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
SF027
1
1.1 Reflection of Plane Mirror and Refraction 1.1.1 Reflection of Plane Mirror Definitio Definition n – is defined 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, shown in figure below.
i equals the angle of reflection, r as
i
Simulation SF027
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 i
i u
'
v
Vertical (extended) object
i ho
r
i r
Object
hi Image
u
v
Simulation SF027
3
The properties of image formed are virtual upright or erect laterally reverse the object distance,
the same size where the linear magnification is given by
M =
u equals the image distance, v
Image height, hi Object height, ho
=1
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?
SF027
4
Image formation by a plane mirror. Point object
where
u : object distance v : image distance ho : object height hi : image height
r i i
i u
'
v
Vertical (extended) object
i ho
r
i r
Object
hi Image
u
v
Simulation SF027
3
The properties of image formed are virtual upright or erect laterally reverse the object distance,
the same size where the linear magnification is given by
M =
u equals the image distance, v
Image height, hi Object height, ho
=1
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?
SF027
4
Solution: By using the ray diagram as shown in figure below. H (head )
E (eyes)
L h 1 AL = HE 2 1 LB = EF 2
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
h = 1.0 m
Height of observer
SF027
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 0.250 m in front of a plane mirror. Nagar Nagar looks into the mirror from 2.00 m in front of it. How far away from Nagar Nagar is the image image of the rose? 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 distance between Nagar and the image of the rose is given by x = 2.00 + v x = 2.25 m SF027
6
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 sin r
=
n2 n1
= constant Snell’ Snell’s law
Or
n1 sin i = n2 sin r where
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)
SF027
7
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 dense than medium 2)
(Medium 1 is denser than medium 2)
Incident ray Incident ray
i
i n1
n1 n2
n2 r
r Refracted ray The light ray is bent toward the normal, thus
r < i
SF027
Simulation-1
Refracted ray The light ray is bent away from the normal, thus
r > i
Simulation-2
8
Refractive index (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 1 n2 =
(Medium containing the incident ray)
velocity of light in medium 2
=
v1 v2
(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) air concerned is given by
n= SF027
velocity of light in medium 1
velocity of light in vacuum velocity of light in medium
=
c v 9
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 refracti ve 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, 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 ea ch medium is given by
v1 = f λ 1
v2 = f λ 2
and
then
v1 v2
=
f λ 1
where
f λ 2
c n1 = λ 1 c λ 2 n2
v1 =
c n1
and
v2 =
c n2
n1λ 1 = n2 λ 2 (Refractive index is inversely proportional to the wavelength)
SF027
11
If medium 1 is vacuum or air, then
n1 = 1. Hence the refractive
index for any medium, n can be expressed as where
n =
λ 0
λ
λ 0 : 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
r
Air (na) Water (nw)
2.00 m
B
D
r i
i
C AB : apparent depth AC : actual depth = 2.00 m
where SF027
12
From the diagram, AD tan r = ABD ACD
tan i =
AB AD
C
By considering only small angles of
tan r ≈ sin r
r and i , hence
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 sin r AB
and
=
=
n1 na
=
nw
AC nw AB = 1.50 m
SF027
Other equation for absolute refractive index in term of depth is given by
n=
real depth apparent depth 13
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 10 8 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
n =
c
v n = 1.55 b. By using the equation below, thus
n =
λ 0
λ λ 0 = nλ SF027
λ 0 = 5.50 x10 −7 m @ 550 nm
14
SF027
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 n water = 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
15
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
SF027
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
f
P
P
f
C F
SF027
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 a way 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.
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.
SF027
18
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
i
i
C
D
F
r
From the figure,
BD
BCD
tan i =
BFD
tan θ =
CD BD
≈i
FD
≈ θ
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
SF027
then
19
BD = 2 FD CD CD = 2 FD BD
Because of AB is paraxial ray, thus point B is too close with pole P then
CD ≈ CP = r
FD ≈ FP = f
Therefore
r = 2 f or
f =
SF027
r
This relationship also valid for convex mirror.
2
20
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 3
O
C
I
2
P
C
P
F
2
2
O
I
F
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
SF027
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, Object
u.
