Transcript Refraction at A Spherical Surface

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
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
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 egual 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.
22.1 Reflection at a spherical surface
Terms and
Definitions
Focal point, F
“point on the principal axis where
rays parallel and close to the
principal axis pass after reflection”.
A concave mirror
8
22.1 Reflection at a spherical surface
Terms and Definitions
F - “ point on the principal axis where
rays parallel to the principal axis
appear to diverge from after
reflection”.
Focal point, F
A convex mirror
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
22.1 Sketch and Use Ray diagram
Images Form by Spherical Mirrors
Information about the image in any case
can be obtained either by drawing a ray
diagram or by calculation using formula.
a) Ray diagram
Ray 1 : A ray
parallel to the
principal axis is
reflected through
the focus (focal
point).
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Images Form by Spherical Mirrors
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
Drawing Compass
13
Images Form by Spherical Mirrors
a)
Ray diagram
14
Image formed by concave mirrors
1) Object beyond C
a) Between C and F
b) Real
c) Inverted
d) Smaller than object
15
Image formed by concave mirrors
2) Object at C
a) At C
b) Real
c) Inverted
d) Same size as object
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Image formed by concave mirrors
3) Object between C and F
a) Beyond C
b) Real
c) Inverted
d) Larger than object
(magnified)
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Image formed by concave mirrors
4) Object between F and P
a) Behind mirror
b) Virtual
c) Upright
d) Larger than object
(magnified)
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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
19
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.
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Image formed by a convex mirror
The image always
• Virtual
• Upright
• Smaller than object21
b) The mirror equation-calculation
using formula
A
P
B
M
v
r
u
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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
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b) The mirror equation-calculation using formula
Linear Magnification, m
height of image h '
m

height of object h
or
image distance
v
m

object distance
u
h'
v
m 
h
u
or
v
m=
u
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b) The mirror equation-calculation using formula
Sign convention
Quantity
Object
distance (u)
Image
distance (v )
Focal
Length (f )
Magnification
(m)
Positive sign (+) Negative sign (-)
Real object
Virtual object
Real image
Virtual image
Concave mirror
Convex mirror
Upright image
Inverted image
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b) The mirror equation-calculation using formula
Example 1.2.1
An object 6 cm high is located 30 cm in
front of a convex spherical mirror of radius
40 cm. Determine the position and height
of its image.
2 1 1
= +
Solution
r u v
2 1 1
= +
r u v
2
1 1
=
+
- 40 30 v
v = -12 cm
v 12
m= =
= 0.40
u 30
h'
m = = 0.40
h
h' = 2.4 cm
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b) The mirror equation-calculation using formula
Example 1.2.2
An object is placed 15 cm from a
a) concave mirror
b) convex mirror
of radius of curvature 20 cm. Calculate the
image position and magnification in each
case.
27
b) The mirror equation-calculation using formula
Solution 1.2.2
a) Concave mirror , u = +15 cm
r = +20 cm
f = +10 cm
Substituting values and signs in the mirror
equation,
1 1 1
 
u v f
1 1 1
 
v f u
1 1
1
1



v 10 15 30
v  30 cm
28
b) The mirror equation-calculation using formula
Solution 1.2.2
The image is real since v is positive and it
is 30 cm in front of the mirror.
Magnification,
v
m
u
30
m
15
 2 -ve (inverted)
The image is twice as high as the object.
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b) The mirror equation-calculation using formula
Solution 1.2.2
b) Convex mirror, u = +15 cm
r = -20 cm
f = -10 cm
1
1 1


v
f
u
1 1
1
 
u v
f
1 1
1
5



v 10 15
30
v  6.0 cm
The image is virtual since v is negative and
it is 6.0 cm behind the mirror.
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b) The mirror equation-calculation using formula
Solution 1.2.2
v
Magnification, m  
u
6 2
m

15 5
The image is two-fifth as high as the object.
31
b) The mirror equation-calculation using formula
Example 1.2.3
What is the focal length of a convex
spherical mirror which produces an image
one-sixth the size of an object located 12
cm from the mirror ?
Solution
v 1
m=- =
u 6
v
1
=
⇒v = -2 cm
12 6
1 1 1 1
1
= + =
+
f u v 12 - 2
f = -2.4 cm
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b) The mirror equation-calculation using formula
Example 1.2.4
When an object is placed 20 cm from a
concave mirror, a real image three times
is formed. Calculate
a) The focal length of the mirror
b) Where the object must be placed to
give a virtual image three times the
height of the object.
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b) The mirror equation-calculation using formula
Solution 1.2.4
a) u = + 20 cm , m = -3
v
  3
u
v  3u
v
m
u
v  3(20)  60 cm
1 1
1
 
