PHYSICS
CHAPTER 1 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 light..
CHAPTER 1: Geometrical optics (5 Hours)
PHYSICS CHAPTER 1 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
1.1 Ref Refle lecti ction on at at a plane plane surfa surface ce (1 hour hour)) At the end of this cchapter, hapter, students should be able to: this chapter, State laws of reflection.
State the characteristics of image formed by a plane mirror. Sketch ray diagrams with minimum two rays.
PHYSICS
CHAPTER 1
1.1 Reflection at a plane surface 1.1 .1.1 .1 Refl flec ecti tion on of li lig ght
is defined as the return of all or part of a beam of light when it encounters the boundary between two media . There are two types of reflection due to the plane surface Specular (regular) reflection is the reflection of light from a smooth shiny surface as shown in Figure 1.1.
Figure 1.1
PHYSICS
CHAPTER 1 All the reflected rays are parallel to each another or move in the same direction. Diffuse reflection is the reflection of light from a rough surface such as papers, flowers, people as shown in Figure 1.2.
Figure 1.2 The reflected rays is sent out in a variety of directions. For both types of reflection, the laws of reflection are obeyed.
PHYSICS
CHAPTER 1
Laws of reflection state : The incident ray, the reflected ray and the normal normal all all lie in the same plane. plane . The angle of incidence, i equals the angle of reflection, r as shown in Figure 1.3.
i
r
i = r Picture 1.1
Plane surface Figure 1.3
Stimulation 1.1
PHYSICS
CHAPTER 1
1.1. 1. 1.2 2 Re Refl flec ecti tion on at a pla plane ne mi mirr rror or
Image formation by a plane mirror as shown in Figures 1.4a and 1.4b. Point object where
u : object distance v : image distance g : glancing angle
g r i A
g i
i u
v Figure 1.4a
A'
PHYSICS
CHAPTER 1
Vertical (extended) object
i ho
i r
Object
hi
r
u
Image
v
Figure 1.4b where
ho : object height hi : image height Stimulation 1.2
PHYSICS
CHAPTER 1
The characteristics of the image formed by the plane mirror are virtual image is seem to form by light coming from the image but light does not actually pass through the image .
would not appear on paper, screen or film placed at the location of the image .
upright or erect image laterally reverse right-hand side of the object becomes the left-hand side of the image.
the object distance, u equals the image distance, v
the same size where the linear magnification, m is given by
m=
Image height, hi Object height, ho
obey the laws of reflection. reflection .
=1 Picture 1.2
PHYSICS
CHAPTER 1
Example 1 : A women is 1.60 m tall and her eyes are 10 cm below the top of her head. She wishes to see the whole length of her body in a vertical plane mirror whilst she herself is standing vertically. a. Sketch and label a ray diagram to show the formation of women’s image. b. What is the minimum length of mirror that makes this possible? c. How far above the ground is the bottom of the mirror?
PHYSICS
CHAPTER 1
Solution : a. The ray diagram to show the formation of the women’s image is
A L
H (head)
0.10 m E(eyes)
h 1.60 m
B y
F(feet )
AL = LB =
1 2 1 2
HE EF
PHYSICS
CHAPTER 1
Solution : b. 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
Height of the women
h=
1
2
(1.60) = 0.80 m
b. The mirror can be placed on the wall with the bottom of the mirror is halved of the distance between the eyes and feet of the women. Therefore
y =
1
(1.60 − 0.10)
2 y = 0.75 m
PHYSICS
CHAPTER 1
Example 2 : A rose in a vase is placed 0.350 m in front of a plane mirror. Ahmad looks into the mirror from 1.00 m in front of it. How far away from Ahmad is the image of the rose? Solution : u 0.350 m
=
1.00 m x
u formedvby the plane mirror, From the characteristic of the image thus
v=u
v = 0.350 m
Therefore,
x = 1.00 + v
x = 1.00 + 0.350
1 350
PHYSICS
CHAPTER 1
Exercise 1.1 : 1.
Figure 1.5 The two mirrors in Figure 1.5 meet at a right angle. The beam be am of light in the vertical plane P strikes mirror 1 as shown. a. Determine the distance of the reflected light beam travels before striking mirror 2. b. Calculate the angle of reflection for the light beam after being reflected from mirror 2. ANS. : 1.95 m ; 40 to the mirror 2.