Ray diagram
distance, u
Image property
u > r
C
I
O
P F
Front
Real Inverted Diminished Formed between point C and F.
back
O
F
u = r
P
C
Real Inverted Same size Formed at point C.
I Front
SF027
Object
back
23
Ray diagram
distance, u
Image property
f < u < r
I
C
P
O
F
Front
back
O
u = f
SF027
C
Front
Real Inverted Magnified Formed at a distance greater than CP.
F
Real Formed at infinity.
P
back
24
Object
Ray diagram
distance, u
Image property
u < f
F P
O
C
I
Front
Virtual Upright Magnified Formed at the back of the mirror
back
Linear (lateral) magnification of the spherical mirror, M is defined as the ratio between image height,
M =
Simulation
hi and object height, ho hi v where
ho
=−
v : image distance from pole
u u : object distance from pole
Negative sign indicates that the object and image are on opposite sides of the SF027
principal axis (refer to the real image), If
ho is positive, hi is negative.
25
1.2.7 Derivation of Spherical mirror equation
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 Figure below shows an object
B which is close to the pole P of the mirror.
B
α O
θ θ φ β I
C
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
By using thus
u
tan α =
(1) (2)
BOD, BCD
BD OD
; tan φ =
BD CD
(3) and
BID
; tanβ =
BD
By considering point B very close to the pole P, hence
tan α ≈ α ; tan φ ≈ φ ; tan β ≈ β OD ≈ OP = u ; CD ≈ CP = r ; ID ≈ IP = v then SF027
α =
BD u
; φ =
BD r
; β =
BD v
Substituting this value in eq. eq. (3)
26
ID
therefore
BD = 2 v r
BD BD
+
u
1 u
1 f
+
1
=
1
v
u
=
2
+
1
where
r
r = 2 f Equation (formula) of spherical mirror
v
Table below shows the sign convention for equation of spherical mirror . Positive sign (+)
Physical Quantity
Real object
Object distance, u Image distance, v
(in front of the mirror)
Negative sign (-)
Virtual object
(at the back of the mirror)
Real image
Virtual image
(same side of the object) (opposite side of the object)
Focal length, f
Concave mirror
Convex mirror
Linear magnification, M SF027
Upright (erect) image
Inverted image
27
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
f 1
= =
1
+
1
u v 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 are v − 30
M = −
M = 3 SF027
u
=−
(
)
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 I
30 cm
a. From the figure above,
u + v = 30 cm
(1) By using the equation of linear magnification, thus
M =
hi
=−
v
ho u v = −2u
(2)
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,
1 f 1
= =
1 u 1
+
1 v
+
1
f u (− 2u ) f = +20 cm and
f =
r 2
therefore
r = 40 cm
The type of spherical mirror is concave because the positive value of focal length.
SF027
30
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. 13 33, 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 f 2 r
= =
1 u 1
+ +
1
and
v 1
f =
r
2
u v v = −9.91 cm
From equation of linear magnification,
M =
hi
ho
=−
v
u
v hi = − ho u hi = 1.35 cm
SF027
31
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 P I 5.0 mm C
35.0 cm
F
u
3.20 m a. By applying the equation of linear magnification,
M =
hi ho
=−
v
− 35.0
u
0.5
=−
3.20 x10 2
u u = 4.57 cm
The position of the filament is 4.57 cm in front of the concave SF027
mirror.
32
b. By applying the equation of spherical mirror, thus
1 f 2
= =
1 u 1
+ +
1
f =
and
v 1
r 2
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. 1 169,Physics for scientists and engineers with modern ph ysics, Serway & Jewett,6th edition.
Ans. : 160 mm, -267 mm
SF027
33
1.3 Refraction of Spherical Surfaces
r forms an interface between two media with refractive indices n1 and n2. Figure below shows a spherical surface with radius,
i
n1
B
θ
α
n2 φ
β
P D
O
C
I
r u
v
The surface forms an image I of a point object above.
