T . Norah Ali Almoneef 1

 We represent light using rays, which are straight lines emanating from an object. This is an idealization, but is very useful for geometric optics.
T . Norah Ali Almoneef 2

When a light ray travels from one medium to another, part of the incident light is reflected and part of the light is transmitted at the boundary between the two media.
The transmitted part is said to be refracted in the second medium.
incident ray reflected ray
3

Law of reflection
: the angle of reflection (that the ray makes with the normal to a surface) equals the angle of incidence.
• The incident ray, the reflected ray and the normal all lie in the same plane.
T . Norah Ali Almoneef 4

(parallel rays will all be reflected in the same directions)
(parallel rays will be reflected in a variety of directions)
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example:
True of false: Reflection of light by a rough surface does not obey the laws of reflection .
example
• If the angle of incidence of a ray of light is 42°, what is each of the following?
• The angle of reflection a. The angle the incident ray makes with the mirror b. The angle between the incident ray and the reflected
T . Norah Ali Almoneef
42°
48°
90°
6

• All the light rays from the source that reflect from the mirror SEEM to come from the same point behind the mirror
T . Norah Ali Almoneef 7

Formation of image in a plane mirror
Stand in front of a looking glass and look at your image.
1. Can you receive your image on a screen ?
2. Is the image erect or inverted?
3. Is the image the same size or larger or smaller?
4. What happens when you tilt your head to the right?
5. How does the image move when you step forward or backward
6. Where is the image formed ?
T . Norah Ali Almoneef 8

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• A real image is one in which light actually passes through the image point
– Real images can be displayed on screens
• A virtual image is one in which the light does not pass through the image point
– The light appears to come (diverge) from that point
– Virtual images cannot be displayed on screens
T . Norah Ali Almoneef 10

• Simplest possible mirror
• Properties of the image can be determined by geometry
• One ray starts at P, follows path PQ and reflects back on itself
• A second ray follows path
PR and reflects according to the Law of Reflection
T . Norah Ali Almoneef
p=q
11

• The object distance is the distance from the object to the mirror or lens
– Denoted by p
• The image distance is the distance from the image to the mirror or lens
– Denoted by q
• The lateral magnification of the mirror or lens is the ratio of the image height (h ’) to the object height (h)
– Denoted by ( M =h’/h)
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• The image is as far behind the mirror as the object is in front ( q = p)
• The image is unmagnified
– The image height is the same as the object height
( h’ = h and M = 1)
• The image is virtual (cannot be picked up on a screen)
• The image is upright
– It has the same orientation as the object
• There is an apparent left-right reversal in the image
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Now you look into a mirror and see the image of yourself.
In front of the mirror.
On the surface of the mirror.
 Behind the mirror.
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A girl can just see her feet at the bottom edge of the mirror.
Her eyes are 10 cm below the top of her head.
(a) What is the distance between the girl and her image in the mirror?
150 m

Distance = 150

2 = 300 cm
T . Norah Ali Almoneef
150 m
17

A boy of height 1.5 m stands 5 m in front of a plane mirror.
T . Norah Ali Almoneef 18

Which of the following descriptions about the image formed by a plane mirror is INCORRECT ?
A Light rays come from the image to our eyes.
B The image cannot be projected on a screen.
C It is as far away from mirror as the object is in front.
D It is virtual.
T . Norah Ali Almoneef 19

• All electromagnetic waves have the same speed c = 3.00×10 8 m/s in a vacuum
• In a material medium (glass, water, etc.), an EM wave travels at a speed v < c
• An EM wave generally travels at different speeds in different materials
• The change in speed as a light ray goes from one material to another causes the ray to deviate from its incident direction
• This change in direction is called refraction
T . Norah Ali Almoneef 20

•Refraction is when waves speed up or slow down due to travelling in a different medium ,and it can cause light rays to change their direction
•A medium is something that light waves will travel through
•Light rays are slowed down by the water
•Causes the ruler to look bent at the surface
•The mediums in this example are water and air
The degree that light bends when it enters a new medium is called the “ index of refraction”
T . Norah Ali Almoneef 21