Using
u v
f
1 1 1
1
1
  

f = +15 cm
34
f u v 60 20
b) The mirror equation-calculation using formula
Solution 1.2.4
b) Given m = 3 , f = +15 cm
v
m
u
v
3
u
v   3u
Using
1 1
1
 
u v
f
1
1
1
 
3u u 15
u  10 cm
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b) The mirror equation-calculation using formula
Example 1.2.5
An object 2.0 cm high is placed 30 cm
from a concave mirror with a radius of
curvature of 10 cm. Find the location and
its characteristics.
Solution
Given : h = 2.0 cm, u = +30 cm,
r = +10 cm, f = r/2= +5 cm
1 1
1
 
u v
f
1 1 1 1
1
  

v f u 5 30
36
v  6.0 cm
b) The mirror equation-calculation using formula
Solution 1.2.5
v
6
m=- == -0.2
u
30
Characteristics :
1) smaller
2) in front of the mirror
3) inverted
4) real
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b) The mirror equation-calculation using formula
Exercise
1) 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 )
2) 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 )
38
b) The mirror equation
Exercise
3) An object is placed 5.0cm in front of a
concave mirror with a 10.0 cm focal
length. Find the location of the image
and its characteristics.( -10cm, m =-2,
virtual, upright and located behind
the mirror )
4) What kind of spherical mirror must be
used, and what must be its radius, in
order to give an erect image one-fifth as
large as an object placed 15 cm in front
39
of it ? (-7.5 cm, convex mirror)
b) The mirror equation
Exercise
5) A concave mirror forms an image , on
the wall 3.0 m from the mirror, of the
filament lamp 10 cm in front of the
mirror.
a) What is the radius of curvature of the
mirror ?
b) What is the height of the image if the
height of the object is 5 mm ?
(19.4 cm, -30)
40
b) The mirror equation
Exercise
6) 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)
7) The magnification of a mirror is -0.333.
Where is the object located if its image is
formed on a card 540 mm from the mirror?
What is the focal length ?
(1.62 m, +405 mm)
41
Refraction at aplane and
spherical surfaces (1 H)
At the end of this chapter, students should be able to:
 22.2.1 State and use the laws of refraction (Snell’s Law)
for layers of materials with different densities.
 22.2.2 Apply
n1 n2 n2  n1 
 
u v
r
for spherical surface.
42
22.2 Refraction at a Plane and
Spherical Surfaces
• Refraction is the change in direction of light
when it passes through matter.
Refraction of light
• When a ray of light traveling through a
transparent medium encounters a
boundary leading into another transparent
medium, part of the ray is reflected and
part enters the second medium.
• The ray that enters the second medium is bent
at the boundary and is said to be refracted.43
22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
Normal
Incident ray
θ1
θ’
Air
Glass
transparent medium
θ2
Reflected ray
θ 1 = θ’
θ 1 > θ2
Refracted ray
44
22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
Law of Refraction
• The incident ray, refracted ray and the
normal all lie in the same plane.
• At the boundary between any two given
materials, the ratio of the sine of the
angle of incidence to the sine of the
angle of refraction is constant for rays
of any particular wavelength (previous
figure). This is known as Snell’s Law.
45
22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
Law of Refraction
sin 1 v1

 constant
sin 2 v2
Snell’s Law
where v1 is the speed of light in medium 1
and v2 is the speed of light in medium 2.
• This relationship shows that the angle of
refraction Ө2 depends on the speed of
light and on the angle of incidence.
46
22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Refraction at a plane surface
nglass > nair
Normal
Normal
v 1 > v2
θ 1 > θ2
θ1
v1
θ1
v 1 < v2
v1
Air
Glass
Glass
Air
θ2
this ray is bent
toward normal
v2
θ 1 < θ2
θ2
v2
this ray is bent
away from normal
47
22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
48
49
22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Refraction at a plane surface
Index of Refraction, n
The speed of light in any material or medium is
less than the speed of light in vacuum or air.
speed of light in vacuum c
n

speed of light in medium v
..(1.1)
• The index of refraction is a dimensionless
and never less than 1 (n ≥ 1) because v is
usually less than c.
• n is equal to unity (n = 1) for vacuum.
50
22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
Index of Refraction, n
• As light travels from one medium to
another, its frequency does not change but
its wavelength does.
Therefore, because the relationship v  f 
must be valid in both medium and because
f1  f 2  f
where v1  f 1 … (1.2) v2  f 2 … (1.3)
51
22.2 Refraction at a Plane and Spherical Surfaces
Index of Refraction, n
1   2 ,
1
v1
c / n1
n2



