the principal axis.
9
22.1 Reflection at a spherical surface
Relation between focal length, f and
radius of curvature, r
FCM is isosceles.(FC=FM)
Consider ray AM is
paraxial (parallel
A
M
and very close to


the
principal
axis).

P
FM = FP
C
F
or FP = 1/2 CP
f
r
r
f 
2
10
Ray diagrams for spherical mirrors
 is defined as the simple graphical method to indicate
the positions of the object and image in a system of
mirrors or lenses.
Images formed by a concave mirror
a) Ray diagram
Ray 1 : A ray
parallel to the
principal axis is
reflected through
the focus (focal
point).
11
a)
Ray diagram
Ray 2 : A ray passing
through the focus is
reflected parallel to
the principal axis.
Ray 3 : A ray passing
through the centre of
curvature is reflected
back through the
centre of curvature.
12
Images formed by a concave mirror
 Concave mirror can be used as a
shaving and makeup mirrors because it
provides an upright and virtual images.
Table 1.1 shows the ray diagrams of
locating an image formed by a concave
mirror for various object distance, u.
* Remember:
At least any two rays for drawing the
ray diagram
13
Object
distance, u
Image
characteristic
Ray diagram

u>r
C

I
P
O

F

Front
back
O
u=r


F
C
P


I
14
Front
back
Real
Inverted
Diminished
Formed
between point
C and F.
Real
Inverted
Same size
Formed at point
C.
Object
distance, u
Image
characteristic
Ray diagram


f<u<r
I C

P
O

F
Front
back
O


u=f
15
C
Front
Real
Inverted
Magnified
Formed at a
distance
greater than
CP.
F
P
back
Real or virtual
Formed at
infinity.
Object
distance, u
Image
characteristic
Ray diagram



u<f

F
C
O
P
Front
Virtual
Upright
Magnified
Formed at the
back of the
mirror
I
back
Table 1.1
16
Image formed by concave mirrors
Notes
i) If the object is at infinity, a real image is
formed at F. Conversely, an object at F
gives a real image at infinity.
5) Object at infinity
F
•
•
•
•
At F
Real
Inverted
Smaller than object
ii) In all cases, the foot of the object is on
the principal axis and its image also lies on
17
this line.
Image formed by a convex mirror
Ray 1 : A ray parallel to the axis is reflected
as though it came from the focal point.
Ray 2 : A ray heading toward the focal point
is reflected parallel to the axis.
Ray 3 : A ray heading toward the centre of
curvature is reflected back on itself.
18
Image formed by a convex mirror
The characteristics of the image formed are
 virtual
 upright
 diminished (smaller than the object)
 formed at the back of the mirror (behind
the mirror)
Object position  any position in front of the
convex mirror.
Convex mirror always being used as a driving
mirror because it has a wide field of view and
providing an upright image.
19
b) The mirror equation-calculation
using formula
A
P
B
M
v
r
u
20
b) The mirror equation-calculation using formula
Object distance
Image distance
Radius of curvature
Object size
Image size
Focal length
=
=
=
=
=
=
OP
IP
CP
OA
IB
f
=
=
=
=
=
u
v
r
h
h’
1 1 1
 