A thin lens is one whose thickness t is small in comparison to distances of optical properties (radius of curvature, focal length, image and object distances. Light is reflected from a mirror. Light is refracted through a lens. For a thin lens, the thickness, t , of the lens can be neglected
• Lenses are commonly used to form images by refraction
• Lenses are used in optical instruments
– Cameras, telescopes, microscopes
• Images from lenses
– Light passing through a lens experiences refraction at two surfaces
– The image formed by one refracting surface serves as the object for the second surface
R focal point t
T . Norah Ali Almoneef 22

• These are examples of converging lenses
• They have positive focal lengths
• They are thickest in the middle
• Used in cameras, telescopes, human eye
• These are examples of diverging lenses
• They have negative focal lengths
• They are thickest at the edges
1.
Used in cameras & telescopes to correct spherical aberation, and also eyeglasses
T . Norah Ali Almoneef 23

• Formed by two curved boundaries between transparent media.
• Lenses may have spherical surfaces (lens-maker’s equation).
Most modern lenses have non-spherical curved surfaces to avoid spherical aberration.
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C
F
• The distance of F from C is the focal length f of the lens.
the two focal lengths are equal focal length
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• The line through the optical centre and 2 foci is called the principal axis .
F' optical centre
C
F principal axis
• Parallel rays are refracted outwards .
principal focus focal length
• Refracted rays appear to spread from a point called the principal focus F .
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• The parallel rays pass through the lens and converge at the focal point
– Lens that converges (brings together) light rays.
– Forms real images and virtual images depending on position of the object
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• The parallel rays diverge after passing through the diverging lens
• The focal point is the point where the rays appear to have originated
– Diverges light rays
– All images are erect and reduced
T . Norah Ali Almoneef 29

Lens Equations
 The equations can be used for both converging and diverging lenses
 A converging lens has a positive focal length
 A diverging lens has a negative focal length
Q
M x
F s f
T h
S
A
’ h
’ f
’
F s’ x’
’
M
’
Q
T . Norah Ali Almoneef 30

Q’TS and F’TA are similar triangles,
QTS and FAS are similar triangles,
 h’ + h
= s’

h’ + h s
= h f’ f h’
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h’ + h h
= s’ f’
Adding the two equations:
h’ + h s’
+
h’ + h s
=
h’ + h s h f’
+
Since f = f’ for a thin lens,
h’ + h s’
+
h’ + h s
Multiplying through by
=
1 h + h’
h’ + h f
= f h’ f h’
1 s’
+
1 s
=
1 f
Lens Formula
T . Norah Ali Almoneef 32

• The lateral magnification of the image is
M  h  h
  s  s
• When M is positive, the image is upright and on the same side of the lens as the object h s

• When M is negative, the image is inverted and on the side of the lens opposite the object

T . Norah Ali Almoneef 34

Light rays converge to a point.
O
Image formation by a lens
A Real images
I
Image can be captured by a screen.
Hence called ‘ real ’.
O
I
Since only convex lenses converge light rays, real images can only be formed by convex lenses screen

Light rays diverge from a point.
I
No rays actually come from the image.
O
Hence called ‘ virtual ’.
convex lens

Light rays diverge from a point.
I
O concave lens

Ray 1: A ray parallel to lens axis passes through the far focus of a converging lens or appears to come from the near focus of a diverging lens.
Converging Lens
Ray 1
F
Diverging Lens
Ray 1
F

Ray 2: A ray passing through the near focal point of a converging lens or proceeding toward the far focal point of a diverging lens is refracted parallel to the lens axis.
Converging Lens
Ray 1
Ray 2
F
Diverging Lens
Ray 1
Ray 2
F

Ray 3: A ray passing through the center of any lens continues in a straight line.
The refraction at the first surface is balanced by the refraction at the second surface.
Converging Lens
Ray 1
Ray 2
F
Ray 3
Diverging Lens
Ray 1
Ray 2
F
Ray 3

The Thin-Lens Equation
This gives us the thin-lens approximation, as well as the magnification:
M
  s ' s magnification
M
  s ' s
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C
1
Converging Lens
Principal axis C
2
C
1
& C
2 are the centres of the spheres of which the surfaces of the lens form a part
The line through C
1
& C
2 form the principal axis
T . Norah Ali Almoneef 42