PHYSICS
CHAPTER 1
Exercise 1.1 : 2.
Figure 1.6 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 s hown in Figure 1.6. Determine x. ANS. : 0.81 m
PHYSICS
CHAPTER 1
Exercise 1.1 : 3.
Stand Standin ing g 2.0 2.00 0 m in fro front nt of a smal smalll vert vertic ical al mir mirro ror, r, you see see the the reflection of your belt buckle, which is 0.70 m below your eyes. a. What is the the vertical location of the mirror relative to the level of your eyes? b. What What is the the angle angle do your your eyes eyes make make with with the hori horizon zontal tal when you look at the buckle?
c. If you you now move move back backwar ward d until until you you are 6.0 6.0 m from from the the mirror, will you still see the buckle? Explain. ANS. : 35 cm below; 9.9 ; U think 4. You You are are 1.80 1.80 m tal talll and and stan stand d 3.0 3.00 0 m from from a pla plane ne mirro mirrorr tha thatt extends vertically upward from the floor. On the floor 1.50 m in front of the mirror is a small table, 0.80 high. What is the minimum height the mirror must have for you to be able to see the top of the table in the mirror? ANS. : 1.13 m
PHYSICS CHAPTER 1 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
1.2 Ref Refle lecti ction on at a spher spherica icall surface surface (1 hou hour) r) At the end of this cchapter, hapter, students should be able to: this chapter, Sketch and use ray diagrams to determine the characteristics of image formed by spherical mirrors. Use
1
f
1
1
= + = u
v
2
r
for real object only.
Use sign convention for focal length: + f for concave mirror and – f for convex mirror.
Sketch ray diagrams with minimum two rays. r = 2 f only applies to spherical mirror.
PHYSICS
CHAPTER 1
1.1 Reflection at a spherical surface 1.2.1 Spherical mi mirror
is defined as a reflecting surface that is part of a sphere. sphere . There are two types of spherical mirror. It is convex (curving outwards) and concave (curving inwards) mirror. Figures 1.7a and 1.7b show the shape of concave and convex mirrors. (a)Con (a) Concav cave e ( Converging Converging)) (b) Convex ( Diverging Diverging)) mirror mirror imaginary sphere A C
r
P
B Picture 1.3
Figure 1.7a
A
silver layer
C
P
r B
reflecting surface
Figure 1.7b
PHYSICS
CHAPTER 1
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. part .
Radius of curvature, r is defined as the radius of the sphere of which a curved mirror forms a part. 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 the mirror . AB is called the aperture of the mirror.
PHYSICS
CHAPTER 1
1.2. 1. 2.2 2 Fo Foca call poi point nt and and fo foca call len lengt gth, h, f
Consider the ray diagram for a concave and convex mirrors as shown in Figures 1.8a and 1.8b. Incident Incident rays rays C F
f
P
P
f
C F
Figure 1.8a Figure 1.8b 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.
PHYSICS
CHAPTER 1
Focal point or focus, F For concave mirror – is defined as a point where the incident parallel rays converge after reflection on the mirror . mirror . Its focal point is real (principal). (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. reflection . Its focal point is virtual virtual.. Focal length, f is defined as the distance between the focal point (focus) F F and pole P of the spherical spherical mirror mirror . The paraxial rays is defined as the rays that are near to and almost parallel to the principal axis. axis .
PHYSICS
CHAPTER 1
1.2.3 1.2. 3 Re Rela latio tions nshi hip p bet betwe ween en fo foca call len lengt gth, h, f and radius of curvature, r
Consider a ray AB parallel to the principal axis of concave mirror as shown in Figure 1.9. A
incident ray
i C
B
i
i
θ D
F
r
f
Figure 1.9
P
PHYSICS
CHAPTER 1
From the Figure 1.9, BD tan i = BCD
tan θ =
BFD
CD BD
FD
≈i
≈ θ
Taken the angles are << small by considering the ray AB is paraxial ray.
By using an isosceles triangle CBF, thus the angle
θ = 2i
then
BD = 2 FD CD CD = 2FD
BD
Because of AB is paraxial ray, thus point B is too close with pole P then CD ≈ CP = r
FD ≈ FP = f
θ is given by
Therefore
r = 2 f
OR
f
=
r 2
This relationship also valid for convex mirror.