f u v
or
2 1 1
= +
r u v
21
Linear Magnification, m
 Linear (lateral) magnification of the spherical mirror, m
is defined as the ratio between image height, hi and
object height, ho
hi v
m 
ho u
where
v : image distance from the pole
u : object distance from the pole
22
• Table 1.2 shows the sign convention for spherical mirror’s
equation .
Physical Quantity
u
Image distance, v
Focal length, f
Object distance,
Positive sign (+)
Negative sign (-)
Real object
Virtual object
(in front of the mirror)
Real image
(same side of the object)
Concave mirror
(at the back of the mirror)
Virtual image
(opposite side of the object)
Convex mirror
Table 1.2
• Note:
– Real image is formed by the actual light rays that pass
through the image.
– Real image can be projected on the screen.
23
Example 22.1.1 :
A dentist uses a small mirror attached to a thin rod to examine one
of your teeth. When the tooth is 1.20 cm in front of the mirror, the
image it forms is 9.25 cm behind the mirror. Determine
a. the focal length of the mirror and state the type of the mirror
used,
b. the magnification of the image.
Solution :
Example 22.1.2:
An upright image is formed 20.5 cm from the real object by using
the spherical mirror. The image’s height is one fourth of object’s
height.
a. Where should the mirror be placed relative to the object?
b. Calculate the radius of curvature of the mirror and describe the
type of mirror required.
c. Sketch and label a ray diagram to show the formation of the
image.
Solution :
25
Solution :
26
Solution :
27
Solution :
28
Example 22.1.3 :
A person of 1.60 m height stands 0.60 m from a surface of a
hanging shiny globe in a garden.
a. If the diameter of the globe is 18 cm, where is the image of the
person relative to the surface of the globe?
b. How large is the person’s image?
c. State the characteristics of the person’s image.
Solution :
29
Solution :
30
Solution :
31
Example 22.1.4 :
A shaving or makeup mirror forms an image of a light bulb on a
wall of a bathroom that is 3.50 m from the mirror. The height of the
bulb is 8.0 mm and the height of its image is 40 cm.
a. Sketch a labeled ray diagram to show the formation of the bulb’s
image.
b. Calculate
i. the position of the bulb from the pole of the mirror,
ii. the focal length of the mirror.
Solution :
32
3
Solution :
Exercise :
1.
a. A concave mirror forms an inverted image four times larger
than the object. Calculate the focal length of the mirror,
assuming the distance between object and image is
0.600 m.
b. A convex mirror forms a virtual image half the size of the
object. Assuming the distance between image and object
is 20.0 cm, determine the radius of curvature of the mirror.
ANS. : 160 mm ; 267 mm
2. a. A 1.74 m tall shopper in a department store is 5.19 m from
a security mirror. The shopper notices that his image in the
mirror appears to be only 16.3 cm tall.
i. Is the shopper’s image upright or inverted? Explain.
ii. Determine the radius of curvature of the mirror.
b. A concave mirror of a focal length 36 cm produces an
image whose distance from the mirror is one third of the
object distance. Calculate the object and image distances.
ANS. : u think, 1.07 m ; 144 cm, 48 cm
3. If a concave mirror has a focal length of
10 cm, find the two positions where an
object can be placed to give, in each
case, an image twice the height of the
object.( 15cm, 5.0cm )
4. A convex mirror of radius of curvature 40
cm forms an image which is half the
height of the object. Find the object and
image position.( 20cm,10cm behind
the mirror )
35
5. What are the nature, size, and location of
the image formed when a 6 cm tall object is
located 15 cm from a spherical concave
mirror of focal length 20 cm ?
(virtual, upright, -60 cm, + 24 cm)
36
22.2 Refraction at a plane and
spherical surfaces (1 H)
At the end of this chapter, students should be able to:
 State and use the laws of refraction (Snell’s Law) for
layers of materials with different densities.
 Apply
n1 n2 n2  n1 
 
u v
r
for spherical surface.
37
22.2 Refraction at a plane and spherical
surfaces
22.2.1 Refraction at a plane surface
• Refraction is defined as the changing of direction of a light
ray and its speed of propagation as it passes from one
medium into another.
• Laws of refraction state :
– The incident ray, the refracted ray and the normal all lie
in the same plane.
– For two given media, Snell’s law states
sin i n2
  constant
sin r n1
where
38
OR
n1 sin i  n2 sin r
n1 : refractive index of the medium 1
(Medium containing the incident ray)
n2 : refractive index of the medium 2
(Medium containing the refracted ray)
r : angle of refraction
• Examples for refraction of light ray travels from one medium to
another medium can be shown in Figures 1.13a and 1.13b.
(a) n1
 n2
(b) n1
(Medium 1 is less dense
medium 2)
 n2
(Medium 1 is denser than
medium 2)
Incident ray
Incident ray
i
i
n1
n2
n1
n2
r
r
Figure 1.13a
39
Refracted ray
The light ray is bent toward the
normal, thus r  i
Figure 1.13b
Refracted ray
The light ray is bent away from
the normal, thus r  i
Refractive index (index of refraction), n
• is defined as the constant ratio sin i for the two given
media.
sin r
• The value of refractive index depends on the type of medium
and the colour of the light.
• It is dimensionless and its value greater than 1.
• Consider the light ray travels from medium 1 into medium 2, the
refractive index can be denoted by
velocity of light in medium 1 v1