C
1
Converging lens
F
Normal
F
A ray of light on entering the lens is refracted towards the normal and on leaving away from the normal. Since surfaces are inclined towards each other the ray is refracted towards the principal axis.
Rays parallel to the principal axis converge towards a point called the principal focus F.
Since light can travel equally well in both directions, there are two foci.
T . Norah Ali Almoneef 43
C
2

To locate the position of an image in a convex lens we use two of the following rays of light
1 parallel to the principal axis emerging through focus
2 striking the centre of the lens passes straight through (if lens is thin)
3 through the focus emerging parallel to principal axis.
F F
F = Focal point
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object
Images in Convex lens
Image formed in convex lens when the object is placed outside twice the focal length
F = Focal point s
F
1 f

1 s

1 s 
F f s
T . Norah Ali Almoneef
Image
Real, inverted & diminished
45

object
Images in Convex lens
Image formed in convex lens when the object is placed at twice focal length
F = Focal point
1 f

1 s

1 s 
F F f s s
Image Real, inverted & same size as object
T . Norah Ali Almoneef 46

object
Images in Convex lens
Image formed in convex lens when the object is placed at the focus
F = Focal point
F F
1 f

1 s

1 s  f s
T . Norah Ali Almoneef
Image at Infinity
47

object
Images in Convex lens
Image formed in convex lens when the object is placed inside the Focus
F = Focal point
F F f s
Image Virtual, magnified & upright
T . Norah Ali Almoneef
48
1 f

1 s

1 s 
48

raybox
Experiment to find the focal length of a convex lens lens s s
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Convave lens
To locate the position of an image in a concave lens we use two of the following rays of light
F
1. A ray which strikes the lens travelling parallel to principal axis is refracted as if it came from focus
F
3. A ray heading for the focus on striking the lens is refracted parallel to principal axis
F = Focal point
T . Norah Ali Almoneef
2. A ray striking the centre of the lens passes straight through
( if lens is thin )
50

object
Images in Concave lens
Image formed in concave lens when the object is placed in front of lens
F s s f
F
1 f

1 s

1 s 
F = Focal point
T . Norah Ali Almoneef 51

Draw an arrow to represent the location of an object, then draw any two of the rays from the tip of the arrow. The image is where lines cross.
1. Is the image erect or inverted?
2. Is the image real or virtual?
• Real images are always on the opposite side of the lens. Virtual images are on the same side.
3. Is it enlarged, diminished, or same size?

2F F
F 2F
Real; inverted; diminished
1. The image is inverted , i.e., opposite to the object orientation.
2. The image is real , i.e., formed by actual light on the opposite side of the lens.
3. The image is diminishe d in size, i.e., smaller than the object.
Image is located between F and 2F

2F F
F 2F
Real; inverted; same size
1. The image is inverted , i.e., opposite to the object orientation.
2. The image is real , i.e., formed by actual light on the opposite side of lens.
3. The image is the same size as the object.
Image is located at 2F on other side

2F
F 2F
Real; inverted; enlarged
F
1. The image is inverted , i.e., opposite to the object orientation.
2. The image is real ; formed by actual light rays on opposite side
3. The image is enlarged in size, i.e., larger than the object.
Image is located beyond 2F

F 2F
Parallel rays; no image formed
2F F
When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is formed.

2F F
F 2F
Virtual; erect; enlarged
1. The image is erect , i.e., same orientation as the object.
2. The image is virtual, i.e., formed where light does NOT go.
3. The image is enlarged in size, i.e., larger than the object.
Image is located on near side of lens

All images formed by diverging lenses are erect , virtual , and diminished . Images get larger as object approaches.
Diverging Lens Diverging Lens
F F

An object is placed 6.0 cm in front of a convex thin lens of focal length 4.0 cm. Where is the image formed and what is its magnification and power?
s = 6.0 cm f = 4.0 cm
1 s’
+
1 s
=
1 f f
1
_
1 s
=
1
4
P =
1
6
1
=
1 s’ s’ = 12 cm
= 25.0 D
M = - 12 / 6 = -2
0.04 m
Negative means real, inverted image
1 s’
T . Norah Ali Almoneef 59

1 s’
+
1 s
=
1 f
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Short-sight defect light from distant object falls short of retina
Corrected light from distant object with help of a diverging (concave) lens
T . Norah Ali Almoneef
62 falls on retina
62

Long-sight defect light from near object falls ‘behind’ retina
Corrected light from near object with help of a converging
(convex) lens
T . Norah Ali Almoneef
63 falls on retina
63

convex lens
( converging lens) concave lens
( diverging lens)

A light ray is incident on a…
A light ray is incident on a convex lens.
Which one represents the path of the light ray?
Z
Y
X
A
B
C
Path X.
Path Y.
Path Z.

