Matrix Physic Note

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.

UNIT 1:Geometrical Optics

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1.1 Reflection of Plane Mirror and Refraction 1.1.1 Reflection of Plane Mirror  Definition – is defined as the return of all or part of a beam of particles or waves when it encounters the boundary between two media.  Laws of reflection state :  The incident ray, the reflected ray and the normal all lie in the same plane. 

The angle of incidence, i equals the angle of reflection, r as shown in figure below.

i

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Simulation

r

Plane mirror

i=r

2



Image formation by a plane mirror.  Point object

where

u : object distance v : image distance ho : object height hi : image height

r i A 

i

i u

v

Vertical (extended) object

i ho

i r

Object

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Simulation

A'

u

hi

r

Image

v 3



The properties of image formed are  virtual  upright or erect  laterally reverse  

the object distance, u equals the image distance, v the same size where the linear magnification is given by

Image height, hi M= =1 Object height, ho 



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obey the laws of reflection.

Example 1 : Find the minimum vertical length of a plane mirror for an observer of 2.0 m height standing upright close to the mirror to see his whole reflection. How should this minimum length mirror be placed on the wall?

4

H (head)

E (eyes)

Solution: By using the ray diagram as shown in figure below.

A L

h

B

F (feet )

The minimum vertical length of the mirror is given by

h = AL + LB 1 1 h = HE + EF 2 2 1 h = ( HE + EF ) 2

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1 AL = HE 2 1 LB = EF 2

Height of observer

h = 1.0 m 5

The mirror can be placed on the wall with the lower end of the mirror is halved of the distance between the eyes and feet of the observer. 

Example 2 : A rose in a vase is placed 0.250 m in front of a plane mirror. Nagar looks into the mirror from 2.00 m in front of it. How far away from Nagar is the image of the rose? Solution: u=0.250

m

2.00 m x From the properties of the image formed by the plane mirror, thus v = u

v = 0.250 m

u

v

Therefore, the distance between Nagar and the image of the rose is given by x = 2.00 + v x = 2.25 m SF027

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1.1.2 Refraction  Definition – is defined as the changing of direction of a light ray and its speed of propagation as it passes from one medium into another.  Laws of refraction state :  The incident ray, the refracted ray and the normal all lie in the same plane.  For two given media,

sin i n2 = = constant sin r n1 Or

Snell’s law

n1 sin i = n2 sin r where

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i : angle of incidence r : angle of refraction n1 : refractive index of the medium 1 (Medium containing the incident ray) n2 : refractive index of the medium 2 (Medium containing the refracted ray)

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

Examples for refraction of light ray travels from one medium to another medium can be shown in figures below. (a) n1 < n2 (b) n1 > n2 (Medium 1 is less (Medium 1 is denser dense than medium 2) than medium 2)

Incident ray

Incident ray

i

i

n1 n2

r

r Refracted ray The light ray is bent toward the normal, thus

r
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Simulation-1

n1 n2

Refracted ray The light ray is bent away from the normal, thus

r >i

Simulation-2

8



Refractive index (index of refraction) refraction sin i  Definition – is defined as the constant ratio for the two given media. sin r  The value of refractive index depends on the type of medium and the colour of the light.  It is dimensionless and its value greater than 1.  Consider the light ray travels from medium 1 into medium 2, the refractive index can be denoted by

velocity of light in medium 1 v1 = 1 n2 = velocity of light in medium 2 v2 (Medium containing the incident ray) 

(Medium containing the refracted ray)

Absolute refractive index, n (for the incident ray is travelling in vacuum or air and is then refracted into the medium concerned) concerned is given by

velocity of light in vacuum c n= = velocity of light in medium v SF027

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

Table below shows the indices of refraction for yellow sodium light having a wavelength of 589 nm in vacuum.

(If the density of medium is greater hence the refractive index is also greater) SF027

10



The relationship between refractive index and the wavelength of light. 





