1t. norah ali almoneef : we represent light using rays, which are straight lines emanating from an...
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T . Norah Ali Almoneef 1
Chapter 24Mirror,lenses,and imaging systems
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T . Norah Ali Almoneef 2
24.1 Mirrors:
We represent light using rays, which are straight lines emanating from an object. This is an idealization, but is very useful for geometric optics.
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T . Norah Ali Almoneef 3
Reflection and RefractionWhen 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
refracted ray
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T . Norah Ali Almoneef 4
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.
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T . Norah Ali Almoneef 5
(parallel rays will all be reflected in the same directions)
(parallel rays will be reflected in a variety of directions)
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T . Norah Ali Almoneef 6
True of false: Reflection of light by a rough surface does not obey the laws of reflection.
(T/F)
• If the angle of incidence of a ray of light is 42°, what is each of the following?
• The angle of reflectiona. The angle the incident ray makes with the mirrorb. The angle between the incident ray and the reflected
example:
example
42°48°
90°
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T . Norah Ali Almoneef 7
• All the light rays from the source that reflect from the mirror SEEM to come from the same point behind the mirror
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T . Norah Ali Almoneef 8
Formation of image in a plane mirrorFormation of image in a plane mirror
Stand in front of a looking glass and look at your image.
2. Is the image erect or inverted?
1. Can you receive your image on a screen ?
3. Is the image the same size or larger or smaller?
6. Where is the image formed ?
4. What happens when you tilt your head to the right?5. How does the image move when you step forward or
backward
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T . Norah Ali Almoneef 9
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T . Norah Ali Almoneef 10
Types of Images for Mirrors and Lenses• 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
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T . Norah Ali Almoneef 11
plane Mirror
• 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
p=q
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T . Norah Ali Almoneef 12
Notations and plane Mirror
• 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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T . Norah Ali Almoneef 13
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T . Norah Ali Almoneef 14
Properties of the Image Formed by a Flat Mirror• 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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T . Norah Ali Almoneef 15
LEFT- RIGHT REVERSAL
AMBULANCE
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T . Norah Ali Almoneef 16
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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T . Norah Ali Almoneef 17
ExampleA girl can just see her feet at the bottom edge of the mirror.
Her eyes are 10 cm below the top of her head.
150 m
150 m
(a) What is the distance between the girl and her image in the mirror? Distance = 150 2 = 300 cm
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T . Norah Ali Almoneef 18
example
A boy of height 1.5 m stands 5 m in front of a plane mirror.
His image is ______ m tall and ______ m from him.1.5 10
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T . Norah Ali Almoneef 19
exampleWhich 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.
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T . Norah Ali Almoneef 20
Light Traveling through Materials• All electromagnetic waves have the same speed c = 3.00×108 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
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T . Norah Ali Almoneef 21
•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”
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T . Norah Ali Almoneef 22
24.2 Thin Lenses
• 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
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
t
Rfocal point
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T . Norah Ali Almoneef 23
• Used in cameras, telescopes, human eye
1. Used in cameras & telescopes to correct spherical aberation, and also eyeglasses
• These are examples of converging lenses
• They have positive focal lengths
• They are thickest in the middle
• These are examples of diverging lenses
• They have negative focal lengths• They are thickest at the edges
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T . Norah Ali Almoneef 24
Lenses• 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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T . Norah Ali Almoneef 25
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• The distance of F from C is the focal length f of the lens.focal length
FC
the two focal lengths are equal
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• The line through the optical centre and 2 foci is called the principal axis.
principal axis
FF'C
• Refracted rays appear to spread from a point called the principal focus F.
• Parallel rays are refracted outwards.
principal focus
optical centre
focal length
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T . Norah Ali Almoneef 28
Focal Length of a Converging Lens
• 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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T . Norah Ali Almoneef 29
Focal Length of a Diverging Lens
• 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.
