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.

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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)
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UNIT 22 GEOMETRICAL OPTICS
22.1 Reflection at a spherical surface 22.2
Refraction at a plane and
spherical surfaces 22.3 Thin lenses
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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
• For real object only

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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.

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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.

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22.1 Reflection at a spherical surface
Terms and Definitions
Imaginary spherical
A
A
P
P
C
C
F
F
B
B
f
f
r
r
A concave mirror
A convex mirror
C centre of curvature of the surface mirror. P
centre of the surface mirror (vertex or
pole). Line CP principal or optical axis.
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Imaginary spherical
22.1 Reflection at a spherical surface
Terms and Definitions
A
A
P
P
C
C
F
F
B
B
f
f
r
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). r radius of curvature of the mirror.
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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
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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.
A convex mirror
Focal point, F
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22.1 Reflection at a spherical surface
Relation between focal length, f and radius of
curvature, r
is isosceles.
(FCFM)
Consider ray AM is paraxial (parallel and very
close to the principal axis).
FM FP
or
FP 1/2 CP
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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.

1. 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

1. 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.
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Drawing Compass
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Images Form by Spherical Mirrors

1. Ray diagram

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Image formed by concave mirrors

• 1) Object beyond C
• Between C and F
• Real
• c) Inverted
• d) Smaller than object

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Image formed by concave mirrors

• 2) Object at C
• At C
• Real
• Inverted
• 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
• Behind mirror
• Virtual
• Upright
• d) Larger than object
• (magnified)

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Image formed by concave mirrors
Notes

• 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

• 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
this line.
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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 object

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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 OP u Image
distance IP v Radius of curvature CP r O
bject size OA h Image size IB h
Focal length f

or
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b) The mirror equation-calculation using formula
Linear Magnification, m
or
or
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b) The mirror equation-calculation using formula
Sign convention
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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.
Solution
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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.
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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,
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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,
-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
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
Magnification,
The image is two-fifth as high as the object.
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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
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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
Using
f 15 cm
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b) The mirror equation-calculation using formula
Solution 1.2.4
b) Given m 3 , f 15 cm
Using
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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
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b) The mirror equation-calculation using formula
Solution 1.2.5
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

• 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 )

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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 of it ? (-7.5 cm, convex mirror)
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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)
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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)
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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
  (Snells Law) for layers of materials with
  different densities.
• 22.2.2 Apply
• for spherical surface.

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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.

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22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
Normal
Reflected ray
Incident ray
?1
?
?1 ?
Air
Glass
?1 gt ?2
?2
transparent medium
Refracted ray
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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 Snells Law.

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22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
Law of Refraction
Snells 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 ?2 depends on the speed of
• light and on the angle of incidence.

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22.2 Refraction at a Plane and Spherical Surfaces
Snells Law
Refraction at a plane surface
nglass gt nair
Normal
Normal
v1 gt v2
v1 lt v2
?1
v1
v1
?1
Air
Glass
Air
Glass
?1 gt ?2
?1 lt ?2
?2
v2
v2
?2
this ray is bent toward normal
this ray is bent away from normal
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22.2 Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
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CONGRATULATIONS!!!
S11T5
FIRST RANK (Tutorial Physics UPS2 11/12)
S1K1
FIRST RANK (Lecture Physics UPS211/12)
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22.2 Refraction at a Plane and Spherical Surfaces
Snells 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.
..(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.

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22.2 Refraction at a Plane and Spherical Surfaces
Index of Refraction, n
Refraction at a plane surface

• As light travels from one medium to
• another, its frequency does not change but
• its wavelength does.
• Therefore, because the relationship
• must be valid in both medium and because

(1.3)
(1.2)
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22.2 Refraction at a Plane and Spherical Surfaces
Index of Refraction, n
(1.4)
(1.6)
(1.5)
If medium 1 is vacuum, (n1 1) then equation
(1.4) can be written as
?o wavelength of light in vacuum. ?n
wavelength in the medium whose index of
refraction is n.
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22.2Refraction at a Plane and Spherical Surfaces
Refraction at a plane surface
(1.5) into Snells Law
This equation is the most widely used and
practical form of Snells Law.
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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snells 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.

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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snells Law
Solution 1.3.1
, n11
, ?2 26.0 o
?1 40.0 o
Using
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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snells Law
Solution 1.3.1
Using
?1 500 nm
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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snells 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.
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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snells Law
Solution 1.3.2
, n2 1.52
, n1 1
?1 30.0 o
Using
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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snells 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.
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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Snells Law
Solution 1.3.3
noil 1.45
nwater 1.33
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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Exercise

• A light ray of wavelength 589 nm in
• vacuum passes through a piece of silica
• (n 1.458).

• Find the speed of light in silica.
• (2.06 x 108 ms-1)
• What is the wavelength of this light in
• silica ? (404 nm)
• (c) Find the frequency of the light.
• (5.09 x 1014Hz)

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Refraction at a plane surface
22.2 Refraction at a Plane and Spherical Surfaces
Exercise

