Refraction at A Spherical Surface

$Refraction at A Spherical Surface$

The study of light based on the assumption that light travels in straight lines and is concerned with the laws controlling the reflection and refraction of rays of light .

CHAPTER 22:

Geometrical optics

(4 Hours)

1

UNIT 22 : GEOMETRICAL

OPTICS

22.1 Reflection at a spherical surface

22.2 Refraction at a plane and spherical surfaces

22.3 Thin lenses

2

At the end of this topic, students should be able to: ( 1 H)

• a) State laws of reflection.

• b) Sketch and use ray diagrams to determine the characteristics of image formed by spherical mirrors.

• c) Use



2 f

1 1 u

1 v r

For real object only

3

The Law of reflection

The law of reflection states that

• The incident ray, the reflected ray and normal, all lie in the same plane

• The angle of incidence i is egual to the angle of reflection r.

4

22.1 Reflection at a spherical surface

Terms and Definitions

• A spherical mirror is a reflecting surface with spherical geometry .

• Two types : i) convex , if the reflection takes place on the outer surface of the spherical shape.

ii) concave , if the reflecting surface is on the inner surface of the sphere.

5


Imaginary spherical

A A

Terms and

Definitions

C F P P F C

B r

A concave mirror f

B f r

A convex mirror

C ~ centre of curvature of the surface mirror.

P ~ centre of the surface mirror ( vertex or pole ).

6 Line CP ~ principal or optical axis .


Imaginary spherical

A A

Terms and

Definitions

C F P P F C

B B f r

AB ~ aperture of the mirror.

f r

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.

7


Focal point, F

“point on the principal axis where

Terms and

Definitions rays parallel and close to the principal axis pass after reflection”.

A concave mirror

8

22.1 Reflection at a spherical surface Terms and Definitions

F “ point on the principal axis where rays parallel to the principal axis appear to diverge from after reflection”.

Focal point, F

A convex mirror

9


A

Relation between focal length, f and radius of curvature, r



FCM is isosceles.( FC=FM )







M

P

Consider ray AM is paraxial (parallel and very close to the principal axis).

C F

FM = FP or FP = 1/2 CP f r f

 r

2

10

22.1 Sketch and Use Ray diagram

Images Form by Spherical Mirrors

Information about the image in any case can be obtained either by drawing a ray diagram or by calculation using formula .

a) Ray diagram

Ray 1 : A ray parallel to the principal axis is reflected through the focus (focal point) .

11


Ray 2 : A ray passing through the focus is reflected parallel to the principal axis.

Ray 3 : A ray passing through the centre of curvature is reflected back through the centre of curvature .


12

Drawing Compass

13



14

Image formed by concave mirrors

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

15


2) Object at C a) At C b) Real c) Inverted d) Same size as object

16


3) Object between C and F a) Beyond C b) Real c) Inverted d) Larger than object

(magnified)

17


4) Object between F and P a) Behind mirror b) Virtual c) Upright d) Larger than object

(magnified) 18

Image formed by concave mirrors Notes i) If the object is at infinity , a real image is formed at F . Conversely, an object at F gives a real image at infinity .

F

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

19 this line.

Image formed by a convex mirror

Ray 1 : A ray parallel to the axis is reflected as though it came from the focal point.

Ray 2 : A ray heading toward the focal point is reflected parallel to the axis.

Ray 3 : A ray heading toward the centre of curvature is reflected back on itself.

20

Image formed by a convex mirror

The image always

• Virtual

• Upright

• Smaller than object

21

b) The mirror equation-calculation

A using formula u

P

B r v

M

22

b) The mirror equation-calculation using formula

Object distance = OP = u

Image distance = I P = v

Radius of curvature = CP = r

Object size

Image size

= OA = h

= I B = h’

Focal length = f

1 1 1 f u v or

2 r

=

1

+

1 u v

23


Linear Magnification,

m m

 height of image height of object

 h ' h or m

 

image distance

object distance

  v u m

 h '

  v h u or v m = u

24


Sign convention

Quantity Positive sign (+) Negative sign (-)

Object distance ( u )

Image distance ( v )

Focal

Length ( f )

Magnification

( m )

Real object Virtual object

Real image Virtual image

Concave mirror

Upright image

Convex mirror

Inverted image

25


Example 1.2.1

An object 6 cm high is located 30 cm in front of a convex spherical mirror of radius

40 cm. Determine the position and height of its image .