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T . Norah Ali Almoneef 30
The equations can be used for both converging and diverging lensesA converging lens has a positive focal lengthA diverging lens has a negative focal length
Lens Equations
Q
M
Q’
M’
F F’
x x’f f’s s’
T
A
S
h
h’
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Q’TS and F’TA are similar triangles, h’ + hs’
=hf’
QTS and FAS are similar triangles, h’ + h
sh’f
=
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T . Norah Ali Almoneef 32
h’ + hs’
=hf’
h’ + hs
h’f
=
Adding the two equations:
h’ + hs’
+hf’
h’ + hs
h’f
= +
Since f = f’ for a thin lens,
h’ + hs’
+h’ + h
s=
h’ + hf
Multiplying through by 1h + h’
1s
1s’
1f
+ = Lens Formula
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T . Norah Ali Almoneef 33T . Norah Ali Almoneef 33
Magnification of Images Through A Thin Lens• The lateral magnification of the image is
• When M is positive, the image is upright and on the same side of the lens as the object
• When M is negative, the image is inverted and on the side of the lens opposite the object
ss
hh
M
ss
hh
M
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T . Norah Ali Almoneef 34
Images by Refraction
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Image can be captured by a screen.Image can be captured by a screen.
screen
O IHence called ‘real’.Hence called ‘real’.
Image formation by a lensA Real images
O I
Light rays converge to a point.Light rays converge to a point.
Since only convex lenses converge light rays, real images can only be formed by convex lenses
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Image formation by a lens
Light rays diverge from a point.Light rays diverge from a point.
b Virtual images
O
I
convex lens
No rays actually come from the image.No rays actually come from the image.
Hence called ‘virtual’.Hence called ‘virtual’.
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Image formation by a lens
Light rays diverge from a point.Light rays diverge from a point.
b Virtual images
OI
concave lens
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Image Construction: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.
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 Diverging Lens
F
Ray 1
F
Ray 1
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Image Construction: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.
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 Diverging Lens
F
Ray 1
F
Ray 1
Ray 2
Ray 2
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Image Construction: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.
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 Diverging Lens
F
Ray 1
F
Ray 1
Ray 2
Ray 2
Ray 3
Ray 3
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T . Norah Ali Almoneef 41
The Thin-Lens Equation
This gives us the thin-lens approximation, as well as the magnification:
s
sM
'
s
sM
'
magnification
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T . Norah Ali Almoneef 42
C1 C2
C1 & C2 are the centres of the spheres of which the surfaces of the lens form a part
The line through C1 & C2 form the principal axis
Converging Lens
Principal axis
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T . Norah Ali Almoneef 43
C1 C2
Converging lens
FF
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.
Normal
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F = Focal point
To locate the position of an image in a convex lens we use two of the following rays of light
FF
3 through the focus emerging parallel to principal axis.
1 parallel to the principal axis emerging through focus
2 striking the centre of the lens passes straight through (if lens is thin)
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T . Norah Ali Almoneef 45
F = Focal point
object
Images in Convex lens
ss
f
ImageReal, inverted & diminished
Image formed in convex lens when the object is placed outside twice the focal length
FF
ssf
1
1
1
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T . Norah Ali Almoneef 46
Image formed in convex lens when the object is placed at twice focal length
F = Focal point
object
Image Real, inverted & same size as object
Images in Convex lens
ss
f
FF
ssf
1
1
1
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T . Norah Ali Almoneef 47
Image formed in convex lens when the object is placed at the focus
F = Focal point
object
Image at Infinity
Images in Convex lens
s
f
FF
ssf
1
1
1
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T . Norah Ali Almoneef 4848
Image formed in convex lens when the object is placed inside the Focus
F = Focal point
object
Images in Convex lens
s
f
Image Virtual, magnified & upright
FF
ssf
1
1
1
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T . Norah Ali Almoneef 49
raybox
lens
screen
s s
Experiment to find the focal length of a convex lens
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T . Norah Ali Almoneef 50
F = Focal point
1. A ray which strikes the lens travelling parallel to principal axis is refracted as if it came from focus
2. A ray striking the centre of the lens
passes straight through (if lens is thin)
3. A ray heading for the focus on striking the
lens is refracted parallel to principal
axis
To locate the position of an image in a concave lens we use two of the following rays of light
Convave lens
FF
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T . Norah Ali Almoneef 51
Image formed in concave lens when the object is placedin front of lens
F = Focal point
object
Images in Concave lens
s
s
f
FF
ssf
1
1
1
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Images Tracing PointsDraw 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.