Q2 A light ray is incident on a…
A light ray is incident on a concave lens.
Which one represents the path of the light ray?
Z
Y
X
A
B
C
Path X.
Path Y.
Path Z.

• For a converging lens , when the object distance is greater than the focal length
( s > ƒ )
– The image is real and inverted
• For a converging lens , when the object is between the focal point and the lens, ( s < ƒ )
– The image is virtual and upright
• For a diverging lens , the image is always virtual and upright
– This is regardless of where the object is placed
T . Norah Ali Almoneef 67

• The lens has an index of refraction n and two spherical surfaces with
– R
1 radii of R
1 and is the radius of curvature of
R the lens surface that the light of
2 the object reaches first
– R
2 is the radius of curvature of the other surface
• The object is placed at point O at a distance of s in front of the first surface
T . Norah Ali Almoneef 68

R
1 and R
2 are interchangeable
R
1 R
2
-
+
R
1
, R
2
= Radii n= index of glass f = focal length
1 f

( n

1)


1

1
R R
1 2


1.
R
1 and surface.
R
2 are positive for convex outward surface and negative for concave
2.
Focal length f is positive for converging and negative for diverging lenses.
T . Norah Ali Almoneef 69

The focal length f for a lens.
1 f

( n

1)


1

1
R R
1 2


The Lensmaker’s Equation: Positive
(Convex)
R
1
R
2
Negative
(Concave)
R
Surfaces of different radius
T . Norah Ali Almoneef
Sign convention
70

R
1 and R
2 are interchangeable
R
1
, R
2
= Radii
+
R
1 n= index of glass f = focal length
R
2
-
1 f

( n

1)


1

1
R R
1 2


1.
R
1 and surface.
R
2 are positive for convex outward surface and negative for concave
2.
Focal length f is positive for converging and negative for diverging lenses.

R
1
R
2 positive negative
R
1
R
2 n =1 n≠ 1 n =1
The quantity, f , is the focal length of the lens. It ’s the single most important parameter of a lens. It can be positive or negative.
f > 0
R
1
R
2 positive positive
The NORMAL (convex) lens above is considered to have R
1 and R
2 both positive, and so f must be positive f < 0
R
1
R
2 negative negative
The (concave) lens is considered to have R
1 and R
2 both negative, and so f must be negative

R
1
R
2 positive negative
R
1
R
2 n =1 n≠ 1 n =1
The quantity, f , is the focal length of the lens. It ’s the single most important parameter of a lens. It can be positive or negative.
f > 0
R
1
R
2 positive positive
The NORMAL (convex) lens above is considered to have R
1 and R
2 both positive, and so f must be positive
T . Norah Ali Almoneef f < 0
R
1
R
2 negative negative
The (concave) lens is considered to have R
1 and R
2 both negative, and so f must be negative
73

• The focal length of a lens is related to the curvature of its front and back surfaces and the index of refraction of the material
{ n lens n medium
1} x {
1
R
1

1
R
2
} =
1 f
• If the medium is air
{ n lens n air
1} x {
1
R
1

1
R
2
} =
1 s

1 s
• ( n for the lens)
1 f
 ( n  1 )


1
R
1
This is called the lens maker’s equation

1
R
2


T . Norah Ali Almoneef
74

T . Norah Ali Almoneef 75

R
1 and R
2 are interchangeable
R
1
, R
2
= Radii
+
R
1 n= index of glass f = focal length
R
2
-
1 f

( n

1)


1

1
R R
1 2


1.
R
1 and surface.
R
2 are positive for convex outward surface and negative for concave
2.
Focal length f is positive for converging and negative for diverging lenses.