As light travels from one medium to another, its wavelength, λ

changes but its frequency, f remains constant. constant The wavelength changes because of different material. material The frequency remains constant because the number of wave cycles arriving per unit time must equal the number leaving per unit time so that the boundary surface cannot create or destroy waves. waves By considering a light travels from medium 1 (n1) into medium 2 (n2), the velocity of light in each medium is given by

v1 = fλ1

then

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and

v1 fλ1 = v2 fλ2 c    n1  = λ1  c  λ2    n2 

v 2 = f λ2

where

c v1 = n1

and

c v2 = n2

n1λ1 = n2 λ2 (Refractive index is inversely proportional to the wavelength) 11



If medium 1 is vacuum or air, then n1 = 1. Hence the refractive index for any medium, n can be expressed as where

λ0 n= λ



λ0 : wavelength of light in vacuum λ : wavelength of light in medium

Example 3 : A fifty cent coin is at the bottom of a swimming pool of depth 2.00 m. The refractive index of air and water are 1.00 and 1.33 respectively. What is the apparent depth of the coin? Solution:

na=1.00, nw=1.33 Air (na)

A

Water (nw)

2.00 m

B

r

r i

i

D

C AB : apparent depth AC : actual depth = 2.00 m

where SF027

12

From the diagram, tan r ABD ACD

AD AB AD tan i = AC =

By considering only small angles of r and i , hence

tan r ≈ sin r

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tan i ≈ sin i  AD  then   tan i sin i  AC  AB = = = tan r sin r  AD  AC    AB  From the Snell’s law, Note : (Important) sin i n2 na = = Other equation for absolute sin r n1 nw refractive index in term of depth is given by AB na = real depth AC nw n= apparent depth AB = 1.50 m 13 and



Example 4 : A light beam travels at 1.94 x 108 m s-1 in quartz. The wavelength of the light in quartz is 355 nm. a. Find the index of refraction of quartz at this wavelength. b. If this same light travels through air, what is its wavelength there? (Given the speed of light in vacuum, c = 3.00 x 108 m s-1) No. 33.3, pg. 1278, University Physics with Modern Physics,11th edition, Young & Freedman.

Solution: v=1.94 x 108 m s-1, λ=355 x 10-9 m a. By applying the equation of absolute refractive index, hence

c n= v n = 1.55

b. By using the equation below, thus

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λ0 n= λ λ0 = nλ λ0 = 5.50 x10 −7 m @ 550 nm

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



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Example 5 : (exercise) We wish to determine the depth of a swimming pool filled with water by measuring the width (x = 5.50 m) and then noting that the bottom edge of the pool is just visible at an angle of 14.0° above the horizontal as shown in figure below. (Gc.835.60)

Calculate the depth of the pool. (Given nwater = 1.33 and nair = 1.00) Ans. : 5.16 m Example 6 : (exercise) A person whose eyes are 1.54 m above the floor stands 2.30 m in front of a vertical plane mirror whose bottom edge is 40 cm above the floor as shown in figure below. (Gc.832.10)

Find x. Ans. : 0.81 m

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1.2 Reflection of Spherical Mirrors 1.2.1 Spherical mirror  Definition – is defined as a reflecting surface that is part of a sphere.  There are two types of spherical mirror. It is convex (curving outwards) and concave (curving inwards) mirror.  Figures below show the shape of concave and convex mirrors. (a) Concave (Converging) Converging mirror (b) Convex (Diverging) Diverging mirror imaginary sphere A C

r

P

B



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A

silver layer P

r

C

B

reflecting surface Some terms of spherical mirror  Centre of curvature (point C)  is defined as the centre of the sphere of which a curved mirror forms a part. 16









Radius of curvature, r  is defined as the radius of the sphere of which a curved mirror forms a part. Pole or vertex (point P)  is defined as the point at the centre of the mirror. Principal axis  is defined as the straight line through the centre of curvature C and pole P of the mirror. AB is called the aperture of the mirror.

1.2.2 Focal point and focal length, f  Consider the ray diagram for concave and convex mirror as shown in figures below. Incident Incident rays rays C F

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f

P

P

f

C F

17



From the figures,  Point F represents the focal point or focus of the mirrors. Distance f represents the focal length of the mirrors.  The parallel incident rays represent the object infinitely far away from the spherical mirror e.g. the sun. Focal point or focus, F  for concave mirror – is defined as a point where the incident parallel rays converge after reflection on the mirror.  Its focal point is real (principal).  for convex mirror – is defined as a point where the incident parallel rays seem to diverge from a point behind the mirror after reflection.  Its focal point is virtual. 







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Focal length, f  Definition – is defined as the distance between the focal point (focus) F and pole P of the spherical mirror. The paraxial rays is defined as the rays that are near to and almost parallel to the principal axis.