1 n2 
velocity of light in medium 2 v2
(Medium containing
the incident ray)
40
(Medium containing the
refracted ray)
• Absolute refractive index, n (for the incident ray travels from
vacuum or air into the medium) is given by
velocity of light in vacuum
c
n

velocity of light in medium
v
• Table 1.3 shows the refractive indices for common substances.
Substance
Table 1.3
Solids
(If the density of
Diamond
medium is
Flint glass
greater hence
Crown glass
Fused quartz (glass)
the refractive
Ice
index is also
Liquids
greater)
Benzene
Ethyl alcohol
Water
Gases
Carbon dioxide
41
Air
Refractive index, n
2.42
1.66
1.52
1.46
1.31
1.50
1.36
1.33
1.00045
1.000293
Relationship between refractive index and the wavelength of
light
• As light travels from one medium to another, its wavelength, 
changes but its frequency, f remains constant.
• The wavelength changes because of different material. The
frequency remains constant because the number of wave
cycles arriving per unit time must equal the number leaving
per unit time so that the boundary surface cannot create or
destroy waves.
• By considering a light travels from medium 1 (n1) into medium 2
(n2), the velocity of light in each medium is given by
then
42
v1  f1 and v2  f2
c
v1
f1
where v1 

n1
v 2 f2
and
c
v2 
n2
c
 
 n1   1
 c  2
 
 n2 
n11  n22
(Refractive index is inversely
proportional to the wavelength)
• If medium 1 is vacuum or air, then n1 = 1. Therefore the
refractive index for any medium, n can be expressed as
0
n

where
43
0 : wavelengt h of light in vacuum
 : wavelengt h of light in medium
Example 22.2.1 :
A fifty cent coin is at the bottom of a swimming pool of depth
3.00 m. The refractive index of air and water are 1.00 and 1.33
respectively. Determine the apparent depth of the coin.
Solution :
44
Solution :
45
Solution :
46
Example 22.2.2 :
A pond with a total depth (ice + water) of 4.00 m is covered by a
transparent layer of ice of thickness 0.32 m. Determine the time
required for light to travel vertically from the surface of the ice to
the bottom of the pond. The refractive index of ice and water are
1.31 and 1.33 respectively.
(Given the speed of light in vacuum is 3.00  108 m s-1.)
Solution :
47
Solution :
48
22.2.2 Refraction at a spherical surface
• Figure 1.14 shows a spherical surface with radius, r forms an
interface between two media with refractive indices n1 and n2.
i
n1
B


n2


PD
O
C
I
r
u
v
Figure 1.14
• The surface forms an image I of a point object O.
• The incident ray OB making an angle i with the normal and is
refracted to ray BI making an angle  where n1 < n2.
49
• Point C is the centre of curvature of the spherical surface and
BC is normal.
• From the figure,
BOC
BIC
i   
   
  
(1)
(2)
• From the Snell’s law
n1 sin i  n2 sin 
By using BOD, BCD and BID thus
BD
BD
BD
tan  
; tan  
; tan 
OD
CD
ID
By considering point B very close to the pole P, hence
sin i  i ; sin    ; tan    ; tan    ; tan  
OD  OP  u ; CD  CP  r ; ID  IP  v
then Snell’s law can be written as
n1i  n2
50
(3)
• By substituting eq. (1) and (2) into eq. (3), thus
then
n1 (   )  n2 (   )
n1  n2   (n2  n1 )
 BD 
 BD 
 BD 
n1 
  n2 
  (n2  n1 )

 u 
 v 
 r 
n1 n2 (n2  n1 )
 