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.
3. Is it enlarged, diminished, or same size?
2. Is the image real or virtual?
1. Is the image erect or inverted?
• Real images are always on the opposite side of the lens. Virtual images are on the same side.
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Object Outside 2F
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 diminished in size, i.e., smaller than the object. Image is located between F
and 2F
Image is located between F and 2F
F
F
2F
2F
Real; inverted; diminished
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Object at 2F
F
F
2F
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
Image is located at 2F on other side
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Object Between 2F and F
F
F
2F
2F
Real; inverted; enlarged
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 2FImage is located beyond 2F
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Object at Focal Length F
F
F
2F
2F
When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is formed.
When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is formed.
Parallel rays; no image formed
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Object Inside F
F
F
2F
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
Image is located on near side of lens
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Diverging Lens Imaging
Diverging Lens
F
Diverging Lens
F
All images formed by diverging lenses are erect, virtual, and diminished. Images get larger as object approaches.
All images formed by diverging lenses are erect, virtual, and diminished. Images get larger as object approaches.
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T . Norah Ali Almoneef 59
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
P = 1
0.04 m= 25.0 D
1s
1s’
1f
+ =1s
1 1f
=
s’
_
16
1 14
=
s’
- s’ = 12 cm
Negative means real, inverted image
M = - 12 / 6 = -2
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T . Norah Ali Almoneef 60
1s
1s’
1f
+ =
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T . Norah Ali Almoneef 61
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T . Norah Ali Almoneef 6262
light from
distant object
falls short of retina
light from
distant object
falls on retina
with help of a diverging (concave) lens
Short-sight defect
Corrected
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T . Norah Ali Almoneef 6363
light from
near object
falls ‘behind’ retina
Long-sight defect
Corrected
with help of a converging (convex) lens
falls on retina
light from
near object
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Convex and concave lenses
a Converging or Diverging?
convex lens
(converging lens)
concave lens
(diverging lens)
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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?
A Path X.
B Path Y.
C Path Z.
X
Y
Z
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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?
A Path X.
B Path Y.
C Path Z.
X
Y
Z
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T . Norah Ali Almoneef 67
Image Summary • 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
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T . Norah Ali Almoneef 68
Image Formed by a Lens
• The lens has an index of refraction n and two spherical surfaces with radii of R1 and R2
– R1 is the radius of curvature of the lens surface that the light of the object reaches first
– R2 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
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T . Norah Ali Almoneef 69
Signs for Lensmaker’s Equation
1. R1 and R2 are positive for convex outward surface and negative for concave surface.
2. Focal length f is positive for converging and negative for diverging lenses.
1. R1 and R2 are positive for convex outward surface and negative for concave surface.
2. Focal length f is positive for converging and negative for diverging lenses.
R1 R2
+
-
R1 and R2 are interchangeable
1 2
1 1 1( 1)n
f R R
R1, R2 = Radii
n= index of glass
f = focal length
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T . Norah Ali Almoneef 70
Lensmaker’s Equation
R1 R2
Surfaces of different radius
The Lensmaker’s Equation:
1 2
1 1 1( 1)n
f R R
The focal length f for a lens.The focal length f for a lens.