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1.2.3 Relationship between focal length, f and radius of curvature, r  Consider a ray AB parallel to the principal axis of concave mirror as shown in figure below. incident ray B A

i

C

i

i

D

F

r 

From the figure, BCD BFD



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BD tan i = ≈i CD BD tan θ = ≈θ FD

θ P

f

Taken the angles are << small by considering the ray AB is paraxial ray.

By using an isosceles triangle CBF, thus the angle θ is given by

θ = 2i

19

then



BD  BD  = 2  FD  CD  CD = 2 FD

Because of AB is paraxial ray, thus point B is too close with pole P then

CD ≈ CP = r FD ≈ FP = f



Therefore

r=2f or

r f = 2

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This relationship also valid for convex mirror.

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O

1.2.4 Ray Diagrams for Spherical Mirrors  Definition – is defined as the simple graphical method to indicate the positions of the object and image in a system of mirrors or lenses.  Ray diagrams below showing the graphical method of locating an image formed by concave and convex mirror. (a) Concave mirror (b) Convex mirror 1 1 1 3 2 2 3 2 C P I C P F O I F 2 3 1

At least any two rays for drawing the ray diagram. SF027







Ray 1 - Parallel to principal axis, after reflection, passes through the focal point (focus) F of a concave mirror or appears to come from the focal point F of a convex mirror. Ray 2 - Passes or directed towards focal point F reflected parallel to principal axis. Ray 3 - Passes or directed towards centre of curvature C, reflected back along the same path. 21

1.2.5 Images formed by a convex mirror  Ray diagrams below showing the graphical method of locating an image formed by a convex mirror.

C

P

O

I u Front





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F

v back

Properties of image formed are  virtual  upright  diminished (smaller than the object)  formed at the back of the mirror Object position → any position in front of the convex mirror. 22

1.2.6 Images formed by a concave mirror  Table below shows the ray diagrams of locating an image formed by a concave mirror for various object distance, u. Object

Ray diagram

distance, u

Image property

 

u>r

O

C

I F



P



Front

back

O  

F

u=r

C



P



I

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Front

Real Inverted Diminished Formed between point C and F.

back

Real Inverted Same size Formed at point C. 23

Object

Ray diagram

distance, u

Image property

 

f
I

C

O



P



F

Front

back

O

 

u=f

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C

Front

Real Inverted Magnified Formed at a distance greater than CP.

F

Real Formed at infinity.

P

back

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Object

Ray diagram

distance, u

Image property

  

u


F

O

C

Front 

P

Virtual Upright Magnified Formed at the back of the mirror

I back

Linear (lateral) magnification of the spherical mirror, M is defined as the ratio between image height, hi and object height, ho h v where

Simulation

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M=

i

ho

=−

v : image distance from pole u u : object distance from pole 25

1.2.7 Derivation of Spherical mirror equation 

Figure below shows an object O at a distance u and on the principal axis of a concave mirror. A ray from the object O is incident at a point B which is close to the pole P of the mirror. 

O

α C

θ θ φ β I

u 

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B

v

D P

From the figure, φ = α +θ BOC β = φ +θ BCI then, eq. (1)-(2) :

φ − β = α −φ α + β = 2φ

(1) (2)

(3)

By using BOD, BCD and BID thus

BD BD BD tan α = ; tan φ = ; tanβ = OD CD ID

By considering point B very close to the pole P, hence

tan α ≈ α ; tan φ ≈ φ ; tanβ ≈ β OD ≈ OP = u ; CD ≈ CP = r ; ID ≈ IP = v then BD BD BD Substituting this α= ;φ= ; β= value in eq. (3) u r v

26

therefore



BD BD  BD  + = 2  u v  r  1 1 2 where r = 2 f + = u v r 1 1 1 Equation (formula) = + of spherical mirror f u v

Table below shows the sign convention for equation of spherical mirror .