u v
r
where
51
Equation of spherical
refracting surface
v : image distance from pole
u : object distance from pole
n1 : refractive index of medium 1
(Medium containing the incident ray)
n2 : refractive index of medium 2
(Medium containing the refracted ray)
• Note :
– If the refracting surface is flat (plane) :
r 
then
n1 n2
 0
u v
– The equation (formula) of linear magnification for refraction
by the spherical surface is given by
hi n1v
m 
ho n2u
52
– Table 1.4 shows the sign convention for refraction or thin
lenses:
Physical Quantity
Object distance,
Image distance,
Focal length,
Radius of
curvature,
r
f
u
v
Positive sign (+)
Real object
(in front of the refracting
surface)
Real image
(opposite side of the
object)
Converging lens
Centre of curvature
is located in more
dense medium
(convex surface)
53
Table 1.4
Negative sign (-)
Virtual object
(at the back of the
refracting surface)
Virtual image
(same side of the object)
Diverging lens
Centre of curvature
is located in less
dense medium
(concave surface)
Example 22.2.3 :
A cylindrical glass rod in air has a refractive index of 1.52. One end
is ground to a hemispherical surface with radius, r =3.00 cm as
shown in Figure 1.15.
air
glass
P
C
O
I
10.0 cm
Figure 1.15
Calculate,
a. the position of the image for a small object on the axis of the rod,
10.0 cm to the left of the pole as shown in figure.
b. the linear magnification.
(Given the refractive index of air , na= 1.00)
54
Solution :
55
Example 22.2.4:
Figure 1.16 shows an object O placed at a distance 20.0 cm from
the surface P of a glass sphere of radius 5.0 cm and refractive
index of 1.63.
Glass sphere
air
P
O
5.0 cm
20.0 cm
Figure 1.16
Determine
a. the position of the image formed by the surface P of the glass
sphere,
b. the position of the final image formed by the glass sphere.
56
(Given the refractive index of air , na= 1.00)
Solution :
57
Solution :
58
Solution :
59
Exercise :
1.
A student wishes to determine the depth of a swimming pool
filled with water by measuring the width (x = 5.50 m) and then
noting that the bottom edge of the pool is just visible at an
angle of 14.0 above the horizontal as shown in Figure 1.17.
Figure 1.17
Calculate the depth of the pool.
(Given nwater = 1.33 and nair = 1.00)
ANS. : 5.16 m
60
2.
A small strip of paper is pasted on one side of a glass sphere
of radius 5 cm. The paper is then view from the opposite
surface of the sphere. Determine the position of the image.
(Given the refractive index of glass =1.52 and the refractive
index of air =1.00)
ANS. : 20.83 cm in front of the 2nd refracting surface.
3.
A point source of light is placed at a distance of 25.0 cm from
the centre of a glass sphere of radius 10 cm. Determine the
image position of the source.
(Given the refractive index of glass =1.52 and the refractive
index of air =1.00)
ANS. : 25.2 cm at the back of the 2nd refracting surface.
61
22.3 Thin lenses (2 hours)
At the end of this chapter, students should be able to:
•
Sketch and use ray diagrams to determine the
characteristics of image formed by diverging and
converging lenses.
•
Use thin lens equation,
1 1 1
 
u v f
•
for real object only.
Use lensmaker’s equation:
1 1 
1
 n  1  
f
 r1 r2 
62
•
Use the thin lens formula for a combination of
converging lenses.
22.3 Thin lenses
• is defined as a transparent material with two spherical
refracting surfaces whose thickness is thin compared to the
radii of curvature of the two refracting surfaces.
• There are two types of thin lenses. It is converging and
diverging lenses.
• Figures 1.18a and 1.18b show the various types of thin lenses,
both converging and diverging.
(a) Converging (Convex) lenses
r1
(+ve)
r 2 r1
(+ve)(+ve)
Biconvex
r2 r 1
() (+ve)
Plano-convex
63
Figure 1.18a
r2
(ve)
Convex meniscus
(b) Diverging (Concave) lenses
r1
(ve)
r2
r1
(ve) (ve)
r2 r1
()(+ve)
Plano-concave
Figure 1.18b
Biconcave
r2
(ve)
Concave meniscus
17.4.1 Terms of thin lenses
• Figures 1.19 show the shape of converging (convex) and
diverging (concave) lenses.
(a) Converging lens
(b) Diverging lens
r1
r1
C1
64
O
r2
C2
C1
Figure 1.19
O
r2
C2
• Centre of curvature (point C1 and C2)
– is defined as the centre of the sphere of which the surface
of the lens is a part.
• Radius of curvature (r1 and r2)
– is defined as the radius of the sphere of which the surface
of the lens is a part.
• Principal (Optical) axis
– is defined as the line joining the two centres of curvature
of a lens.
• Optical centre (point O)
– is defined as the point at which any rays entering the lens
pass without deviation.
65
Focal point and focal length, f
• Consider the ray diagrams for converging and diverging lenses
as shown in Figures 1.20a and 1.20b.
O
F1
f
F2
f
O
F1
f
F2
f
Figure 1.20a
Figure 1.20b
• From the figures,
– Points F1 and F2 represent the focus of the lenses.
– Distance f represents the focal length of the lenses.
66
Focus (point F1 and F2)
• For converging (convex) lens – is defined as the point on the
principal axis where rays which are parallel and close to the
principal axis converges after passing through the lens.
– Its focus is real (principal).
• For diverging (concave) lens – is defined as the point on the
principal axis where rays which are parallel to the principal
axis seem to diverge from after passing through the lens.
– Its focus is virtual.
Focal length ( f )
• is defined as the distance between the focus F and the
optical centre O of the lens.
67
Image Formation by Thin Lenses
A thin lens is one whose thickness is small
compared to its focal length, f .
Ray diagrams can be drawn to determine
the location and size of the image.
68
Ray Diagrams
Image Formation by Thin Lenses
Ray 1: A ray entering a converging lens parallel to
its axis passes through the focal point F of the
lens on the other side.
Ray 2 : A ray entering a converging lens through
its focal point exits parallel to its axis.
Ray 3 : A ray passing through the center of the
lens does not change direction.
69
Ray Diagrams
Image Formation by Thin Lenses
Ray 1: A ray entering a diverging lens parallel to
its axis seems to come from the focal point F.
Ray 2 : A ray that enters a diverging lens by
heading toward the focal point on the opposite
side exits parallel to its axis.
Ray 3 : A ray passing through the center of the
lens does not change direction.
70
Image Formation by Thin Lenses
• If the image is real, the position of the
image point is determined by intersection
of any two rays 1, 2 and 3.
• Real image is formed on the back side
of the lens.(OPPOSITE SIDE)
• If the image is virtual, we extend the
diverging outgoing rays backward to their
intersection point to seek the image point.
• Virtual image is formed on the front side
of the lens. (SAME SIDE)
71
Images formed by a converging lens
• Table 1.5 shows the ray diagrams of locating an image formed
by a converging lens for various object distance, u.
Object
distance, u
Image
characteristic
Ray diagram
Real
 Inverted
 Diminished
 Formed between
point F2 and 2F2.
(at the back of
the lens)