…(1.4)
2
v2
c / n2
n1
v1 n2
…(1.5)

v2 n1
n11  n22 …(1.6)
If medium 1 is vacuum, (n1 = 1) then
equation (1.4) can be written as
o
n
n
λo = wavelength of light in vacuum.
λn = wavelength in the medium
whose index of refraction is n.
52
53
22.2Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
(1.5) into Snell’s Law
n1 sin 1  n2 sin  2
This equation is the most widely used and
practical form of Snell’s Law.
54
Refraction at a plane surface 22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Example 1.3.1
A beam of light of wavelength 500 nm
traveling in air incident on a slab of
transparent material. The incident beam
makes an angle of 40.0 o with the normal,
and the refracted beam makes an angle of
26.0 o with the normal. Calculate the index
of refraction of the material and the
wavelength of light in the material.
55
Refraction at a plane surface 22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Solution 1.3.1
Ө1 =40.0 o , Ө2 =26.0 o , n1=1
Using
n1 sin 1  n2 sin  2
n1 sin 1
n2 
sin 2
0
(1)(sin 40.0 )


1
.
47
0
sin 26.0
56
Refraction at a plane surface 22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Solution 1.3.1
Using
n11  n22
λ1= 500 nm
n11 (1)(500)
2 

 374 nm
n2
1.47
57
Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Example 1.3.2
A light ray of wavelength 589 nm traveling
through air is incident on a smooth, flat slab
of crown glass (n = 1.52) at an angle of
30.0o to the normal. Find the angle of
refraction.
58
Refraction at a plane surface 22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Solution 1.3.2
Ө1 =30.0 o , n1 =1 , n2 = 1.52
Using
n1 sin 1  n2 sin  2
n1
sin 2  sin 1
n2
(1)
0

 (sin 30.0 )  0.329
1.52
1
 2  sin (0.329)  19.2
0
59
Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
Example 1.3.3
40o
Air
Oil
Water
A layer of oil (n = 1.45) floats on water (n
= 1.33). A ray of light shines onto the oil
with an incidence angle of 40.0o. Find
the angle the ray makes in the water.
60
Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snell’s Law
40o
Solution 1.3.3
noil = 1.45
nwater = 1.33
Air
Oil
Water
nair sin 400 = noil sin  oil ... (1)
noil sin  oil = nwater sin  water ... (2)
(1) = (2)
nair sin 40 = nwater sin  water
0
 water = sin-1
(n
0
o
)
(
)
sin
40
1
sin
40
air
-1
= sin
= 28.90
nwater
1.33
61
Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Exercise
1. A light ray of wavelength 589 nm in
vacuum passes through a piece of silica
(n = 1.458).
(a) Find the speed of light in silica.
(2.06 x 108 ms-1)
(b) What is the wavelength of this light in
silica ? (404 nm)
(c) Find the frequency of the light.
(5.09 x 1014Hz)
62
Refraction at a plane surface 22.2 Refraction at a Plane and Spherical Surfaces
Exercise
2. A light ray of wavelength 589 nm moves
from inside the glass ( n = 1.52 ) toward
the glass-air interface at an angle of
30.0 to the normal. Determine the angle
of refraction.
(49.5 away from the normal)
3. A beam of light traveling in air is incident
on a transparent plastic material at an
angle of incidence of 50o. The angle of
refraction is 35o. What is the index of
63
refraction of the plastic ? (1.34)
Refraction at a plane surface 22,2 Refraction at a Plane and Spherical Surfaces
Exercise
4. A ray of light in water (n = 1.33) is
incident upon a plate of glass (n = 1.5) at
an angle of 40 o . What is the angle of
refraction into the glass ?
(34.7o)
5. Light of wavelength 50 nm in a particular
glass has a speed of 1.7 x 108 m/s.
What is the index of refraction for this
glass? What is the wavelength of this
light in air ? (1.76, 1146 nm)
64
22.2 Refraction at a Plane and Spherical Surfaces
Refraction at A Spherical Surface
n2  n1
A
r
u
v
O = an object point
I = an image point
C = center of curvature
65
Refraction at A Spherical Surface
n2  n1
A
r
u
n1 n2 n2  n1
 