Negative (Concave)
Positive (Convex)
Sign convention
R
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Signs for Lensmaker’s Equation
1. R1 and R2 are positive for convex outward surface and negative for concave surface.
2. Focal length f is positive for converging and negative for diverging lenses.
1. R1 and R2 are positive for convex outward surface and negative for concave surface.
2. Focal length f is positive for converging and negative for diverging lenses.
R1 R2
+
-
R1 and R2 are interchangeable
1 2
1 1 1( 1)n
f R R
R1, R2 = Radii
n= index of glass
f = focal length
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T . Norah Ali Almoneef 72T . Norah Ali Almoneef 72
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 f < 0
R1 positiveR2 positive
R1 negativeR2 negative
n=1
R1 R2
n≠1n=1
R1 positiveR2 negative
The (concave) lens is considered to have R1 and R2 both negative, and so f must be negative
The NORMAL (convex) lens above is considered to have R1 and R2 both positive, and so f must be positive
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T . Norah Ali Almoneef 73
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 f < 0
R1 positiveR2 positive
R1 negativeR2 negative
n=1
R1 R2
n≠1n=1
R1 positiveR2 negative
The (concave) lens is considered to have R1 and R2 both negative, and so f must be negative
The NORMAL (convex) lens above is considered to have R1 and R2 both positive, and so f must be positive
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T . Norah Ali Almoneef 74
Focal Length for a Lens(Lens Makers’ Equation)• 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
• If the medium is air
• ( n for the lens)This is called the lens maker’s equation
21
11)1(
1RR
nf
f1
= }R1
R1
{ x 1}-nn
{21medium
lens
s1
s1
= }R1
R1
{ x 1}-nn
{21air
lens
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T . Norah Ali Almoneef 75
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Summary: Lensmaker’s Equation
1. R1 and R2 are positive for convex outward surface and negative for concave surface.
2. Focal length f is positive for converging and negative for diverging lenses.
1. R1 and R2 are positive for convex outward surface and negative for concave surface.
2. Focal length f is positive for converging and negative for diverging lenses.
R1 R2
+
-
R1 and R2 are interchangeable
1 2
1 1 1( 1)n
f R R
R1, R2 = Radii
n= index of glass
f = focal length
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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.
R1 = 20 cm, R2 = -40 cm
-40 cm
+20 cm
n = 1.51 2
1 1 1( 1)n
f R R
1 1 1 2 1(1.5 1)
20 cm ( 40 cm 40 cmf
f = 80.0 cmf = 80.0 cm Converging (+) lens.
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Example: What must be the radius of the curved surface in a plano-convex lens in order that the focal length be 25 cm?
R1 = , f= 25 cm
2
1 1 1( 1)n
f R
R1= R2=?
f = ?
0
2 2
1 1 0.500(1.5 1)
25 cm R R
R2 = 12.5 cmR2 = 12.5 cm Convex (+) surface.
R2 = 0.5(25 cm)
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Be careful with substitution of signed numbers!Be careful with substitution of signed numbers!
Alternative SolutionsIt might be useful to solve the lens equation algebraically for each of the parameters:
fss
1
'
11
fs
sfs
'fs
fss
ss
ssf
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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
s = -37.5 cm
The fact that S is negative means that the image is virtual (on same side as object).
The fact that S is negative means that the image is virtual (on same side as object).
fss
1
'
11
cmcmcmcmcm
fssf
s 5.3725152515
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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.
F
F
s = 15 cm; s = -37.5 cm
h’ = +20 mm
The fact that h’ is positive means that the image is erect. It is also larger than object.
The fact that h’ is positive means that the image is erect. It is also larger than object.
y’ y
s
s
h
hM
''
cmcm
mmh
155.37
8
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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
s = +12.7 cm
M = +0.364
fss
1
'
11
cmcmcmcmcm
fssf
s 7.12)20(35
2035
s
s
h
hM
''
364.035
)7.12('
cmcm
ss
M
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SummaryA Converging lens is one that refracts and converges parallel light to a real focus beyond the lens. It is thicker near the middle.
A Converging lens is one that refracts and converges parallel light to a real focus beyond the lens. It is thicker near the middle.
FF
A diverging lens is one that refracts and diverges parallel light which appears to come from a virtual focus in front of the lens.
A diverging lens is one that refracts and diverges parallel light which appears to come from a virtual focus in front of the lens.