u > 2f
72
I
O2F1
F1
F2
Front
back
2F2
Object
distance, u
Image
characteristic
Ray diagram
Real
 Inverted
 Same size
 Formed at point
2F2. (at the back
of the lens)

u = 2f
O
2F2
F1
F2
Front
back
2F1
I
Real
 Inverted
 Magnified
 Formed at a
distance greater

f < u < 2f 2F1 O F1
Front
73
I
F2
back
2F2
than 2f at the
back of the lens.
Object
distance, u
Image
characteristic
Ray diagram
Real or virtual
 Formed at infinity.

u=f
O
2F1
F1
F2
Front
back
2F2
Virtual
 Upright
 Magnified
 Formed in front
of the lens.

u<f
I
74
2F1
F1 O
Front
Table 1.5
F2
back
2F2
Diverging Lens
The image always ,
Image Formation by Thin Lenses
a) Between lens and F1
a) Virtual
b) Upright
d) Smaller than object
F1
75
Thin lens formula, lens maker’s and
linear magnification equations
Thin lens formula and lens maker’s equation
• Considering the ray diagram of refraction for two spherical
surfaces as shown in Figure 1.23.
u1
r1
v2
D
A
n1
n2
C1
O
76
u2  t  v1
r2
v1
Figure 1.23
P1
I1
B
n1
C2
P2
E
t
I2
In general,
1 1 1
 