Physical Quantity

Object distance, u

Positive sign (+)

Real object

(in front of the mirror)

Negative sign (-)

Virtual object

(at the back of the mirror)

Real image

Virtual image

Focal length, f

Concave mirror

Convex mirror

Linear magnification, M SF027

Upright (erect) image

Inverted image

Image distance, v

(same side of the object) (opposite side of the object)

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

Example 7 : An object is placed 10 cm in front of a concave mirror whose focal length is 15 cm. Determine a. the position of the image. b. the linear magnification and state the properties of the image. Solution:

u=+10 cm, f=+15 cm

a. By applying the equation of spherical mirror, thus

1 1 1 = + f u v 1 1 1 = + 15 10 v v = −30 cm

The image is 30 cm from the mirror on the opposite side of the object (or 30 cm at the back of the mirror). b. The linear magnification is given by The properties of the image v − 30 are M =− =−

(

u

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M =3

)

10

 Virtual  Upright  Magnified

28



Example 8 : An upright image is formed 30 cm from the real object by using the spherical mirror. The height of image is twice the height of object. a. Where should the mirror be placed relative to the object? b. Calculate the radius of curvature of the mirror and describe the type of mirror required. Solution:

hi=2ho

Spherical

u O

mirror

v

30 cm

u + v = 30 cm

I (1)

hi v M = =− a. From the figure above, ho u (2) By using the equation of linear v =magnification, −2u thus SF027

29

By substituting eq. (2) into eq. (1), hence

u = 10 cm

The mirror should be placed 10 cm in front of the object. b. By using the equation of spherical mirror,

and

r f = 2

1 1 1 = + f u v 1 1 1 = + f u ( − 2u ) f = +20 cm therefore

r = 40 cm

The type of spherical mirror is concave because the positive value of focal length.

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

Example 9 : A mirror on the passenger side of your car is convex and has a radius of curvature 20.0 cm. Another car is seen in this side mirror and is 11.0 m behind the mirror. If this car is 1.5 m tall, calculate the height of the car image . (Similar to No. 34.66, pg. 1333, University Physics with Modern Physics,11th edition, Young & Freedman.)

Solution:

ho=1.5x102 cm, r=-20.0 cm, u=+11.0x102 cm

By applying the equation of spherical mirror,

1 1 1 = + f u v 2 1 1 = + r u v

and

r f = 2

v = −9.91 cm From equation of linear magnification,

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hi v M = =− ho u v hi = − ho u hi = 1.35 cm

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

Example 10 : A concave mirror forms an image on a wall 3.20 m from the mirror of the filament of a headlight lamp. If the height of the filament is 5.0 mm and the height of its image is 35.0 cm, calculate a. the position of the filament from the pole of the mirror. b. the radius of curvature of the mirror. Solution:

hi=-35.0 cm, v=3.20x102 cm, ho=0.5 cm O I 5.0 mm P C

35.0 cm

F

u

3.20 m hi v M = =− ho u SF027

− 35.0 3.20 x10 2 =− 0.5 u u = 4.57 cm

a. By applying the equation of linear magnification,

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b. By applying the equation of spherical mirror, thus

r 1 1 1 and f = = + 2 f u v 2 1 1 = + r u v r = 9.01 cm 

Example 11 : (exercise) a. A concave mirror forms an inverted image four times larger than the object. Find the focal length of the mirror, assuming the distance between object and image is 0.600 m. b. A convex mirror forms a virtual image half the size of the object. Assuming the distance between image and object is 20.0 cm, determine the radius of curvature of the mirror. No. 14, pg. 1169,Physics for scientists and engineers with modern physics, Serway & Jewett,6th edition.

Ans. : 160 mm, -267 mm

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1.3 Refraction of Spherical Surfaces 

Figure below shows a spherical surface with radius, r forms an interface between two media with refractive indices n1 and n2.

i

n1

B

θ

α

φ

PD

O

n2

C

β

I

r u

v



The surface forms an image I of a point object O as shown in figure above.



The incident ray OB making an angle i with the normal and is refracted to ray BI making an angle θ where n1


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Point C is the centre of curvature of the spherical surface and BC is normal. 34



From the figure,

BOC BIC 

i = α +φ φ = β +θ θ =φ −β

(1) (2)