O as shown in figure
OB making an angle i with the normal and is refracted to ray BI making an angle θ where n1
normal. SF027
34
From the figure,
BOC
i = α + φ
(1)
BIC
φ = β + θ θ = φ − β
(2)
From the Snell’s law
n1 sin i = n2 sin θ
By using
BOD, BCD
tan α =
BD OD
and
BID
thus
BD
; tan φ =
CD
; tanβ =
BD ID
By considering point B very close to the pole P, hence
sin i ≈ i ; sin θ ≈ θ ; tan α ≈ α ; tan φ ≈ φ ; tan β ≈ 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
n1 u
+
n2 v
=
( n2 − n1 )
35
Equation of spherical refracting surface
r
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
M = SF027
+
hi ho
=−
n1v n2u 36
Sign convention for refraction :
Positive sign (+)
Physical Quantity
Real object
Object distance, u Image distance, v
Negative sign (-)
(in front of the refracting surface)
Virtual object
(at the back of the refracting surface)
Real image
Virtual image
(opposite side of the object)
(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
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
m 0 c 2 .0
O SF027
C
I 37
8 . 00 cm
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: n g =1.52, u=8.00 cm, r=+2.00 cm a. By applying the equation of spherical refracting surface,
n1 u na
u
+ +
n2 v n g
v
= =
(n2 − n1 )
r (n g − na )
r
v = +11.26 cm The image is 11.26 cm at the back of the con vex surface. b. By using the equation of linear magnification for refracting surface,
M = SF027
hi
ho
=−
n1v
n2 u
M = −
na v
n g u = −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, n g =n2=1.50, u=20.0 cm, r=5.0 cm
By using the equation of spherical refracting surface, thus
na 1.00 20.0
u
+
n g
+
=
v 1.50 v
=
( n g − na )
and
r (1.50 − 1.00 )
r = +5.0 cm
Convex surface (first surface)
The image is real and 30 cm at the back
5.0
v = +30 cm na
of the convex surface.
n g
O
I 1
C
P
r v = 30 cm
u = 20.0 cm SF027
39
b.
n g
na O
P
C
na Q I 2 30 cm 20 cm
First surface
I 1
Second surface From the figure above, the image I 1 formed by the first surface is in glass and 20 cm from the point Q of the second surface. I 1 acts as a virtual object for the second refraction surface and
n g =n1=1.50, na=n2=1.00, u=-20.0 cm, r=-5.0 cm Using
n g u
1.50
(−20.0)
+
+
na
v 1.00 v
= =
(na − n g ) r (1.00 − 1.50 )
Concave surface (second surface)
(−5.0 )
v = +5.71 cm
The image is real and 5.71cm at the back of the concave surface SF027
(5.71 cm from point Q as shown in figure above).
40
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).
SF027
41
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 PlanoPlano-convex (b) Diverging (Concave) lenses
SF027
Biconcave
PlanoPlano-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
r 1
r 1 O
C1
SF027
r 2
C2
O
C1
r 2
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 (r 1 and r 2) 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.
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
SF027
F1
O
F2
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) 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
1
2 3
u
v
SF027
45
(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.
SF027
2
Ray 1 - Parallel to the principal axis, after refraction by the lens, passes through the focal point (focus) F 2 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 F 1 of a diverging lens, after refraction by the lens the ray parallel to the principal axis. 46
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
I
F1
Front
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.
SF027
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, Object
u.
Ray diagram
distance, u
Image property
u > 2f
O 2F1
F1
F2
Front
back
2F2
u = 2f
SF027
O
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) 48
Object
Ray diagram
distance, u
Image property
f < u < 2f
2F1O
2F2
F2
F1
than 2 f at the back of the lens.
back
Front
u = f
2F1
F1
F2
Front
back
2F2
49
Object
Ray diagram
distance, u
Image property
u < f
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,
Simulation
M =
hi ho
=−
hi and object height, ho v where
u
v : image distance from optical centre u : object distance from optical centre
Negative sign indicates that when SF027
Real Formed at infinity.
O
SF027
Real Inverted Magnified Formed at a distance greater
inverted and
u and v are both positive, the image is
ho and hi have opposite signs.