The principal focus is denoted by the red F.FF
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Summary of Math Approach
F
F
2F
2F
p
f
q
y
-y’
Lens Equation: Magnification:
s
s
h
hM
''fss
1'
11
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Terms for Image Construction
Converging Lens Diverging Lens
• 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.
F
Near focus
F
Near focus
F
Far focus
F
Far focus
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T . Norah Ali Almoneef 86
Image PropertiesConvex vs. Concave Lenses
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T . Norah Ali Almoneef 87
Image Properties Sign Conventions-
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Diverging Lens Imaging
Diverging Lens
F
Diverging Lens
F
All images formed by diverging lenses are erect, virtual, and diminished. Images get larger as object approaches.
All images formed by diverging lenses are erect, virtual, and diminished. Images get larger as object approaches.
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Analytical Approach to Imaging
F
F
2F
2F
s
f
h
-h’
Lens Equation: Magnification:
fss
1
'
11 s
s
h
hM
''
s’
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T . Norah Ali Almoneef 90
Parallel ray
focus (f)focus (f) 2 f2 fx x x x
Focal ray
Image is:RealInvertedReducedAppears between f and 2f
Object beyond 2f
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T . Norah Ali Almoneef 91
focus (f)focus (f) 2 f2 fx x x x
Focal ray
Parallel ray
Image is:RealInvertedSame sizeAppears between f and 2f
Object at 2f
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T . Norah Ali Almoneef 92
focus (f)focus (f) 2 f2 fx x x x
Focal ray
Parallel ray
Image is:RealInvertedEnlargedAppears beyond 2f
Object betweenf and 2f
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T . Norah Ali Almoneef 93
focus (f)focus (f) 2 f2 f
x x x x
Image is:VirtualErectEnlargedAppears on sameSide as Object
ApparentConvergence
Of rays
ObjectInsidefocus
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T . Norah Ali Almoneef 94
f2 f f 2 fFocal ray
Parallel ray
Ray thru center
Image is:VirtualErectReducedAppears on sameSide as object
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T . Norah Ali Almoneef 95
Thin Lens Equation: sign conventions
s s’
fss
1
'
11f
object image
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.
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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.
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?
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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.
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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 5 cm
B 10 cm
C 20 cm
D 40 cm
10 cm
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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.
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Both convex and concave...
Both convex and concave lenses can produce _______ images, which must be _________ than the object if convex lenses are used.
virtual larger
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T . Norah Ali Almoneef 101
General Image Trends real images are always inverted virtual images are always upright real images are always in frontof the mirror virtual images are alwaysbehind the mirror negative image distance means virtual image positive image distance means real image
101T .Norah Ali Almoneef
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T . Norah Ali Almoneef 102
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T . Norah Ali Almoneef 103
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.
A lens produces a sharply-focused, inverted image on a screen. What will you see on the screen if the lens is removed?
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T . Norah Ali Almoneef 104
The focal length of a converging lens is
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.
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T . Norah Ali Almoneef 105
ExampleAn 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?
Real image, magnification = -1
cmscmcmcms
cmcmsfs
cmf
cms
2020
120
120
2120
110
1111
10
20
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T . Norah Ali Almoneef 106
ExampleAn 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?
05.0/
0241
41111
4
111
(concave) 4
ssm
cmscmcmsfs
cmsfss
cmf
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T . Norah Ali Almoneef 107
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.
Image is real, inverted.
. .F1 F2s
Virtual side Real side
m 10
10 1
Example
fss111
101
101
511
s
cms 10
ss
hh
m
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T . Norah Ali Almoneef 108
24(e). Given a lens with the properties (lengths in cm) R1 = +30, R2 = +30, s = +10, and n = 1.5, find the following: f, s and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.
cmf 30
m 15
101.5
Image is virtual, upright.
Virtual side Real side
R1. .F1 F2
pR2
21
111
1RR
nf
301
301
301
15.11
f
ss
hh
m
sfs111
151
101
3011
s
cms 15
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T . Norah Ali Almoneef 109
ExampleAn 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?
Virtual image, as viewed from the right, the light appears to be coming from the (virtual) image, and not the object.