From the Snell’s law

n1 sin i = n2 sin θ

By using BOD, BCD and BID thus

BD BD BD tan α = ; tan φ = ; tanβ = OD CD ID

By considering point B very close to the pole P, hence

sin i ≈ i ; sin θ ≈ θ ; tan α ≈ α ; tan φ ≈ φ ; tanβ ≈ β OD ≈ OP = u ; CD ≈ CP = r ; ID ≈ IP = v

then Snell’s law can be written as

n1i = n2θ



(3) By substituting eq. (1) and (2) into eq. (3), thus

then SF027

n1 (α + φ ) = n2 (φ − β ) n1α + n2 β = (n2 − n1 )φ

 BD   BD   BD  n1   + n2   = (n2 − n1 )   u   v   r 

35

n1 n2 (n2 − n1 ) + = u v r

Equation of spherical refracting surface

where

v : image distance from pole u : object distance from pole n1 : refractive index of medium 1 (Medium containing the incident ray) n2 : refractive index of medium 2 (Medium containing the refracted ray) 

Note :  If the refraction surface is flat (plane) : r = ∞ then n1 n2

u 

+

v

=0

The equation (formula) of linear magnification for refraction by the spherical surface is given by

hi n1v M = =− ho n2u SF027

36



Sign convention for refraction :

Physical Quantity

Object distance, u Image distance, v

Positive sign (+)

Real object

(in front of the refracting surface)

Real image

(opposite side of the object)

Negative sign (-)

Virtual object

(at the back of the refracting surface)

Virtual image

(same side of the object)

Focal length, f

Convex surface

Concave surface

Radius of curvature, r Linear magnification, M

Convex surface

Concave surface

Upright (erect) image

Inverted image



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Example 12 : A cylindrical glass rod in air has refractive index of 1.52. One end is ground to a hemispherical surface with radius, r =2.00 cm as shown in figure below. air glass P cm C O 0 I 0 . 2

8.00 cm

37

Find, a. the position of the image for a small object on the axis of the rod, 8.00 cm to the left of the pole as shown in figure. b. the linear magnification. (Given the refractive index of air , na= 1.00) Example 34.5, pg. 1302, University Physics with Modern Physics,11th edition, Young & Freedman.

Solution:

ng=1.52, u=8.00 cm, r=+2.00 cm

a. By applying the equation of spherical refracting surface,

n1 n2 (n2 − n1 ) + = u v r na ng ( ng − na ) + = u v r

v = +11.26 cm The image is 11.26 cm at the back of the convex surface. b. By using the equation of linear magnification for refracting surface,

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hi n1v M = =− ho n2 u

na v M =− ng u M = −0.93

Negative sign indicates the image is inverted. 38



Example 13 : 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. b. the first and second refractive surfaces of spheres. Solution: a. Given na=n1=1.00,

ng=n2=1.50, u=20.0 cm, r=5.0 cm

By using the equation of spherical refracting surface, thus na ng (ng − na ) Convex surface r = + 5 . 0 cm and + = (first surface) u v r

1.00 1.50 (1.50 − 1.00 ) + = The image is real and 30 cm at the back 20.0 v 5.0 v = +30 cm na O

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u = 20.0 cm

of the convex surface.

ng P

C

r v = 30 cm

I1 39

b.

ng

na O

P

C

First surface

na Q I 2 30 cm 20 cm

I1

Second surface From the figure above, the image I1 formed by the first surface is in glass and 20 cm from the point Q of the second surface.I1 acts as a virtual object for the second refraction surface and

ng=n1=1.50, na=n2=1.00, u=-20.0 cm, r=-5.0 cm

na (na − ng ) + = u v r 1.50 1.00 (1.00 − 1.50 ) + = (−20.0 ) v (−5.0 ) Using

ng

Concave surface (second surface)

v = +5.71 cm

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The image is real and 5.71cm at the back of the concave surface 40 (5.71 cm from point Q as shown in figure above).





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Example 14 : (exercise) A small strip of paper is pasted on one side of a glass sphere of radius 5 cm. The paper is then view from the opposite surface of the sphere. Find the position of the image. (Given refractive index of glass =1.52 and refractive index of air=1.00) Ans. : 20.83 cm in front of the concave surface (second refracting surface) Example 15 : (exercise) A point source of light is placed at a distance of 25.0 cm from the centre of a glass sphere of radius 10 cm. Find the image position of the source. (Gc.830.Exam.33-11) (Given refractive index of glass =1.50 and refractive index of air=1.00) Ans. : 28 cm at the back of the concave surface (second refracting surface).