2

Solution r

=

1

+ u

1 v

2

= r

1

+ u

1 v

2

40

=

1

30

+

1 v v = -12 cm v m = u

=

12

30

= 0 .

40 m = h ' h

= 0 .

40 h ' = 2 .

4 cm

26


Example 1.2.2

An object is placed 15 cm from a a) concave mirror b) convex mirror of radius of curvature 20 cm. Calculate the image position and magnification in each case.

27


Solution 1.2.2

a) Concave mirror , u = +15 cm r = +20 cm f = +10 cm

Substituting values and signs in the mirror equation,

1 u



1 v

 f

1

1 v

1 v





 f

1

1

10



1 u





1



1

15 30 v

 

30 cm

28


Solution 1.2.2

The image is real since v is positive and it is 30 cm in front of the mirror.

Magnification, m

  u v m

 

30

 

2

15

-ve (inverted)

The image is twice as high as the object.

29


Solution 1.2.2

b) Convex mirror, u = +15 cm r = -20 cm f = -10 cm

1 v



1 f



1 u

1 u



1 v



1 f

1 v





1

10





1

15

 

5

30 v

 

The image is virtual since v is negative and it is 6.0 cm behind the mirror.

30


Solution 1.2.2

Magnification, m

  v u m

 



6



2

15 5

The image is two-fifth as high as the object.

31


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 m = v

u

1

=

6 v

-

12

=

1

⇒ v

6

= 2 cm f

1

=

1

+ u

1

= v

1

12

+

1

2 f = -2.4

cm

32


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.

33


Solution 1.2.4

a) u = + 20 cm , m = -3 v

3 u

  

3 u v

 

)

 

60 cm m

  v u

Using

1 u



1 v



1 f

1 1 1 1 f u v







1

60 20 f = +15 cm

34


Solution 1.2.4

b) Given m = 3 , f = + 15 cm m

  v u

3 v

  v u

 

3 u

Using

1 u



1 v



1 f



1 1

3 u u



1

15 u

 

10 cm

35


Example 1.2.5

An object 2.0 cm high is placed 30 cm from a concave mirror with a radius of curvature of 10 cm. Find the location and its characteristics .

Solution

Given : h = 2.0 cm, u = +30 cm, r = +10 cm, f = r /2= +5 cm

1 u



1 v



1 f

1 1 1 v f u



1

5





1

30 v

 

.

36


Solution 1.2.5

m = v

u

=

6

-

30

= -0.2

Characteristics :

1) smaller

2) in front of the mirror

3) inverted

4) real

37


Exercise

1) If a concave mirror has a focal length of

10 cm, find the two positions where an object can be placed to give, in each case, an image twice the height of the object.( 15cm, 5.0cm

)

2) A convex mirror of radius of curvature 40 cm forms an image which is half the height of the object. Find the object and image position .( 20cm,10cm behind the mirror )

38

b) The mirror equation

Exercise

3) An object is placed 5.0cm in front of a concave mirror with a 10.0 cm focal length. Find the location of the image and its characteristics .( -10cm, m =-2, virtual, upright and located behind the mirror )

4) What kind of spherical mirror must be used, and what must be its radius, in order to give an erect image one-fifth as large as an object placed 15 cm in front of it ? (-7.5 cm, convex mirror) 39


Exercise

5) A concave mirror forms an image , on the wall 3.0 m from the mirror, of the filament lamp 10 cm in front of the mirror.

a) What is the radius of curvature of the mirror ?

b) What is the height of the image if the height of the object is 5 mm ?

(19.4 cm, -30)

40


Exercise

6) What are the nature, size, and location of the image formed when a 6 cm tall object is located 15 cm from a spherical concave mirror of focal length 20 cm ?

(virtual, upright, -60 cm, + 24 cm)

7) The magnification of a mirror is -0.333.

Where is the object located if its image is formed on a card 540 mm from the mirror?

What is the focal length ?

(1.62 m, +405 mm)

41

Refraction at aplane and spherical surfaces (1 H)

At the end of this chapter, students should be able to:

 22.2.1 State and use the laws of refraction (Snell’s Law) for layers of materials with different densities.

 22.2.2 Apply n

1 u

 n

2 v



 n

2

 n

1

 r for spherical surface.