u v
r
v
(equation of refraction
of a spherical surface)
object-image relationship
66
Refraction at A Spherical Surface
To obtain the magnification of an
image formed by refraction at a
spherical surface.
Q
n1
y
Ø
P
V
n2
P’
y’
C
Ø’
Q’
u
v
67
Refraction at A Spherical Surface
n1 n2 n2  n1
 
u v
r
n1v
y'
m 
y
n2u
These equations can be applied to
both convex and concave surfaces,
provided that sign rules (refer to next
table) are obeyed and whether n2 is
greater or less than n1.
68
Refraction at A Spherical Surface
Quantity
Sign
u
+
u
v
+
v
r
+
r
-
r
∞
Remarks
Real object (object is on the front side
of the surface)
Virtual object (back side of the surface)
Real image (OPPOSITE side with
object)
Virtual image (SAME side with object)
Center of curvature is located in more
dense medium
Center of curvature is located in less
dense medium
Flat surface
69
Refraction at A Spherical Surface
(n2 - n1)
n1 n2
+ =
u v
r
Use sign convention for r :
• +ve if center of curvature is
located in more dense medium.
• -ve if center of curvature is
located in less dense medium.
70
Refraction at A Spherical Surface
Example 1.3.4
A cylindrical glass rod has an index of
refraction 1.50. One end is ground to a
hemispherical surface with radius r = 20 mm.
A point object on the axis of the rod, 80 mm
to the left of the vertex. The rod is in air.
Calculate
a) the image distance
b) magnification of the image.
71
Refraction at A Spherical Surface
Solution 1.3.4
n1 = 1 , n2 = 1.50 , r =+20 mm , u = + 80 mm
n1 n2 n2  n1
a)
 
u v
r
OR
(n2 - n1)
n1 n2
+ =
u v
r
1 1.50 1.50  1


80
v
20
v  120 mm
+ve (to the right of
vertex – back side)
72
Refraction at A Spherical Surface
Solution 1.3.4
n1 = 1 , n2 = 1.50 , r =+20 mm , u = + 80 mm
n1v
y'
b) m   
y
n2u
(1)(120)
m
 1
(1.50(80)
same size but inverted
73
Refraction at A Spherical Surface
Example 1.3.5
The rod in example 1.3.4 is immersed in
water of index 1.33. Other quantities have
the have the same values in example 1.3.
Calculate the image distance and its
magnification.
74
Refraction at A Spherical Surface
Solution 1.3.5
n1 = 1 , n2 = 1.50 , r =+20 mm , u = + 80 mm
n1 n2 n2  n1
a)
 
u v
r
OR
(n2 - n1)
n1 n2
+ =
u v
r
1.33 1.50 1.50  1.33


80
v
20
v  185 mm
+ve (to the left of
vertex – front side)
75
Refraction at A Spherical Surface
Solution 1.3.5
n1 = 1 , n2 = 1.50 , r =+20 mm , u = + 80 mm
n1v
y'
b) m   
y
n2u
(1)(185)
m
 1.54
(1.50(80)
greater and upright
76
Refraction at A Spherical Surface
Example 1.3.6
A set of coins is embedded in a spherical
plastic paper-weight having a radius of
3.0 cm. The index of refraction of the
plastic is n1 = 1.50. One coin is located
2.0 cm from the edge of the sphere
(figure-next page). Find the position of
the image of the coin.
77
78
Refraction at A Spherical Surface
v
(n2 - n1)
n1 n2
+ =
u v
r
v = - 0.017 m
79
Example 1.3.7
Calculate the image distance and
magnification.
Solution
n1 n2 n2  n1
 
u v
r
1.33 1.00 1.00  1.33


8
v

1.00  8 
v
 6 cm
1.33
u = real/actual depth
v = apparent depth
v
8 cm
80
Refraction at A Spherical Surface
Solution 1.3.7
n1v
m
n2u
1.33  6 
m
1.00  8 
m  0.9975
v
8 cm
81
Refraction at A Spherical Surface
Exercise
1.A point object is 25.0 cm from the centre of
a glass sphere of radius 5.0 cm. The
refractive index of glass is 1.50. Find the
position of the image formed due to
refraction by
a. the first spherical glass surface.
82
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
for real object only.
 