Example 1. A glass meniscus lens ( n = 1.5
) has a concave surface of radius –40 cm and a convex surface whose radius is
+20 cm. What is the focal length of the lens.
+20 cm
R
1
= 20 cm, R
2
= -40 cm f
1

( n

1)


1

1
R R
1 2


-40 cm f
1

(1.5 1)


1


1
20 cm ( 40 cm



  40 cm n = 1.5
f = 80.0 cm Converging (+) lens .

Example : What must be the radius of the curved surface in a plano-convex lens in order that the focal length be 25 cm?
R
1
=

, f= 25 cm f
1
 
1)


1

0

1
R
2


1
25 cm

(1.5 1)


1
R
2

 
0.500
R
2
R
2
=?
R
1
=

f = ?
R
2
= 0.5(25 cm)
R
2
= 12.5 cm Convex (+) surface
.

It might be useful to solve the lens equation algebraically for each of the parameters:
1
 s
1 s '

1 f s
 s
 s
 f
 f s '
 s sf
 f f
 s
 s s
  s
Be careful with substitution of signed numbers!

Example. A magnifying glass consists of a converging lens of focal length 25 cm. A bug is 8 mm long and placed 15 cm from the lens. What are the nature, size, and location of image.
F
F
p = 15 cm; f = 25 cm
1
 s
1 s '

1 f s   s sf
 f

15 cm  25 cm
15 cm  25 cm
  37 .
5 cm s = -37.5 cm
The fact that S is negative means that the image is virtual (on same side as object).

Example 3 Cont.) A magnifying glass consists of a converging lens of focal length 25 cm. A bug is 8 mm long and placed
15 cm from the lens. What are size of image.
y’ y
F
s = 15 cm; s = -37.5 cm
F
M
 h '
  s ' h s h 
8 mm
 
 37 .
5 cm
15 cm
h’ = +20 mm
The fact that h’ is positive means that the image is erect. It is also larger than object.

Example : What is the magnification of a diverging lens ( f = -20 cm ) the object is located 35 cm from the center of the lens?
F
First we find q . . . then M
1
 s
1 s '

1 f
M
 h ' h
  s ' s s   sf s  f

35 cm   20 cm
35 cm  (  20 cm )
 12 .
7 cm
M   s ' s

 (  12 .
7 cm )
35 cm
 0 .
364
s = +12.7 cm
M = +0.364

A Converging lens is one that refracts and converges parallel light to a real focus beyond the lens. It is thicker near the middle.
F
F
F F
The principal focus is denoted by the red F .
A diverging lens is one that refracts and diverges parallel light which appears to come from a virtual focus in front of the lens.

y
2F
F 2F
F -y’ f p q
Lens Equation:
1 s

1 s '

1 f
Magnification:
M
 h ' h
  s ' s

• The near focal point is the focus F on the same side of the lens as the incident light.
• The far focal point is the focus F on the opposite side to the incident light.
Converging Lens
Far focus
Diverging Lens
Far focus
F
Near focus
F F
Near focus
F

Image Properties Convex vs. Concave Lenses
T . Norah Ali Almoneef 86

Image Properties Sign Conventions
-
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All images formed by diverging lenses are erect , virtual , and diminished . Images get larger as object approaches.
Diverging Lens Diverging Lens
F F

h
2F
F 2F
F -h’ f s s’
Lens Equation:
1
 s
1 s '

1 f
Magnification:
M
 h '
  s ' h s

Parallel ray x
2 f focus (f) x
Object beyond 2f
T . Norah Ali Almoneef focus (f) x
Image is:
Real
Inverted
Reduced
Appears between f and 2f
2 f x
90

Image is:
Real
Inverted
Same size
Appears between f and 2f
Parallel ray x
2 f focus (f) x
Object at 2f
T . Norah Ali Almoneef focus (f) x
2 f x
91

Image is:
Real
Inverted
Enlarged
Appears beyond 2f
Parallel ray x
2 f focus (f) x
Object between f and 2f focus (f) x
T . Norah Ali Almoneef
2 f x
92

Apparent
Convergence
Of rays x
2 f focus (f) x
Object
Inside focus T . Norah Ali Almoneef
Image is:
Virtual
Erect
Enlarged
Appears on same
Side as Object focus (f) x
2 f x
93

2 f
Parallel ray f
Image is:
Virtual
Erect
Reduced
Appears on same
Side as object
T . Norah Ali Almoneef f 2 f
94

object f
1
 s
1 s '

1 f s s’ s is positive for objects to the left of lens, negative for objects to the right of lens (virtual objects).
s’ is positive for images to the right of lens, negative for images to the left of lens (virtual images).
f is positive for converging lenses, negative for diverging lenses.
T . Norah Ali Almoneef 95

A boy holds a magnifying glass at arm’s lengthHe looks at a poster through the glass and sees a magnified erect image. What happens to the image if he moves the lens closer to his eyes?
A It gets larger till it gets totally blurred at some distance.
B It gets larger, keeping erect all the way.
C It gets smaller and becomes totally blurred at some distance.
D It gets smaller, keeping erect all the way.