42

22.2 Refraction at a Plane and

Spherical Surfaces

• Refraction is the change in direction of light when it passes through matter.

Refraction of light

• When a ray of light traveling through a transparent medium encounters a boundary leading into another transparent medium, part of the ray is reflected and part enters the second medium .

• The ray that enters the second medium is bent at the boundary and is said to be refracted .

43

22.2 Refraction at a Plane and Spherical Surfaces

Refraction at a plane surface

Normal

Reflected ray

Incident ray

θ

1

θ ’

θ

1

= θ

’

Air

Glass transparent medium

θ

2

θ

1

> θ

2

Refracted ray

44



Law of Refraction

• The incident ray, refracted ray and the normal all lie in the same plane.

• At the boundary between any two given materials, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for rays of any particular wavelength (previous figure). This is known as Snell’s Law .

45



Law of Refraction sin sin



1



2

 v

1  v

2 constant Snell’s Law where v

1 and v

2 is the speed of light in medium 1 is the speed of light in medium 2.

• This relationship shows that the angle of refraction Ө

2 depends on the speed of light and on the angle of incidence.

46



Snell’s Law

Normal n glass

> n air

Normal v

1

> v

2

θ

1 v

1

Air

θ

1 v

1

< v

2 v

1

Glass

θ

1

> θ

2 θ

2

Glass v

2 this ray is bent toward normal

θ

1

< θ

2

θ

2

Air v

2 this ray is bent away from normal 47



48

49



Snell’s Law

Index of Refraction, n

The speed of light in any material or medium is less than the speed of light in vacuum or air.

n

 speed of light in vacuum speed of light in medium

 c

..(1.1) v

• The index of refraction is a dimensionless and never less than 1 ( n ≥

1) because v is usually less than c .

• n is equal to unity ( n = 1) for vacuum.

50




• As light travels from one medium to another, its frequency does not change but its wavelength does.

Therefore, because the relationship v f must be valid in both medium and because f

1

 f

2

 where v

1

 

1

… (1.2) f v

2 f

2

… (1.3)

51

2


,




1



2

 v

1  n

2 v

2 n

1 v

1 v

2



…(1.5)

1

2

 n

2 n

1

…(1.4) n

1 1 n …(1.6)

If medium 1 is vacuum, ( n

1

= 1 ) then equation (1.4) can be written as n



 o

 n

λ o

λ n

= wavelength of light in vacuum.

= wavelength in the medium whose index of refraction is n .

52

53

22.2Refraction at a Plane and Spherical Surfaces


(1.5) into Snell’s Law n

1 sin



1

 n

2 sin



2

This equation is the most widely used and practical form of Snell

’ s Law .

54


Example 1.3.1


Snell’s Law

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.

0

55


Solution 1.3.1


Snell’s Law

Ө

1

=40.0 o , Ө

2

=26.0 o , n

1

=1

Using n

1 sin



1

 n

2 sin



2 n

2

 n

1 sin sin



2



1



1 40 0

0

( )(sin .

) sin .

0



56


Solution 1.3.1


Snell’s Law

Using n

1 1 n

2 2

λ

1

= 500 nm n

 n

2



( )( )



374 nm

57


Example 1.3.2


Snell’s Law

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

30.0

o to the normal. Find the angle of refraction .

58


Solution 1.3.2


Snell’s Law

Ө

1

=30.0 o , n

1

=1

Using

, n

2

= 1.52

n

1 sin



1

 n

2 sin



2 sin n

1 sin



1 n

2

 

(sin 30 0

0

)



0 329



1 sin ( .

)



.

0

59


Example 1.3.3


Snell’s Law

40 o

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 40.0

o . Find the angle the ray makes in the water.

60


Solution 1.3.3


Snell’s Law

40 o

Air n oil

= 1.45

Oil n water

= 1.33

Water n air sin 40 0

= n oil sin

 oil

...

(1) n oil sin

 oil

= n water sin

 water

...

(2)

(1) = (2) n air sin 40

0

= n water sin

 water

 water

= sin

1

( n air sin 40

0

) n water

= sin

1

(

1 sin 40 o )

1 .