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1.4 Thin Lenses 

 

Definition – is defined as a transparent material with two spherical refracting surfaces whose thickness is thin compared to the radii of curvature of the two refracting surfaces. There are two types of thin lens. It is converging and diverging lens. Figures below show the various types of thin lenses, both converging and diverging. (a) Converging (Convex) lenses

Biconvex Plano-convex (b) Diverging (Concave) lenses

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Biconcave

Plano-concave

Convex meniscus

Concave meniscus 42

1.4.1 Terms of lens  Figures below show the shape of converging (convex) and diverging (concave) lenses. (b) Diverging lens (a) Converging lens

r1

r1 O

C1



r2

C2

C1

O

r2

C2

Centre of curvature (point C1 and C2) is defined as the centre of the sphere of which the surface of the lens is a part. Radius of curvature (r1 and r2) 



is defined as the radius of the sphere of which the surface of the lens is a part. Principal (Optical) axis  is defined as the line joining the two centres of curvature of a lens. Optical centre (point O)  is defined as the point at which any rays entering the lens 43 pass without deviation. 



 SF027

1.4.2 Focus (Focal point) and focal length  Consider the ray diagrams for converging and diverging lens as shown in figures below.

O

F1



f

F2

F1

O

f

From the figures, f  Point F1 and F2 represent the focus of the lens.

F2

f

Distance f represents the focal length of the lens. Focus (point F1 and F2)

 





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For converging (convex) lens – is defined as the point on the principal axis where rays which are parallel and close to the principal axis converges after passing through the lens.  Its focus is real (principal). For diverging (concave) lens – is defined as the point on the principal axis where rays which are parallel to the principal axis seem to diverge from after passing through the lens.  Its focus is virtual. 44

Focal length ( f )  Definition – is defined as the distance between the focus F and the optical centre O of the lens. 1.4.3 Ray Diagrams for Lenses  Ray diagrams below showing the graphical method of locating an image formed by converging (convex) and diverging (concave) lenses. (a) Converging (convex) lens 

1 2 3

O

F2

I

F1 2

1

3

u

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v

45

(b) Diverging (concave) lens 1 1

3

2

O

F2

u

At least any two rays for drawing the ray diagram.

I

F1

v 2



Ray 1 - Parallel to the principal axis, after refraction by the lens, passes through the focal point (focus) F2 of a converging lens or appears to come from the focal point F2 of a diverging lens.



Ray 2 - Passes through the optical centre of the lens is undeviated. Ray 3 - Passes through the focus F1 of a converging lens or appears to converge towards the focus F1 of a diverging lens, after refraction by the lens the ray parallel to the principal axis. 46



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3

1.4.4 Images formed by a diverging lens  Ray diagrams below showing the graphical method of locating an image formed by a diverging lens.

O

F2

Front 



SF027

I

F1

back

Properties of image formed are  virtual  upright  diminished (smaller than the object)  formed in front of the lens. Object position → any position in front of the diverging lens. 47

1.4.5 Images formed by a converging lens  Table below shows the ray diagrams of locating an image formed by a converging lens for various object distance, u. Object

Ray diagram

distance, u

Image property

 

I u > 2f

O 2F1

F1

F2

Front

back

2F2

 

(at the back of the lens) 

u = 2f

SF027

O

2F2 2F1

F1

F2

Front

back

Real Inverted Diminished Formed between point F2 and 2F2.

I

  

Real Inverted Same size Formed at point 2F2. (at the back of the lens) 48

Object

Ray diagram

distance, u

Image property

 

f < u < 2f

I

2F1 O

F1

F2

Front

back

2F2

 

 

u=f

SF027

O 2F1

F1

F2

Front

back

Real Inverted Magnified Formed at a distance greater than 2f at the back of the lens.

Real Formed at infinity.

2F2

49

Object

Ray diagram

distance, u

Image property

  

u


I 

2F1

F1 O

F2

Front

back

Virtual Upright Magnified Formed in front of the lens.