If you can capture an image...
If you can capture an image of a doll on a screen using a lens, which of the following may NOT be correct?
A The lens you use is a convex lens.
B The image is magnified.
C The image is real.
D The image is erect.

In the diagram, a ray parallel...
In the diagram, a ray parallel to the principal axis of the lens is reflected backwards.
What is the focal length of the cylindrical convex lens?
A
B
C
D
5 cm
10 cm
20 cm
40 cm

What happens to the image…
What happens to the image when the plane mirror is moved backwards?
A The image becomes blurred.
B The image becomes smaller.
C The image does not change.

Both convex and concave...
than the object if convex lenses are used.


 real images are always inverted
 virtual images are always upright
 real images are always in front
 of the mirror
 virtual images are always
 behind the mirror
 negative image distance means virtual image
 positive image distance means real image
T . Norah Ali Almoneef
T .Norah Ali Almoneef
101
101

T . Norah Ali Almoneef 102

A lens produces a sharply-focused, inverted image on a screen. What will you see on the screen if the lens is removed?
1. The image will be as it was, but much dimmer.
2. The image will be right-side-up and sharp.
3. The image will be right-side-up and blurry.
4. The image will be inverted and blurry.
5. There will be no image at all.
T . Norah Ali Almoneef 103

1. the distance at which an image is formed.
2. the distance at which an object must be placed to form an image.
3. the distance at which parallel light rays are focused.
4. the distance from the front surface to the back surface.
T . Norah Ali Almoneef 104

Example
An object is placed 20 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?
f s   20 cm
  10 cm
1 s s s
1

1

1
 f s


2

20
20 cm cm

1
10 cm
20
1 cm


1
20 cm
20
1 cm
Real image, magnification =
1
T . Norah Ali Almoneef 105

f
An object is placed 8 cm in front of a diverging lens of focal length 4 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the
 magnification of the image?
 4 cm (concave) s s
1


1 s
 4

1 f cm s s
1


1 f
 2

1 s cm

 0

1
4 cm

1
4 cm m   s / s  0 .
5  0
T . Norah Ali Almoneef 106

Example
24(b). Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.
Virtual side Real side
1 s
1 s


1
5
1 s

1
10


1 f

1
10 s   10 cm
F
1 s
.
.
F
2 m  h h m  
10
10
 
  1 s s
T . Norah Ali Almoneef
Image is real, inverted.
107


24(e). Given a lens with the properties (lengths in cm) R
1
= +30, R
= +30, s = +10, and n = 1.5, find the following: f, s and m. Is the
2 image real or virtual? Upright or inverted? Draw 3 rays.
Virtual side
F
1
.
R
2 p
Real side
R
1
F
2
.
m  h h
  s s f
1
  n  1 


 1
R
1

1
R
2


 f
1
  1 .
5  1 


 1
30

1
30 



1
30 f

30 cm
1 s

1 f

1 s
1 s

1
30

1
10
 
1
15 s   15 cm

T . Norah Ali Almoneef m  
 15
10
  1.5
Image is virtual, upright.
108

Example
An object is placed 5 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?
f s 

 5 cm
 10 cm s s
1 s
1

1

1
 f s



10
1
10 cm cm

10
1 cm
2
10 cm

1

5 cm
 1
10 cm
1 s

1 s

1 f
Virtual image , as viewed from the right, the light appears to be coming from the (virtual) image, and not the object.
Magnification = +2
T . Norah Ali Almoneef 109
109

1 s
A diverging lens with f = -20 cm h = 2 cm, s = 30 cm

1 s '