33

= 28 .

9

0

61

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

Exercise

1. A light ray of wavelength 589 nm in vacuum passes through a piece of silica

( n = 1.458).

(a) Find the speed of light in silica.

(2.06 x 10 8 ms -1 )

(b) What is the wavelength of this light in silica ?

(404 nm)

(c) Find the frequency of the light.

(5.09 x 10 14 Hz)

62

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

Exercise

2. A light ray of wavelength 589 nm moves from inside the glass ( n = 1.52 ) toward the glass-air interface at an angle of

30.0 to the normal. Determine the angle of refraction .

(49.5 away from the normal)

3. A beam of light traveling in air is incident on a transparent plastic material at an angle of incidence of 50 o . The angle of refraction is 35 o . What is the index of refraction of the plastic ? (1.34)

63

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 ?

(34.7

o )

5. Light of wavelength 50 nm in a particular glass has a speed of 1.7 x 10 8 m/s.

What is the index of refraction for this glass? What is the wavelength of this light in air ? (1.76, 1146 nm)

64


Refraction at A Spherical Surface n

2

 n

1

A r u v

O = an object point

I = an image point

C = center of curvature

65


2

 n

1

A u r v n

1  n u v

2  n

2 r

 n

1

(equation of refraction of a spherical surface) object-image relationship

66

Refraction at A Spherical Surface y

Q

To obtain the magnification of an image formed by refraction at a spherical surface.

P

Ø

V n

1 n

2

C

Ø’

P’ y’

Q’ u v

67


1  n u v

2  n

2 r

 n

1 m

 y '

  y n u

These equations can be applied to both convex and concave surfaces, provided that sign rules (refer to next table) are obeyed and whether greater or less than n

1

.

n

2 is

68

Refraction at A Spherical Surface

Quantity Sign Remarks u u v v r r r

+ Real object (object is on the of the surface) front side

Virtual object ( back side of the surface)

+ Real image ( OPPOSITE side with object)

Virtual image ( SAME side with object)

+ Center of curvature is located in more dense medium

Center of curvature is located in less dense medium

∞ Flat surface

69


1 u

+ n

2 v

=

( n

2

n

1

) r

Use sign convention for r :

• +ve if center of curvature is located in more dense medium.

• -ve if center of curvature is located in less dense medium.

70


Example 1.3.4

A cylindrical glass rod has an index of refraction 1.50

. One end is ground to a hemispherical surface with radius r = 20 mm .

A point object on the axis of the rod, 80 mm to the left of the vertex. The rod is in air .

Calculate a) the image distance b) magnification of the image .

71


Solution 1.3.4

n

1

= 1 , n

2

= 1.50 , r =+ 20 mm , u = + 80 mm a) n

1  n u v

2  n

2 r

 n

1

OR

1



.

.

80 v





20

 v

 

120 mm

+ve (to the right of vertex – back side) n

1 u

+ n

2 v

=

( n

2

n

1

) r

72


Solution 1.3.4

n

1

= 1 , n

2

= 1.50 , r =+ 20 mm , u = + 80 mm b) m

 y '

  y n u m

 

( )( )

( .

( )

 

1 same size but inverted

73


Example 1.3.5

The rod in example 1.3.4 is immersed in water of index 1.33. Other quantities have the have the same values in example 1.3.

Calculate the image distance and its magnification .

74


Solution 1.3.5

a) n

1

= 1 , n

2

= 1.50 , r =+ 20 mm , u = + 80 mm n

1  n u v

2  n

2 r

 n

1

OR v

.



.

.

80 v



 

185 mm





20

+ve (to the left of vertex – front side)

.

n

1 u

+ n

2 v

=

( n

2

n

1

) r

75


Solution 1.3.5

n

1

= 1 , n

2

= 1.50 , r =+ 20 mm , u = + 80 mm b) m

 y '

  y n u m

 

( )(



185

( .

( )

)

 greater and upright

76


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 n

1

= 1.50

. One coin is located

2.0 cm from the edge of the sphere

(figure-next page). Find the position of the image of the coin.

77

78

Refraction at A Spherical Surface v n

1 u

+ n

2 v

=

( n

2

n

1

) r v = - 0.017 m

79

Example 1.3.7

Calculate the image distance and magnification .