Magnification = +2109
cmscmcmcms
cmcmsfs
cmf
cms
1010
110
210
115
110
1111
10
5
fss111
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T . Norah Ali Almoneef 110
Examples
A diverging lens with f = -20 cmh = 2 cm, s = 30 cm
cms 12
The image is virtual and uprightcmh
hM
8.0
4.030
12
2
A converging lens with f = 10 cm
(a) S = 30 cm
cmss
151011
301
(b) S = 10 cm
(c) s = 5 cm
2510
ss
M
The image is real and inverted
The image is virtual and upright
The image is at infinity
fss
1
'
11
2011
301
s
ss
hh
M
5.03015
ss
M
s
cmss10
1011
51
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T . Norah Ali Almoneef 111
The thin lens equation,, can be used to find the image distance:
5.03015
ss
M
fss
1
'
11
cmsscmcmcm
ssf
15
130
230
110
1
1
1 -
1
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T . Norah Ali Almoneef 112
cmsscmcmcm
ssf
10
110
15
110
1
1
1 -
1
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T . Norah Ali Almoneef 113
Repeat Example for a diverging lens of focal length 10.0 cm.Solution(A) We begin by constructing a ray diagram as in Figure36.31a taking the object distance to be 30.0 cm. The diagramshows that we should expect an image that is virtual,smaller than the object, and upright. Let us now apply thethin lens equation with p = 30.0 cm:
cmsscmcmcm
ssf
5.7
130
430
1101
1
1 -
1
25.030
)5.7(
cmss
M
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T . Norah Ali Almoneef 114
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T . Norah Ali Almoneef 115
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T . Norah Ali Almoneef 116
QUICK QUIZAn 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 2f > s > f, and a magnified, virtual image is formed if s
> f.
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T . Norah Ali Almoneef 117
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.
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T . Norah Ali Almoneef 118
An object is placed in front of a plano-concave lens at r1/2. 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.
r1
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119
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.
T .Norah Ali Almoneef
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T . Norah Ali Almoneef 120
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.
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T . Norah Ali Almoneef 121
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
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T . Norah Ali Almoneef 122
Example• 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)
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T . Norah Ali Almoneef 123
Example
• 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)
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T .Norah Ali Almoneef 124
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T .Norah Ali Almoneef 125
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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.)
126T .Norah Ali Almoneef
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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.
127T .Norah Ali Almoneef
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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.)
128T .Norah Ali Almoneef
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When the ray in the diagram is continued through the diverging lens, it passes through which point? (F marks the two focal points.)
129T .Norah Ali Almoneef
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The image of the encircled point on the object formed in the positive lens is at which circle? (F
marks the two focal points.)
130T .Norah Ali Almoneef
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The image produced by the converging lens is at which point? (F marks the two focal points.)
131T .Norah Ali Almoneef
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The image of the object formed by the diverging lens is located at which point? (F marks the two
focal points.)
132T .Norah Ali Almoneef
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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.
133T .Norah Ali Almoneef
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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. 134T .Norah Ali Almoneef
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T . Norah Ali Almoneef 135
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
• Meaning of P• 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)
P (in diopters) 1
f (meters)
• 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
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T . Norah Ali Almoneef 136
24.4 the power of a Lens, aberations
• For lens in contact (separation is negligible)
power)each of sum is(power
111
:or 111
21
21
21
PPP
fff
fff
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T . Norah Ali Almoneef 137
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T . Norah Ali Almoneef 138
Lens and Mirror Aberrations
• 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
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T . Norah Ali Almoneef 139
Chromatic Aberration• 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
T . Norah Ali Almoneef 139
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
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Chromatic aberration can be minimized using additional lensesIn 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.
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Spherical Aberration
Results from the focal points of light rays farfrom the principle axis are different from thefocal points of rays passing near the axis For a mirror, parabolic shapes can be used tocorrect for spherical aberration
Spherical aberration can be also minimized using additional lenses
The additional lenses cancel the spherical aberration of the first.
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Multi-element lenses• Are used to reduce aberration.