u
v
f
 Use lensmaker’s equation:
1 1 
1 
  n  1   
f 
  r1 r2 
83
THIN
LENSES
22.3
Thin
lenses
Introduction
• A lens is a transparent object with two
refracting surfaces whose principal axes
coincide.
• This lens is usually circular, and its two
faces are portions of a sphere.
• Two types of lenses :
i) converging lens (convex)
ii) diverging lens (concave)
84
Introduction
R2
22.3 Thin lenses
R1
Converging lens
R1
R2
Diverging lens
Converging lens- a
lens causes incident
parallel rays to
converge after
exiting the lens.
Diverging lens- a
lens causes incident
parallel rays to
diverge after exiting
the lens.
85
Introduction
22.3 Thin lenses
Terms and definition
Converging lens
86
Introduction
22.3 Thin lenses
Terms and definition
Diverging lens
87
Introduction
22.3 Thin lenses
Terms and definition
• Principal axis – a straight line passing through
the very center of the lens and perpendicular to
its two surfaces.
• Focal point – the point at which the rays cross.
• Focal length – the distance between the focal
point and the center of the lens.
88
Several types of diverging lenses
22.3 Thin lenses
Converging lenses are thicker at the
center than at the edges whereas
diverging lenses are thinner at the
center.
89
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.
90
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.
91
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.
92
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)
93
Image Formation by Thin Lenses
Converging Lens
1) Object at 2F1 , the image is
a) At 2F2
b) Real
c) Inverted
d) Same size as object
94
Image Formation by Thin Lenses
Converging Lens
2) Object beyond 2F1 , the image is
a) Between F2 and 2F2
b) Real
c) Inverted d) Smaller than object
95
Image Formation by Thin Lenses
Converging Lens
3) Object between F1 and 2F1 , the image is
a) Beyond 2F2
b) Real
c) Inverted d) Larger than object
96
Image Formation by Thin Lenses
Converging Lens
4) Object at F1 , the image is at infinity
F2
O
F1
u
I at infinity
97
Image Formation by Thin Lenses
Converging Lens
5) Object between lens and F1, the image is
a) Behind the object
b) Virtual
c) Upright
d) Larger than object
98
Diverging Lens
The image always ,
Image Formation by Thin Lenses
a) Between lens and F1
a) Virtual
b) Upright
d) Smaller than object
F1
99
The Lens Equation
thin-lens equation
(1) = (2)
v f v
1 1 1


 
f
u
u v f
100
u
v
Thin lens formula
1 1 1
 
f u v
101
Sign Convention
Q
Positive (+)
f
Negative (-)
Converging (convex) Diverging (concave)
u In front of the lens
Behind the lens
v Behind the lens (real)
OPPOSITE SIDE
hi Upright image with
respect to the object
m Upright image with
respect to the object
In front of the lens
SAME SIDE
Inverted image with
respect to the object
Inverted image with
respect to the object
102
1.4 Thin lenses
The Lens Equation
Example 1.4.1
Linear Magnification, m
An object is placed 10 cm from the a 15
cm focal length converging lens. Find the
image position and its characteristics.
Solution
1 1 1
 
u v f
1
1
1
 
10 v 15
v  30 cm
v
30
m 
 3.0
u
10
-Magnified
- upright
- virtual (in front of the lens)
103
1.4 Thin lenses
Example 1.4.2
The Lens Equation
Linear Magnification, m
Where must a small insect be placed if a
25 cm focal length diverging lens is to form
a virtual image 20 cm in front of the lens?
Solution
1 1 1
 
u v f
1
1
1


u 20 25
u  100 cm
104
1.4 Thin lenses
Example 1.4.3
The Lens Equation
Linear Magnification, m
A diverging meniscus lens has a focal
length of -16 cm. If the lens is held 10 cm
fro an object, where is the image located ?
What is the magnification of the lens ?
Solution
1 1 1
 