Solution n

1  n u v

2  n

2 r

 n

1 v

.



.

.

8 v



 

 





 

6 cm

.

8 cm v u = real/actual depth v = apparent depth

80


Solution 1.3.7

m

  m

 

.

 

  m



8 cm v

81


Exercise

1.A point object is 25.0 cm from the centre of a glass sphere of radius 5.0 cm. The refractive index of glass is 1.50. Find the position of the image formed due to refraction by a. the first spherical glass surface.

82

22.3

Thin lenses (2 hours)

At the end of this chapter, students should be able to:

 Sketch and use ray diagrams to determine the characteristics of image formed by diverging and converging lenses.

 Use thin lens equation,

1 u



1 v



1 f for real object only.

 Use lensmaker’s equation: f

1







 n



1



 

 



1



1 r r

1 2







83

Introduction

• A lens is a transparent object with two refracting surfaces whose principal axes coincide.

• This lens is usually circular, and its two faces are portions of a sphere .

• Two types of lenses : i) converging lens (convex) ii) diverging lens (concave)

84

Introduction

R

2

R

1

Converging lens

R

1

R

2

Diverging lens

22.3 Thin lenses

Converging lensa lens causes incident parallel rays to converge after exiting the lens.

Diverging lensa lens causes incident parallel rays to diverge after exiting the lens.

85

Introduction

Terms and definition


Converging lens

86

Introduction 22.3 Thin lenses


Diverging lens

87

Introduction 22.3 Thin lenses


• Principal axis – a straight line passing through the very center of the lens and perpendicular to its two surfaces.

• Focal point – the point at which the rays cross.

• Focal length – the distance between the focal point and the center of the lens.

88

Several types of diverging lenses


Converging lenses are thicker at the center than at the edges whereas diverging lenses are thinner at the center.

89

Image Formation by Thin Lenses

A thin lens is one whose thickness is small compared to its focal length, f .

Ray diagrams can be drawn to determine the location and size of the image .

90

Ray Diagrams


Ray 1 : A ray entering a converging lens parallel to its axis passes through the focal point F of the lens on the other side.

Ray 2 : A ray entering a converging lens through its focal point exits parallel to its axis.

Ray 3 : A ray passing through the center of the lens does not change direction.

91

Ray Diagrams


Ray 1 : A ray entering a diverging lens parallel to its axis seems to come from the focal point F.

Ray 2 : A ray that enters a diverging lens by heading toward the focal point on the opposite side exits parallel to its axis.

Ray 3 : A ray passing through the center of the lens does not change direction.

92


• If the image is real , the position of the image point is determined by intersection of any two rays 1, 2 and 3 .

• Real image is formed on the back side of the lens.

(OPPOSITE SIDE)

• If the image is virtual , we extend the diverging outgoing rays backward to their intersection point to seek the image point.

• Virtual image is formed on the front side of the lens. (SAME SIDE)

93

Converging Lens Image Formation by Thin Lenses

1) Object at 2F

1

, the image is a) At 2F

2 b) Real c) Inverted d) Same size as object

94

Converging Lens


2) Object beyond 2F

1

, the image is a) Between F

2 and 2F

2 b) Real c) Inverted d) Smaller than object

95


3) Object between F

1 and 2F

1

, the image is a) Beyond 2F

2 b) Real c) Inverted d) Larger than object

96


Converging Lens

4) Object at F

1

, the image is at infinity

O

F

1 u

F

2

I at infinity

97


5) Object between lens and F

1

, the image is a) Behind the object b) Virtual c) Upright d) Larger than object

98

Diverging Lens

The image always , a) Between lens and F

1 a) Virtual b) Upright d) Smaller than object


F

1

99

The Lens Equation thin-lens equation v

 f

 v f u



1 1 u v f

1 u v

100

Thin lens formula

f

1



1 u



1 v

101

Q

Sign Convention

Positive (+) Negative (-) f Converging (convex) Diverging (concave) u In front of the lens Behind the lens v Behind the lens (real) h i

OPPOSITE SIDE

Upright image with respect to the object m Upright image with respect to the object

In front of the lens

SAME SIDE

Inverted image with respect to the object

Inverted image with respect to the object

102

1.4 Thin lenses

Example 1.4.1

The Lens Equation

Linear Magnification, m

An object is placed 10 cm from the a 15 cm focal length converging lens. Find the image position and its characteristics .