1 f
1
30

1 s 

1
 20 s    12 cm
M  h  h
  s  s
M
 h

 
2 h
 
0 .
8 cm

12
30

0 .
4
The image is virtual and upright
A converging lens with f = 10 cm
(a) S = 30 cm
1
30 s  

1 s
15

 cm
1
10
M   s  s
 
15
30
  0 .
5
The image is real and inverted
(b) S = 10 cm s   
The image is at infinity
(c) s = 5 cm
1
5 s 


1 s 

 10
1
10 cm
T . Norah Ali Almoneef
M   s  s
 
 10
5
 2
The image is virtual and upright 110

The thin lens equation,, can be used to find the image distance:
1 s

1 s '

1 f
1 f
-
1 s

1 s 
1
10 s  cm
 15

30 cm
1 cm

2
30 cm

1 s 
M   s  s
 
15
30
  0 .
5
T . Norah Ali Almoneef 111

1 f
-
1 s

1

10 s   cm
 10
1 s 
5
1 cm cm
 
1
10 cm

1 s 
T . Norah Ali Almoneef 112

Repeat Example for a diverging lens of focal length 10.0 cm.
Solution
(A) We begin by constructing a ray diagram as in Figure
36.31a taking the object distance to be 30.0 cm. The diagram shows that we should expect an image that is virtual, smaller than the object, and upright. Let us now apply the thin lens equation with p = 30.0 cm:
1 f
-
1 s

1 s 
1
 s 
10
 cm
 7 .
5

30 cm
1 cm
 
30
4 cm

1 s 
M   s  s
 
(  7 .
5 cm )
30
 0 .
25
T . Norah Ali Almoneef 113

T . Norah Ali Almoneef 114

T . Norah Ali Almoneef 115

QUICK QUIZ
An object is placed to the left of a converging lens. Which of the following statements are true and which are false? (a) The image is always to the right of the lens. (b) The image can be upright or inverted. (c) The image is always smaller or the same size as the object..
(a) False. A virtual image is formed on the left side of the lens if s < f .
(b) True. An upright, virtual image is formed when s < f , while an inverted, real image is formed when s > f .
(c) False. A magnified, real image is formed if 2 f > s > f , and a magnified, virtual image is formed if s > f .
T . Norah Ali Almoneef 116

When a real object is placed just inside the focal point F of a diverging lens, the image is
A.virtual, erect, and diminished.
B.real, inverted, and enlarged.
C.real, inverted, and diminished.
D.virtual, erect, and enlarged.
E. virtual, inverted, and diminished.
T . Norah Ali Almoneef 117

An object is placed in front of a plano-concave lens at r
1
/2. The image produced by the lens is
A.inverted, real and reduced in size.
B.inverted, virtual and enlarged in size.
C.upright, virtual and reduced in size.
D.upright, virtual and enlarged in size.
E. upright, real and enlarged in size.
T . Norah Ali Almoneef r
1
118

Which of the following statements is false?
A.The image produced by a diverging lens is always virtual, upright and reduced in size.
B.The image produced by a converging lens can be virtual, upright and magnified in size.
C.The image produced by a converging lens cannot be virtual, upright and reduced in size.
D.The image produced by a converging lens cannot be real, inverted and reduced in size.
119
T .Norah Ali Almoneef

To project an image onto a screen using a lens,
A. the lens must be diverging and the object must be farther from the lens than the second focal point.
B. the lens must be converging and the object must be between the first focal point and the lens.
C. the lens must be diverging and the image must be farther from the lens than the second focal point.
D. the lens must be converging and the object must be farther from the lens than the first focal point.
E. the lens must be diverging and the object must be between the first focal point and the lens.
T . Norah Ali Almoneef 120

A real image is formed by a converging lens. If a weak diverging lens is placed between the converging lens and the image, where is the new image located?
A. farther from the converging lens than the original image
B. closer to the converging lens than the original image
C. at the original image position
T . Norah Ali Almoneef 121

• An object is 32 cm to the left of a convex lens of +8.0 cm focal length.
– Where is the image located?
– Is the image
• real or virtual
• upright or inverted
• magnified or reduced
• Image is located +11 cm from lens
• Real reduced (m = -.33)
• inverted (m is negative)
T . Norah Ali Almoneef 122