u v f
1 1
1
+ =
10 v - 16
v = -6.15 cm
6.15
m=
= 0.615
10
105
1.4 Thin lenses
Example 1.4.4
The Lens Equation
Linear Magnification, m
An object 450 mm from a converging lens
forms a real image 900 mm from the lens.
What is the focal length of the lens.
Solution
1 1 1
 
u v f
1
1
1
+
=
450 900 f
f = 30 cm
106
1.4 Thin lenses
Example 1.4.5
The Lens Equation
Linear Magnification, m
An object 6 cm high is held 4 cm from a
diverging meniscus lens of focal length -24 cm.
What are the nature, size, and location of the
image ?
Solution
1 1 1
 
u v f
3.43
1 1
1
m=
= 0.86
+ =
4
4 v - 24
v = -3.43 cm hi = m ho = 0.86 × 6
hi
m=
hi = 5.16 cm
ho virtual, upright
5.16 cm, -.3.43 cm
107
1.4 Thin lenses
Exercise
The Lens Equation
Linear Magnification, m
1)An object 8 cm high is placed 30 cm from
a thin converging lens of focal length 12
cm. What are the nature, size, and location
of the image formed ?
(real, inverted, -5.33 cm, +20 cm)
2) A 1.70 m tall person is standing in front of
a camera. The camera uses a converging
lens whose focal length is 0.0500 m. Find
the characteristics of the image formed on
the film.
(real,inverted, smaller(-.0204),-0.0347 m)
108
1.4 Thin lenses
Exercise
The Lens Equation
Linear Magnification, m
3) An object is 18 cm in front of a diverging
lens that has a focal length of -12 cm. How
far in front of the lens should the object be
placed so that the size of its image is
reduced by a factor of 2.0 ? (48 cm)
4) How far from a 50.0 mm focal length lens
must an object be placed if its image is to be
magnified 3.00 x and be real ? What if the
image is to be virtual and magnified 3.00 x ?
(66.7 mm, 33.3 mm)
109
1.4 Thin lenses
Exercise
The Lens Equation
Linear Magnification, m
5) A certain lens focuses an object 22.5 cm
away as an image 33.0 cm on the other side
of the lens. What type of lens is it and what is
its focal length ? Is the image real or virtual ?
(converging, 13.4 cm, real)
6) a) A 2.70 cm high insect is 2.20 m from a
135 mm focal length lens. Where is the
image, how high is it, and what type is it ?
b) What if f = -135 mm ?
(144 mm behind lens,real,inverted, 1.77 mm ;
110
127 mm in front of lens,virtual,upright, 1.56 mm)
Two Lenses System
111
Two Lenses System
f = +20 cm
B
A
p1
v1
p2
v2
112
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.
113
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.
114
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 )
115
Example 1.4.6
Two Lenses System
Two converging lenses, with focal lengths f1
= 10.0 cm and f2 = 20.0 cm, are placed 20.0
cm apart. An object is placed 30.0 cm in front
of the first. Calculate the position and the
magnification of the final image formed by
the combination of the two lenses.
116
Solution 1.4.6
Two Lenses System
Given : u1 =30.0 cm, f1 = 10.0 cm , f2 = 20.0
cm, are placed 20.0 cm apart.
1 1
1
1 1 1


 
u2 v2 f 2
u1 v1 f1
1
1
1
1
1
1
 


30.0 v1 10.0
5.0 v2 20.0
v1  15.0 cm
v2  6.67 cm
Magnification, m = m1 x m2
v1
v2  15.0
6.67 
m   
 