2F2

Linear (lateral) magnification of the thin lenses, M is defined as the ratio between image height, hi and object height, ho h v where

Simulation

SF027

M=

i

ho

=−

u

v : image distance from optical centre u : object distance from optical centre 50

Equation



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

v1

u1

r1 A

n1

v2 D

n2 C1

O

I1

n1 C2 P2

P1

B

SF027

t − v1 r2

t

I2

E

51



By using the equation of spherical refracting surface, the refraction by first surface AB and second surface DE are given by 



Convex surface AB (r = +r1)

n1 n2 (n2 − n1 ) + = u1 v1 r1 Concave surface DE (r = -r2) n2 n1 (n1 − n2 ) + = ( t − v1 ) v2 − r2 Assuming the lens is very thin thus t

(1)

= 0,

n2 n1 (n1 − n2 ) + = − v1 v2 − r2  n1 − n2  n1  n2  −  = −  v1  − r2  v2 

n2 n1  n2 − n1   = −  v1 v2  r2  SF027

(2) 52



By substituting eq. (2) into eq. (1), thus

n1  n1  n2 − n1  (n2 − n1 )  = +  −  u1  v2  r2  r1 n1 n1 (n2 − n1 ) (n2 − n1 ) + = + u1 v 2 r1 r2 then



If

 1 1  1 1  n2 + =  − 1  +  u1 v2  n1  r1 r2  u1 = ∞ and v2 = f hence eq. (3) becomes

 1 1  1  n2 =  − 1 +  f  n1  r1 r2  where

SF027

(3)

Lens maker’s equation

f : focal length r1 : radius of curvature of first refracting surface r2 : radius of curvature of second refracting surface n1 : refractive index of the medium 53 n2 : refractive index of the lens material



By equating eq. (3) with lens maker’s equation, hence

1 1 1 + = u1 v 2 f therefore in general,

1 1 1 = + f u v 

Thin lens formula

Note : 

If the medium is air (n1= will be

nair=1) thus the lens maker’s equation

1 1 1 = ( n − 1)  +  f  r1 r2  where 



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n : refractive index of the lens material

For thin lens formula and lens maker’s equation, Use the sign convention for refraction. refraction Very Important

The radius of curvature of flat refracting surface is infinity, r = ∞ . 54



Example 16 : A biconvex lens is made of glass with refractive index 1.52 having the radii of curvature of 20 cm respectively. Calculate the focal length of the lens in a. water, b. carbon disulfide. (Given nw = 1.33 and nc=1.63) Solution: r1=+20

cm, r2=+20 cm, ng=n2=1.52

a. Given the refractive index of water, nw

= n1

By using the lens maker’s equation, thus

 1 1  1  ng =  − 1  +  f  nw  r1 r2  f = +70 cm

b. Given the refractive index of carbon disulfide, nc

= n1

By using the lens maker’s equation, thus

 1 1  1  ng =  − 1  +  f  nc  r1 r2 

SF027

f = −148.18 cm

55



Example 17 : A converging lens with a focal length of 90.0 cm forms an image of a 3.20 cm tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Find a. the object position from the lens. b. the image position from the lens. Is the image real or virtual? No. 34.26, pg. 1331, University Physics with Modern Physics,11th edition, Young & Freedman.

Solution: f=+90.0

cm, ho=3.20 cm, hi=-4.50 cm

a. By using the linear magnification equation, hence

hi v M = =− ho u v = 1.41u

(1)

By applying the thin lens formula,

1 1 1 = + f u v 1 1 1 = + 90.0 u v

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(2) 56

By substituting eq. (1) into eq. (2),hence

u = 154 cm

The object is placed 154 cm in front of the lens. b. By substituting u

= 154 cm into eq. (1),therefore

v = 217 cm

The image forms 217 cm at the back of the lens (at the opposite side of the object placed) and the image is real. 

Example 18 : An object is placed 90.0 cm from a glass lens (n=1.56) with one concave surface of radius 22.0 cm and one convex surface of radius 18.5 cm. Determine a. the image position. b. the linear magnification. (Gc.862.28) Solution: u=+90.0

cm, n=1.56, r1=-22.0 cm, r2=+18.5 cm

a. By applying the lens maker’s equation in air,

1 1 1 = ( n − 1)  +  f  r1 r2 

SF027

f = +208 cm

57

By applying the thin lens formula, thus

1 1 1 = + f u v v = −159 cm

The image forms 159 cm in front of the lens (at the same side of the object placed) b. By applying equation of linear magnification for thin lens, thus

v M =− u





SF027

M = 1.77

Example 19 : (exercise) A glass (n=1.50) plano-concave lens has a focal length of 21.5 cm. Calculate the radius of the concave surface. (Gc.862.26) Ans. : -10.8 cm Example 20 : (exercise) An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens. a. Calculate the focal length of the lens and state the type of the lens. b. If the object is 8.00 mm tall, find the height of the image. c. Sketch the ray diagram for the case above. (UP. 1332.34.34) Ans. : +11.1 cm, -1.8 cm 58

1.6 Optical Devices  

There are 3 optical devices that extend human vision. It is magnifier, compound microscope and telescope. telescope

1.6.1 Angular magnification (magnifying power), Ma 

The angular magnification of an optical device is defined as the ratio of the angle subtended at the eye by the image , β to the angle subtended at the unaided eye by the object (without lens), α.