f u v
Thin lens formula
1  n2  1 1 
Lens maker’s
   1  
equation
f  n1  r1 r2 
where f : focal length
r1 : radius of curvature for 1stndrefracting surface
r2 : radius of curvature for 2 refracting surface
n1 : refractive index of the medium
n2 : refractive index of the lens material
77
• Note :
– If the medium is air (n1= nair=1) thus the lens maker’s
equation can be written as
1 1
1
 n  1  
f
 r1 r2 
– For thin lenses and lens maker’s equations, use the sign
convention for refraction.
where
n : refractive index of the lens material
Use sign convention for R :
• +ve for convex surface
• -ve for concave surface
78
78
Linear magnification, m
• is defined as the ratio between image height, hi and object
height, ho.
•
79
hi
v
m 
ho u
where v : image distance from optical centre
u : object distance from optical centre
1 1 1
Since
  the linear magnification equation can be
f u v
written as  1
1 1
     v
 f u v
v v
v
 1
m  1
f u
f
Example 22.3.1 :
A person of height 1.75 m is standing 2.50 m in from of a camera.
The camera uses a thin biconvex lens of radii of curvature
7.69 mm. The lens made from the crown glass of refractive index
1.52.
a. Calculate the focal length of the lens.
b. Sketch a labeled ray diagram to show the formation of the
image.
c. Determine the position of the image and its height.
d. State the characteristics of the image.
Solution :
80
Solution :
81
Solution :
82
Example 22.3.2 :
A thin plano-convex lens is made of glass of refractive index 1.66.
When an object is set up 10 cm from the lens, a virtual image ten
times its size is formed. Determine
a. the focal length of the lens,
b. the radius of curvature of the convex surface.
Solution :
83
Solution :
84
Example 22.3.4 :
The radii of curvature of the faces of a thin concave meniscus lens
of material of refractive index 3/2 are 20 cm and 10 cm. What is the
focal length of lens
a. in air,
b. when completely immersed in water of refractive index 4/3?
Solution :
85
Solution :
86
Combination of lenses
• Many optical instruments, such as microscopes and
telescopes, use two converging lenses together to produce
an image.
• In both instruments, the 1st lens (closest to the object )is called
the objective and the 2nd lens (closest to the eye) is referred to
as the eyepiece or ocular.
• The image formed by the 1st lens is treated as the object for
the 2nd lens and the final image is the image formed by the 2nd
lens.
• The position of the final image in a two lenses system can be
determined by applying the thin lens formula to each lens
separately.
• The overall magnification of a two lenses system is the
product of the magnifications of the separate lenses.
where
m  m1m2 m : overall magnificat ion
87
m1 : magnificat ion due to the 1stndlens
m2 : magnificat ion due to the 2 lens
Two Lenses System
f = +20 cm
B
A
p1
v1
p2
v2
88
First Step
Two Lenses System
• Let p1 represent the distance of object O
from lens A.
• Then find the distance v1 the image
produced by lens A, either by using
equation or drawing rays.
89
Second Step
Two Lenses System
• Ignore the presence of lens A, treat the
image found in first step I1 as the object for
lens B.
• If this new object is located beyond lens
B, the object distance p2 is taken to be
negative.
•If this new object is located in front of the
lens B, the object distance p2 is taken to be
positive.
90
Second Step
Two Lenses System
• The distance v2 of the final image (I2)
produced by lens B can be found by using
equation or drawing rays.
• The overall linear magnification m
produced by a system of two lenses is the
product of the overall magnifications mA
and mB produced by two lenses,
m  (mA )(mB )
91
Example 22.3.5 :
The objective and eyepiece of the compound microscope are both
converging lenses and have focal lengths of 15.0 mm and 25.5 mm
respectively. A distance of 61.0 mm separates the lenses. The
microscope is being used to examine a sample placed 24.1 mm in
front of the objective.
a. Determine
i. the position of the final image,
ii. the overall magnification of the microscope.
b. State the characteristics of the final image.
Solution :
92
Solution :
93
Solution :
94
Solution :
95
Exercise:
1.
a. A glass of refractive index 1.50 plano-concave lens has a
focal length of 21.5 cm. Calculate the radius of the
concave surface.
b. A rod of length 15.0 cm is placed horizontally along the
principal axis of a converging lens of focal length 10.0 cm.
If the closest end of the rod is 20.0 cm from the lens
calculate the length of the image formed.
ANS. : 10.8 cm; 6.00 cm
2. An object is placed 16.0 cm to the left of a lens. The lens
forms an image which is 36.0 cm to the right of the lens.
a. Calculate the focal length of the lens and state the type of
the lens.
b. If the object is 8.00 mm tall, calculate the height of the
image.
c. Sketch a labelled ray diagram for the case above.
ANS. : 11.1 cm; 1.8 cm
96
3.
97
When a small light bulb is placed on the left side of a
converging lens, a sharp image is formed on a screen placed
30.0 cm on the right side of the lens. When the lens is moved
5.0 cm to the right, the screen has to be moved 5.0 cm to the
left so that a sharp image is again formed on the screen. What
is the focal length of the lens?
ANS. : 10.0 cm
4. A converging lens of focal length 8.00 cm is 20.0 cm to the left
of a converging lens of focal length 6.00 cm. A coin is placed
10.0 cm to the left of the 1st lens. Calculate
a. the distance of the final image from the 1st lens,
b. the total magnification of the system.
ANS. : 24.6 cm; 0.924
5. A converging lens with a focal length of 4.0 cm is to the left of
a second identical lens. When a feather is placed 12 cm to the
left of the first lens, the final image is the same size and
orientation as the feather itself. Calculate the separation
between the lenses.
ANS. : 12.0 cm