• An object is located 4.0 cm to the left of a convex lens, the focal length is 6.0 cm. Is the object
– real or virtual
– magnified or reduced
– upright or inverted
• Virtual (s = -12 cm)
• Magnified (m = 3)
• Upright (m is positive)
T . Norah Ali Almoneef 123

T .Norah Ali Almoneef 124

T .Norah Ali Almoneef 125

One ray is shown as it leaves an object placed before a positive lens. If this ray were continued to show its path through the lens, it would pass through which point? (F marks the two focal points.)
T .Norah Ali Almoneef 126

A point object O is placed in front of a thin converging lens.
F marks the two focal points. Observers are at 1, 2, and 3.
The image of point O is seen by the
A.observer at 1 only.
B.observer at 2 only.
C.observer at 3 only.
D.observers at 1, 2, and 3.
E. observers at 1 and 2.
T .Norah Ali Almoneef 127

A ray of light leaves point O and passes through a thin positive lens. It crosses the principal axis at which point? (F marks the two focal points.)
T .Norah Ali Almoneef 128

When the ray in the diagram is continued through the diverging lens, it passes through which point? (F marks the two focal points.)
T .Norah Ali Almoneef 129

The image of the encircled point on the object formed in the positive lens is at which circle? (F marks the two focal points.)
T .Norah Ali Almoneef 130

The image produced by the converging lens is at which point? (F marks the two focal points.)
T .Norah Ali Almoneef 131

The image of the object formed by the diverging lens is located at which point? (F marks the two focal points.)
T .Norah Ali Almoneef 132

A concave (diverging) lens can produce an image that is
A.virtual, inverted, and magnified.
B.real, erect, and magnified.
C.diminished, erect, and virtual.
D.magnified, erect, and virtual.
E. diminished, real, and erect.
T .Norah Ali Almoneef 133

A converging lens and a screen are so arranged that an image of the sun falls on the screen. The distance from the lens to the screen is
A.the focal length.
B.the object distance.
C.the magnifying power.
D.one-half the radius of curvature of one of the lens faces.
E. the average radius of curvature of the two lens faces.
T .Norah Ali Almoneef 134


The "power," P, of a lens is equivalent to the focal length, f. One can be found from the other.
•
• Definition of P in terms of f is
P (in diopters)
Meaning of P
 f
1
(meters)
• P is a measure of the ray bending power of the lens
• Large P means the lens bends rays more than if P were small
• Your eyeglass or contact lens prescription is usually given in diopters (P)
T . Norah Ali Almoneef
• The power of a converging lens is always positive because f is a positive number for a converging lens
– The converging lens always bends rays towards the 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 rays away from the axis behind the lens
135

• For lens in contact (separation is negligible)

1 f
1

1 f

1 f
P


P
1
1 f
1

 f
1
2
P
2
1 f
2 or :
(power is sum of each power)
T . Norah Ali Almoneef
136

T . Norah Ali Almoneef 137

• One of the basic problems is the imperfect quality of the images
– Largely the result of defects in shape and form
• Two common types of aberrations exist
• (a) Spherical aberration: Rays passing through different regions of a lens and do not come together in a common focal plane
• (B)Chromatic Aberration: Different dispersion of red and blue
T . Norah Ali Almoneef 138

• Different wavelengths of light refracted by a lens focus at different points
• Violet rays are refracted more than red rays so the focal length for red light is greater than the focal length for violet light
• Chromatic aberration can be minimized by the use of a combination of converging and diverging lenses
Chromatic aberration can be minimized using additional lenses
In an A chromat, the second lens cancels the dispersion of the first.
A chromats use two different materials, and one has a negative focal length.
This adds to the expense and is one reason why good cameras are so expensive

Chromatic aberration can be minimized using additional lenses
In an A chromat, the second lens cancels the dispersion of the first.
A chromats use two different materials, and one has a negative focal length.
T . Norah Ali Almoneef 140

Spherical Aberration
Results from the focal points of light rays far from the principle axis are different from the focal points of rays passing near the axis
For a mirror, parabolic shapes can be used to correct for spherical aberration
Spherical aberration can be also minimized using additional lenses
The additional lenses cancel the spherical aberration of the first.
T . Norah Ali Almoneef 141

• Are used to reduce aberration.
T . Norah Ali Almoneef 142