50
1.5 Thin Lenses Formula and Lens maker’s Equation
Considering the ray diagram of refraction for 2 spherical surfaces as shown in figure below.
v1
u1
t − v1 r 2
r 1 A
n1
v2 D
n2 I 1
C1
O
n1 C2 P2
P1
B
E
t
SF027
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 = +r 1)
n1
u1
+
n2
+
n1
(t − v1 )
v1
=
( n2 − n1 )
=
( n1 − n2 )
(1)
r 1 Concave surface DE (r = -r 2) n2
v2
− r 2
Assuming the lens is very thin thus t
n2
− v1
+
n1
=
= 0,
( n1 − n2 )
− r 2 n − n n n2 = − 1 2 − 1 v1 − r 2 v2 v2
n2 v1 SF027
I 2
=
n − n − 2 1 v2 r 2 n1
(2) 52
By substituting eq. (2) into eq. (1), thus
n + 1 u1 v2 n1
n − n (n − n ) − 2 1 = 2 1 r 1 r 2 n1 n1 ( n2 − n1 ) ( n2 − n1 ) + = + u1
then
If
1
v2
+
r 1
r 2
n 1 1 = 2 − 1 + v2 n1 r 1 r 2 1
(3)
u1 u1 = ∞ and v2 = f hence eq. (3) becomes
n 1 1 = 2 − 1 + f n1 r 1 r 2 1
Lens maker ’s equation
where
f : focal length r 1 : radius of curvature of first refracting surface r 2 : radius of curvature of second refracting surface n1 : refractive index of the medium n2 : refractive index of the lens material
SF027
53
By equating eq. (3) with lens maker’s equation, hence
1 u1
+
1 v2
=
1 f
therefore in general,
1 f
=
1 u
+
1 v
Thin lens formula
Note :
If the medium is air (n1= will be
nair =1) thus the lens maker’s equation
1 1 = (n − 1) + f r 1 r 2 1
SF027
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= 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: r 1=+20
cm, r 2=+20 cm, n g =n2=1.52
a. Given the refractive index of water, nw = n1 By using the lens maker’s equation, thus
n g 1 1 = − 1 + f nw r 1 r 2 1
f = +70 cm
b. Given the refractive index of carbon disulfide, nc =
n1
By using the lens maker’s equation, thus
n 1 1 = g − 1 + f nc r 1 r 2 1
f = −148.18 cm
SF027
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
M =
hi ho
=−
v 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) 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, r 1=-22.0 cm, r 2=+18.5 cm
a. By applying the lens maker’s equation in air,
1 1 = (n − 1) + f r 1 r 2 1
f = +208 cm
SF027
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
M = −
SF027
v
u
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), M a
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 .
M a =
β α
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).
SF027
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,
tan α =
ho D
By making small angle approximation, we get
tan α ≈ α = SF027
ho D 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
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,
tan β =
hi
=
D
ho u
By making small angle approximation, we get
tan β ≈ β =
hi
D
=
ho
The properties of the image are
SF027
u u < f
Virtual, upright and magnified
61
The angular magnification in terms of D and f can be evaluated by derivation below. By applying the thin lens formula,
1 f
=
1
+
1
where
u v Df u= D + f
v = −D (1)
From the definition of angular magnification,
ho β u M a = = α ho D D M a = (2) u
By substituting eq. (1) into eq. (2), thus where
M a = SF027
D
f
+ 1 f : focal length
D : distance of distinct vision = 25 cm 62
The relationship between linear magnification,
with angular
magnification, M a
From the definition of angular magnification,
then
hi β D M a = = α ho D M a =
hi ho
= M
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
ho f β f Therefore, since M a = then M a = α ho D
M a =
D 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. The properties of final image are Virtual, inverted and magnified
u
L
f e
Objective lens
v >(f o+ f e )
F o' F e O F o f o
I 1 Eyepiece lens acts as a magnifier. The properties of first image are Real, inverted and magnified
SF027
I 2
v >2f o
64
The properties of the compound microscope are
The distance between two lenses, L > (f o+f e ) f o < f e The final image is I 2. The angular magnification formula is given by where
L D f : focal length of the eyepiece lens M a = − e f o f e f o : focal length of the objective lens D : distance of distinct v ision = 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.
f o
Parallel rays from object at infinity
f e
f e
F e' F o
F e
I 1 Objective lens
Eyepiece lens acts as a magnifier. The properties of first image are Real, inverted and diminished
I 2
v =f o
The properties of final image are Virtual, inverted and magnified SF027
v >(f o+ f e )
66