• 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
refraction of the plastic ? (1.34)
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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)
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22.2 Refraction at a Plane and Spherical Surfaces
Refraction at A Spherical Surface
A
r
u
v
O an object point I an image point C
center of curvature
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Refraction at A Spherical Surface
A
r
u
v
(equation of refraction of a spherical surface)
object-image relationship
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Refraction at A Spherical Surface
To obtain the magnification of an image formed
by refraction at a spherical surface.
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Refraction at A Spherical Surface
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.
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Refraction at A Spherical Surface
Quantity Sign Remarks
u Real object (object is on the front side of the surface)
u - Virtual object (back side of the surface)
v Real image (OPPOSITE side with object)
v - Virtual image (SAME side with object)
r Center of curvature is located in more dense medium
r - Center of curvature is located in less dense medium
r 8 Flat surface
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Refraction at A Spherical Surface
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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.
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Refraction at A Spherical Surface
Solution 1.3.4
n1 1 , n2 1.50 , r 20 mm , u 80 mm
a)
OR
ve (to the right of vertex back side)
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Refraction at A Spherical Surface
Solution 1.3.4
n1 1 , n2 1.50 , r 20 mm , u 80 mm
b)
same size but inverted
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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.
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Refraction at A Spherical Surface
Solution 1.3.5
n1 1 , n2 1.50 , r 20 mm , u 80 mm
a)
OR
ve (to the left of vertex front side)
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Refraction at A Spherical Surface
Solution 1.3.5
n1 1 , n2 1.50 , r 20 mm , u 80 mm
b)
greater and upright
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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.
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Refraction at A Spherical Surface
v
v - 0.017 m
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Example 1.3.7
Calculate the image distance and magnification.
Solution
v
8 cm
u real/actual depth v apparent depth
81
Refraction at A Spherical Surface
Solution 1.3.7
v
8 cm
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Refraction at A Spherical Surface
Exercise

• 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.

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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,
• for real object only.
• Use lensmakers equation
• Use the thin lens formula for a combination of
  converging lenses.

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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)

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22.3 Thin lenses
Introduction
Converging lens- a lens causes incident parallel
rays to converge after exiting the lens.
Converging lens
Diverging lens- a lens causes incident parallel
rays to diverge after exiting the lens.
Diverging lens
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22.3 Thin lenses
Introduction
Terms and definition
Converging lens
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22.3 Thin lenses
Introduction
Terms and definition
Diverging lens
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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.

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22.3 Thin lenses
Several types of diverging lenses
Converging lenses are thicker at the center than
at the edges whereas diverging lenses are thinner
at the center.
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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.
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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.
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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.
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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)

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Converging Lens
Image Formation by Thin Lenses

• 1) Object at 2F1 , the image is
• At 2F2
• Real
• Inverted
• Same size as object

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Converging Lens
Image Formation by Thin Lenses

• 2) Object beyond 2F1 , the image is
• Between F2 and 2F2
• Real
• Inverted d) Smaller than object

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Converging Lens
Image Formation by Thin Lenses

• 3) Object between F1 and 2F1 , the image is
• Beyond 2F2
• Real
• Inverted d) Larger than object

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Image Formation by Thin Lenses
Converging Lens
4) Object at F1 , the image is at infinity
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Converging Lens
Image Formation by Thin Lenses

• 5) Object between lens and F1, the image is
• Behind the object
• Virtual
• Upright
• d) Larger than object

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Diverging Lens
Image Formation by Thin Lenses

• The image always ,
• a) Between lens and F1
• Virtual
• Upright
• d) Smaller than object

F1
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The Lens Equation

thin-lens equation
u
v
101
Thin lens formula
102
Sign Convention
Q Positive () Negative (-)
f Converging (convex) Diverging (concave)
u In front of the lens Behind the lens
v Behind the lens (real) OPPOSITE SIDE In front of the lens SAME SIDE
hi Upright image with respect to the object Inverted image with respect to the object
m Upright image with respect to the object Inverted image with respect to the object
103
The Lens Equation
1.4 Thin lenses
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

• Magnified
• - upright
• - virtual (in front of the lens)

104
The Lens Equation
1.4 Thin lenses
Linear Magnification, m
Example 1.4.2
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
105
The Lens Equation
1.4 Thin lenses
Example 1.4.3
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
106
The Lens Equation
1.4 Thin lenses
Example 1.4.4
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
107
The Lens Equation
1.4 Thin lenses
Example 1.4.5
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
virtual, upright 5.16 cm, -.3.43 cm
108
The Lens Equation
1.4 Thin lenses
Linear Magnification, m
Exercise

• 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)

109
The Lens Equation
1.4 Thin lenses
Linear Magnification, m
Exercise
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)
110
The Lens Equation
1.4 Thin lenses
Exercise
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 127 mm in
front of lens,virtual,upright, 1.56 mm)
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Two Lenses System
112
Two Lenses System
f 20 cm
B
A
p1
v2
p2
v1
113
Two Lenses System
First Step

• 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.

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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.

115
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,

116
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.
117
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.
Magnification, m m1 x m2
118
Solution 1.4.6
Two Lenses System
119
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)
120
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)

121
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)

122
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)

123
Thin Lenses Formula and Lens makers Equation
Considering the ray diagram of refraction for 2
spherical surfaces as shown in figure below.
124

where
125

• Note
• If the medium is air (n1 nair1) thus the lens
  makers equation will be

Where,
126
Lensmakers Equation
ve sign
n for lens n1 for medium
127
Sign Convention
Lensmakers Equation
Q Sign Remarks
R Convex surface
R - Concave surface
R 8 Flat surface
f Converging lens
f - Diverging lens
128
Lensmakers Equation
OR
n for lens, the lens in air
Use sign convention for R

• ve for convex surface
• -ve for concave surface

129
Example 1.4.8
Lensmakers 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
130
Solution 1.4.8
Lensmakers Equation
Given n 1.52 , n1 1, R1 22.4 cm, R2
46.2 cm
46.2 cm
Converging lens ? f
131
Lensmakers 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)
132
Solution 1.4.9
Lensmakers Equation
Given R1 20.0 cm,R2 -20.0 cm,n 1.50
(n 1.63)
a) In air (n1 1)
b) In carbon disulfide (n1 1.63)
133
Example 1.4.10
Lensmakers 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
134
Exercise
Lensmakers Equation

• 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)

135
Exercise
Lensmakers 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 material ?
(2.52)