  0.67
u1
u2  30.0
5.0 
117
Solution 1.4.6
Two Lenses System
118
Example 1.4.7
Two Lenses System
Two converging lenses, with focal lengths f1
= 20.0 cm and f2 = 25.0 cm, are placed 80.0
cm apart. An object is placed 60.0 cm in front
of the first. Calculate the position and the
magnification of the final image formed by
the combination of the two lenses.
(v2 = +50.0 cm , m= +0.500)
119
Exercise
Two Lenses System
1.A converging lens has a focal length of
0.080 m. An object is located 0.040 m to
the left of this lens. A second converging
lens has the same focal length as the first
one and is located 0.120 m to the right of
it. Relative to the second lens, where is
the final image located ? (0.133 m)
120
Exercise
Two Lenses System
2. A coin is located 20.0 cm to the left of a
converging lens ( f = 16.0 cm) A second,
identical lens is placed to the right of the
first lens, such that the image formed by
the combination has the same size and
orientation as the original coin. Find the
separation between the lenses.
(160 cm)
121
Exercise
Two Lenses System
3. Two thin converging lenses are placed
60 cm apart and have the same axis.
The first lens has a focal length 10 cm,
and the second has a focal length of
15.0 cm. If an object 6.0 cm high is
placed 20 cm in front of the first lens,
what are the location and size of the
final image ? Is it real or virtual ?
(24 cm, 3.6 cm, real)
122
Thin Lenses Formula and Lens
maker’s Equation
Considering the ray diagram of refraction for 2
spherical surfaces as shown in figure below.
t  v1
r2
v1
u1
r1
A
n1
v2
D
n2
C1
O
I1
n1
C2
P2
P1
B
E
t
I2
123
Lens maker’s
equation
 1 1 
1  n2
   1  
f  n1
 r1 r2 
where
f : focal length
r1 : radius of curvature of first refracting surface
r2 : radius of curvature of second refracting surface
n1 : refractive index of the medium
n2 : refractive index of the lens material
124
Note :
– If the medium is air (n1= nair=1) thus the lens
maker’s equation will be
1 1
1
 n  1  
f
 r1 r2 
Where,
n : refractive index of the lens material
125
Lensmaker’s Equation
 1 1 
1 n
   1  
f  n1  R1 R2 
n for lens
+ve sign
n1 for medium
126
Sign Convention
Lensmaker’s Equation
Q Sign
Remarks
R
+ Convex surface
R
-
Concave surface
R
f
f
∞
+
-
Flat surface
Converging lens
Diverging lens
127
Lensmaker’s Equation
OR
( )
1
1 1
= (n - 1)
+
f
R1 R2
n for lens, the lens in air
Use sign convention for R :
• +ve for convex surface
• -ve for concave surface
128
Example 1.4.8
Lensmaker’s Equation
Suppose the lens in the figure below has
n= 1.52 and is placed in air. Calculate its
focal length. What type of this lens ?
46.2 cm
129
Solution 1.4.8
Lensmaker’s Equation
Given: n= 1.52 , n1 = 1, R1 = +22.4 cm,
R2 = +46.2 cm
 1 1 
1 n
   1  
f  n1  R1 R2 
1  1.52  1
1 

 1


f  1
 22.4 46.2 
f  0.67 cm
46.2 cm
Converging lens → f +
130
Lensmaker’s Equation
Example 1.4.9
Each face of a double convex lens has a
radius of 20.0 cm. The index of refraction of
the glass is 1.50. Calculate the focal length
of this lens
a) in air and
b) when it is immersed in carbon disulfide
(n= 1.63)
131
Solution 1.4.9
Lensmaker’s Equation
Given : R1 = +20.0 cm,R2 = -20.0 cm,n = 1.50
(n = 1.63) 1  n  1 1 
a) In air (n1 = 1)
   1  
f  n1  R1 R2 
1  1.50  1
1 

 1


f  1
 20.0 20.0 
f  20.0 cm
b) In carbon
disulfide
(n1 = 1.63)
1  1.50  1
1 

 1


f  1.63  20.0 20.0 
f  125 cm  diverging lens
132
Example 1.4.10
Lensmaker’s Equation
A plano-convex is to be constructed out of
glass so that it has a focal length of 40 cm.
What is the radius of curvature of the curve
surface.
Solution
( )
1
1 1
= (n - 1)
+
f
R1 R2
( )
1
1 1
= (1.5 - 1)
+
40
∞ R2
R2 = 20 cm
133
Exercise
Lensmaker’s Equation
1.The curve surface of a plano-concave
lens has a radius of -12 cm. What is the
focal length if the lens is made from a
material with a refractive index of 1.54 ?
(-22.2 cm)
2. A plastic lens (n = 1.54) has a convex
surface of radius 25 cm and a concave
surface of -70 cm. What is the focal
length ? Is it diverging or converging ?
(72.0 cm, converging)
134
Exercise
Lensmaker’s Equation
3. A plano-convex lens is ground from
crown glass(n = 1.52). What should be
the radius of the curved surface if the
desired focal length is to be 400 mm ?
(208 mm)
4. A thin meniscus lens is formed with a
concave of radius -40 cm and a convex
surface of radius +30 cm. If the resulting
focal length is 79 .0 cm, what was the
index of refraction of the transparent
135
material ? (2.52)