β Ma = α





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In order to determine the angle α it is necessary to specify the position of the object.  For microscope, microscope the best object position is at the near point. point  For telescope, the object position is not meaningful because the telescope is used for viewing distant object. Near point is defined as the nearest point at which an object is seen most clearly by the human eye.  The distance between the near point to the eye is 25 cm and is known as distance of distinct vision (D). 59

1.6.2 Magnifier  It also known as magnifying glass or simple microscope. microscope  It is an optical device used for viewing near object.  It consists of single converging (biconvex) lens.  Suppose a leaf is viewed at near point of the human eye as shown in figure below.

ho

α D



From the figure,

ho tan α = D

By making small angle approximation, we get

ho tan α ≈ α = D SF027

60



To increase the apparent size of the leaf, a converging lens can be placed in front of the eye as shown in figure below.

hi

β I



ho F O u v

β

The apparent size of the leaf is maximum when the image is at the near point where

v = − D = −25 cm



From the figure above,

hi ho tan β = = D u

By making small angle approximation, we get

hi ho tan β ≈ β = = D u

The properties of the image are SF027



Virtual, upright and magnified

u
61



The angular magnification in terms of D and f can be evaluated by derivation below.  By applying the thin lens formula,

1 1 1 = + f u v Df u= D+ f



where

v = −D (1)

From the definition of angular magnification,

 ho    β u Ma = = α  ho    D D Ma = (2) u 

SF027

By substituting eq. (1) into eq. (2), thus where

D M a = + 1 f : focal length f D : distance of distinct vision = 25 cm 62



The relationship between linear magnification, M with angular magnification, Ma 

From the definition of angular magnification,

then



 hi    β D Ma = = α  ho    D hi Ma = = M ho

Note:  If the object placed at the focal point of the converging lens, the image formed at infinity. infinity Thus The eye is relax. ho

β=



SF027

f β Therefore, since M a = α

then

 ho    f   Ma =  ho    D

D Ma = f 63

1.6.3 Compound Microscope  Because it makes use of two lenses, the magnifying power of the compound microscope is much greater than that of the magnifier.  The two lenses are converging lens and is known as objective lens (close to the object) and eyepiece lens (close to the eye).  The figure below shows the schematic diagram of the compound microscope. u L The properties of final image are fe  Virtual, inverted and magnified Objective lens 

v >(fo+ fe)

O Fo

Fo'

Fe I1

fo

Eyepiece lens acts as a magnifier.

SF027

I2

The properties of first image are  Real, inverted and magnified 

v >2fo

64





The properties of the compound microscope are 

The distance between two lenses, L > (fo+fe)



fo < fe



The final image is I2.



The angular magnification formula is given by where

L  D  f e : focal length of the eyepiece lens M a = −   f o  f e  f o : focal length of the objective lens D : distance of distinct vision = 25 cm

The negative sign indicates that the image is inverted. It is used for viewing small objects that are very close to the objective lens.

1.6.4 Astronomical (refracting) Telescope  This telescope consists of two converging lenses.  Like compound microscope, the two lenses are objective and eyepiece lens.  It is used to magnify objects that are very far away (considered to be at infinity). SF027

65



The figure below shows the schematic diagram of the telescope.

fo

Parallel rays from object at infinity

fe

fe

Fe' Fo

Fe

I1 Objective lens

Eyepiece lens acts as a magnifier. The properties of first image are  Real, inverted and diminished

I2



v =fo

The properties of final image are  Virtual, inverted and magnified  SF027

v >(fo+ fe)

66



The properties of the telescope are 

The distance between two lenses, L <(fo+fe)



fo > fe



The final image is I2.



The angular magnification formula is given by

fo Ma = − fe The negative sign indicates that the image is inverted.

SF027

67