Solution

1 1 1 u v f m u



30

10





1

10

1 v



1

15 v

 

30 cm

Magnified

- upright

- virtual (in front of the lens)

103

1.4 Thin lenses

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

1 1 1 u v f

1



1 u







1

20 25 u

 

100 cm

104

1.4 Thin lenses The Lens Equation

Linear Magnification, m Example 1.4.3

A diverging meniscus lens has a focal length of -16 cm. If the lens is held 10 cm fro an object, where is the image located ?

What is the magnification of the lens ?

Solution

1 1 1 u v f

1

10

+

1

= v

1

16 v = -6.15

cm m =

6 .

15

10

= 0 .

615

105

1.4 Thin lenses

Example 1.4.4



An object 450 mm from a converging lens forms a real image 900 mm from the lens.

What is the focal length of the lens.

Solution

1 1 1 u v f

1

450

1

+

900

=

1 f f = 30 cm

106

1.4 Thin lenses

Example 1.4.5



An object 6 cm high is held 4 cm from a diverging meniscus lens of focal length -24 cm.

What are the nature, size, and location of the image ?

Solution

1 1 1 u v f

1

+

4

1

= v

1

24 m =

3 .

43

4

= 0 .

86 v = -3.43

cm h i h i

= mh o

= 0 .

86 × 6 m = h o virtual, upright

5.16 cm, -.3.43 cm h i

= 5 .

16 cm

107



Exercise

1)An object 8 cm high is placed 30 cm from a thin converging lens of focal length 12 cm. What are the nature, size, and location of the image formed ?

(real, inverted, -5.33 cm, +20 cm)

2) A 1.70 m tall person is standing in front of a camera. The camera uses a converging lens whose focal length is 0.0500 m. Find the characteristics of the image formed on the film.

(real,inverted, smaller(-.0204),-0.0347 m)

108



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)

109

1.4 Thin lenses

Exercise



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 ;

Two Lenses System

111

Two Lenses System f = +20 cm

A B p

1 v

1 p

2 v

2 112


First Step

• Let p

1 represent the distance of object O from lens A.

• Then find the distance v

1 the image produced by lens A, either by using equation or drawing rays.

113

Second Step Two Lenses System

• Ignore the presence of lens A, treat the image found in first step lens B.

I

1 as the object for

• If this new object is located beyond lens

B, the object distance p

2 negative. is taken to be

•If this new object is located in front of the lens B, the object distance positive.

p

2 is taken to be

114

Second Step Two Lenses System

• The distance v

2 of the final image ( I

2

) 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 m

A and m

B produced by two lenses, m



( m

A

)( m

B

)

115

Example 1.4.6


Two converging lenses, with focal lengths f

1

= 10.0 cm and f

2

= 20.0 cm, are placed 20.0 cm apart. An object is placed 30.0 cm in front of the first. Calculate the position and the magnification of the final image formed by the combination of the two lenses.

116

Solution 1.4.6


Given : u

1

=30.0 cm, f

1

= 10.0 cm , cm, are placed 20.0 cm apart. f

2

= 20.0

1



1



1 u

1 v

1 f

1



1



1

 v

1 v

1

 



1

1



1



1 u

2 v

2 f

2

 v

2

1



1

 v

2

  m

Magnification, m = m

1 x m

2 v

1 v

2 u

1 u

2

 





.







.



 

1

117

Solution 1.4.6


118

Example 1.4.7


Two converging lenses, with focal lengths f

1

= 20.0 cm and f

2

= 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.

( v

2

= +50.0 cm , m= +0.500)

119

Exercise


1.A converging lens has a focal length of

0.080 m. An object is located 0.040 m to the left of this lens. A second converging lens has the same focal length as the first one and is located 0.120 m to the right of it. Relative to the second lens, where is the final image located ? (0.133 m)

120

Exercise


2. A coin is located 20.0 cm to the left of a converging lens ( f = 16.0 cm) A second, identical lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Find the separation between the lenses.

(160 cm)

121

Exercise


3. Two thin converging lenses are placed

60 cm apart and have the same axis.

