Pdt Unit9 (students)

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 9:Geometrical Optics

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- Law of reflection

Plane surface Reflection Spherical surface Geometrical Optics

- Ray diagram (plane mirror) - Characteristic of image

- Term in used: principal axis. Centre of curvature( c ),radius of curvature( r ), focal point( F ),focal length( f ),pole of the spherical mirror - Ray diagram (spherical mirror) - Characteristic of image - Sign convention( u, v , f )

Refraction

Plane surface

-

Snell’s Law

- Refractive index( n ) - Material of different n - Convex and concave lenses

Thin lenses

- Term in used: principal axis. focal point( F ),focal length( f ),optical centre of lens - Ray diagram (lenses) - Characteristic of image - Sign convention( u, v , f )

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- Thin lens formula:

1 1 1 + = u v f

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9.1 Reflection of Plane Mirror 9.1.1 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 is equal to the angle of reflection, r as shown in figure below.

i Simulation SF027

r

Plane mirror

i=r

3




Image formation by a plane mirror. (ray diagrams)  Point object where

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

r i A 

i

i u

v

A'

Vertical (extended) object

i hi

ho

i r

Object

u

r Image

v

Simulation SF027

4



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



the object distance, u is equal to the image distance, v same size as the object 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?

5

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

H (head )

E (eyes)

A L h

B

1 AL = HE 2 1 LB = EF 2

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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Height of observer

h = 1 .0 m 6

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

u

v

Therefore, the distance between Nagar and the image of the rose is given by SF027

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9.2 Reflection at a Spherical surface 9.2.1 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. 8













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.

9.2.2 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

9




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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9.2.3 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 ≈θ tan 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

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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

9.2.4 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.



 

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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. 13

9.2.5 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. 14

9.2.6 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

C

I




P

O

F




Front

Real Inverted Diminished Formed between point C and F.

back

O



F

u=r

C

P





Real Inverted Same size Formed at point C.

I SF027

Front

back

15

Object

Ray diagram

distance, u

Image property





f
I

C




P

O




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 P

O

C

Front


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

=

u

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

Negative sign indicates that the object and image are on opposite sides of the principal axis (refer to the real image), If ho is positive, hi is negative. 17

9.2.7 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.


α O

B

θ θ φ β I

C

u

v

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

φ − β = α −φ α + β = 2φ

D P

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(3)

By using
BOD,
BCD and
BID thus

tan α =


(1) (2)

BD BD BD ; 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

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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. v = −30 cm b. the linear magnification and state the properties of the image. Solution:

u=+10 cm, f=+15 cm

M =3

a. By applying the equation of spherical mirror, thus

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

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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 mirror

u O

v 30 cm

I

a. From the figure above, By using the equation of linear magnification, thus

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

b. By using the equation of spherical mirror,

and

therefore

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

hi = 1.35 cm ho=1.5x102 cm, r=-20.0 cm, u=+11.0x102 cm

Modern Physics,11th edition, Young & Freedman.)

Solution:

By applying the equation of spherical mirror, and

From equation of linear magnification,

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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.u = 4.57 cm b. the radius of curvature of the mirror. r = 9.01 cm Solution:

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

35.0 cm

F

u

3.20 m a. By applying the equation of linear magnification,

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b. By applying the equation of spherical mirror, thus and




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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9.3 Refraction on plane surface





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

n1 n2 r

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

r
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Simulation-1

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

r >i

Simulation-2

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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

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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

v1 fλ1 = v 2 f λ2

c    n1  = λ1  c  λ2    n2  SF027

and

v2 = fλ2

where

c v1 = n1

and

c v2 = n2

n1λ1 = n2 λ2 (Refractive index is inversely proportional to the wavelength) 30



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, Air (na)

nw=1.33

A

Water (nw)

2.00 m

r

B

r

i

D

i

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

where SF027

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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 32 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. n = 1.55 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)

λ0 = 5.50 x10 −7 m @ 550 nm 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

b. By using the equation below, thus

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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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9.4 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 35

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

r1

r1 C1











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O

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 centre's of curvature of a lens. Optical centre (point O)
is defined as the point at which any rays entering the lens 36 pass without deviation.

9.4.2 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

F2

f

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

f

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




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Focal length ( f )  Definition – is defined as the distance between the focus F and the optical centre O of the lens. 9.4.3 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

38

(b) Diverging (concave) lens 1 1 2

3 3

O

F2

I

F1

v u 

At least any two rays for drawing the ray diagram.

 

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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 not deviated. Ray 3 - Passes through the focus point 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. 39

9.4.4 Images formed by a diverging (concave) lens
Ray diagrams below showing the graphical method of locating an image formed by a diverging lens.

O

F2

Front





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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. 40

9.4.5 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








O u = 2f

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2F2 2F1

F1

F2

Front

back

I






Real Inverted Diminished Formed between point F2 and 2F2. (at the back of the lens)

Real Inverted Same size Formed at point 2F2. (at the back of the lens) 41

Object

Ray diagram

distance, u

Image property





f < u < 2f

I 2F1O

F1

F2

Front

back

2F2









u=f

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

Real Formed at infinity.

O 2F1

F1

F2

Front

back

2F2

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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

M=

i

ho

=−

u

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

Negative sign indicates that when u and v are both positive, the image is SF027

inverted and ho and hi have opposite signs.

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Thin Lenses Formula and Lens maker’s Equation n2 second surface first surface n1 Io

I

O v u

vo

A ray from an object O in medium of refractive index n1 passes through the first surface of thin lens with refractive index n2. An image is formed (Io). Io acts as the virtual object for the second surface of the lens and finally form a final image SF027

44




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



Convex surface (r = +r1)

n1 n2 ( n2 − n1 ) + = u vo r1 Concave surface (r = -r2) n2 n1 ( n1 − n2 ) + = − vo v − r2

(1)

(2)

adding up eq. (1) and eq. (2):

n1 n2 n2 n1 n2 − n1 ( n1 − n2 ) + + + = + u vo − vo v r1 − r2

n1 n1 n2 − n1 n2 − n1 + = + u v r1 r2

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45

1 1  1 1 n1  +  = ( n2 − n1 ) +  u v  r1 r2  1 1  n2 − n1  1 1   +  + =  u v  n1  r1 r2 
If

(3)

u at infinity then v = f, hence eq. (3) becomes

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

Lens maker’s equation

where

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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 46 n2 : refractive index of the lens material




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

1 1 1 + = u v f therefore in general,

1 1 1 = + f u v


Thin lens formula

Note : 

If the medium is air (n1= nair=1) thus the lens maker’s equation will be

1 1 1 = (n − 1) +  f  r1 r2  



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where 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 = ∞. 47




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

48




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 SF027

(2) 49

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

cm, r2=+18.5 cm

1 1 1 = (n − 1) +  f  r1 r2  SF027

f = +208 cm

50

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







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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 51