The first lens has a focal length 10 cm, and the second has a focal length of

15.0 cm. If an object 6.0 cm high is placed 20 cm in front of the first lens, what are the location and size of the final image ? Is it real or virtual ?

(24 cm, 3.6 cm, real)

122

Thin Lenses Formula and Lens maker’s Equation

Considering the ray diagram of refraction for 2 spherical surfaces as shown in figure below.

u

1

A r

1 v

1 t

 v

1 r

2

D v

2 n

1 n

2 n

1

C

1

I

1

C

2

O

P

1

P

2

I

2

B E t 123

Lens maker’s equation f

1





 n

2 n

1



1









1 r

1

 r

1





2 where f : focal length r

1

: radius of curvature of first refracting surface r

2

: radius of curvature of second refracting surface n

1

: refractive index of the medium n

2

: refractive index of the lens material

124

Note :

– If the medium is air ( n

1

= n maker’s equation will be air

=1 ) thus the lens

Where, f

1



 n



1





 r

1

1



1 r

2



 n : refractive index of the lens material

125

Lensmaker’s Equation

1 f





 n n

1



1



1



1

R

1

R

2



 n for lens n

1 for medium

+ve sign

126

Sign Convention


Q Sign Remarks

R +

Convex surface

R -

Concave surface

R

∞ Flat surface f +

Converging lens

f -

Diverging lens

127

OR f

1

= ( n 1 )

1

R

1

+ n for lens, the lens in air

Use sign convention for R :

• +ve for convex surface

• -ve for concave surface


128

Example 1.4.8


Suppose the lens in the figure below has n= 1.52 and is placed in air. Calculate its focal length . What type of this lens ?

46.2 cm

129

Solution 1.4.8

Given: n = 1.52 , n

1

R

2

= +46.2 cm

1 f





 n n

1



1



1



= 1, R

1

R

1

R

1

2






= +22.4 cm, f

1





.

1



1





1

.

f



.





1

.

46.2 cm

Converging lens → f +

130


Example 1.4.9

Each face of a double convex lens has a radius of 20.0 cm. The index of refraction of the glass is 1.50. Calculate the focal length of this lens a) in air and b) when it is immersed in carbon disulfide

( n = 1.63)

131

Solution 1.4.9


Given : R

( a) In air ( n

1 n

1

= +20.0 cm, R

2

= 1.63)

= 1)

= -20.0 cm,

1 f





 n n

1



1



1



1

R

1

R

2 n





= 1.50

f f

1



.

1

 



1





1

.





1

.

b) In carbon disulfide

( n

1

= 1.63) f f

1



.



1





1

.





1

.

 

125 cm

 diverging lens

132


Example 1.4.10

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.

f

Solution

1

= ( n 1 )

( )

R

1

R

2

1

40

= ( 1 .

5 1 )

( )

∞ R

2

R

2

= 20 cm

133

Exercise


1.The curve surface of a plano-concave lens has a radius of -12 cm. What is the focal length if the lens is made from a material with a refractive index of 1.54 ?

(-22.2 cm)

2. A plastic lens ( n = 1.54) has a convex surface of radius 25 cm and a concave surface of -70 cm. What is the focal length ? Is it diverging or converging ?

(72.0 cm, converging)

134

Exercise


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) 135

Refraction at A Spherical Surface

CHAPTER 22:

Geometrical optics

(4 Hours)

UNIT 22 : GEOMETRICAL

OPTICS

The Law of reflection

Drawing Compass

Linear Magnification,

Refraction at aplane and spherical surfaces (1 H)

22.3

Thin lenses (2 hours)

Thin lens formula

Thin Lenses Formula and Lens maker’s Equation

Sign Convention

Convex surface

Concave surface

Converging lens

Diverging lens

Related documents

Products

Support

Refraction at A Spherical Surface

CHAPTER 22:

Geometrical optics

(4 Hours)

UNIT 22 : GEOMETRICAL

OPTICS

The Law of reflection

Drawing Compass

Linear Magnification,

Refraction at aplane and spherical surfaces (1 H)

22.3

Thin lenses (2 hours)

Thin lens formula

Thin Lenses Formula and Lens maker’s Equation

Sign Convention

Convex surface

Concave surface

Converging lens

Diverging lens

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