PHYSICS
CHAPTER 1
1.2. 1. 2.4 4 Ra Ray y dia diagr gram ams s for for sp sphe heri rica call mir mirro rors rs is defined as the simple graphical method to indicate the positions of the object and image in a system sy stem of mirrors or lenses.. lenses Figures 1.10a and 1.10b show 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 I P C P F O I F 2
O
3 1 Figure 1.10a
Figure 1.10b
PHYSICS
CHAPTER 1
Ray 1 - Parallel to principal axis, after reflection, passes through the focal point (focus) F of a concave At least any mirror or appears to come from the focal point F two rays of a convex mirror. for drawing Ray 2 - Passes or directed towards towards focal focal point point F reflected the ray parallel to principal axis. diagram. Ray 3 - Passes or directed towards centre of curvature C, reflected back along the same path. Images formed by a convex mirror Figure 1.11 shows the graphical method of locating an image formed by a convex mirror.
C
P
O
I u
Figure 1.11
front
v
F Picture 1.4
PHYSICS
CHAPTER 1
The characteristics of the image formed are virtual upright diminished (smaller than the object) formed at the back of the mirror (behind the mirror) Object position → any position in front of the convex mirror. Convex mirror always being used as a driving mirror because mirror because it has a wide field of view and providing an upright image. image. Images formed by a concave mirror Concave mirror can be used as a shaving and makeup mirrors because it provides an upright and virtual images. images . Table 1.1 shows the ray diagrams of locating an image formed by a concave mirror for various object distance, u.
PHYSICS
CHAPTER 1
Object distance, u
Ray diagram
Image characteristic
u > r
O
C
I F
P
Front
back
O
u = r
F C
P
I Front
back
Real Inverted Diminished Formed between point C and F.
Real Inverted Same size Formed at point C.
Object PH Y SICS distance, u
Ray diagram
CHAPTEImRag1e characteristic
f < u < r
I C
O F
P
Front
back
O
u = f
C
Front
Real Inverted Magnified Formed at a distance greater than CP.
F
P
back
Real or virtual Formed at infinity.
PHObject YSICS
Ray diagram
distance, u
ImR age1characteristic CHAPTE
u < f
F
O
C
P
Front
Virtual Upright Magnified Formed at the back of the mirror
I back
Table 1.1 Linear (lateral) magnification of the spherical mirror, m is defined as the ratio between image height, hi and object height, ho
m=
hi ho
=
v u
where
v : image distance from the pole u : object distance from the pole
PHYSICS
CHAPTER 1
1.2. 1. 2.5 5 De Deri riva vati tion on of of Sphe Spheri rica call mirro mirrorr equa equatio tion n
Figure 1.12 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.
B
θ
α O
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
By using thus
tan α = Figure 1.12
(1) (2)
(3)
BOD, BCD
BD OD
; tan φ =
BD CD
and
BID
; tanβ =
BD ID
PHYSICS
CHAPTER 1
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
α =
therefore
BD
; φ =
u
BD u
1
f
BD r
; β =
BD v
Substituting this value in eq. (3)
BD + = 2 v r 1 1 2 + = where r = 2 f BD
u
v
r
1
1
2
u
v
r
= + =
Spherical mirror’s equation
PHYSICS
CHAPTER 1
Table 1.2 shows the sign convention for spherical mirror’s equation .
Physical Quantity Object distance, u Image distance, v Focal length, f
Positive sign (+)
Real object
(in (i n fr front of of th the mir mirrror)
Real image
Negative sign (-)
Virtual object
(att the (a the back of of the the mir irrror)
Virtual image
(same (sam e side of the object object)) (oppos (opposite ite side side of the object) object)
Concave mirror
Convex mirror
Table 1.2
Note: Real image is formed by the actual light rays that pass through the image. image . Real image can be projected on the screen. screen .
PHYSICS
CHAPTER 1
Example 3 : A dentist uses a small mirror attached to a thin rod to examine one of your teeth. When the tooth is 1.20 cm in front of the mirror, the image it forms is 9.25 cm behind the mirror. Determine a. the focal length of the mirror and state the type of the mirror used, b. the magnification of the image. Solution :
u = +1.20 cm; v = −9.25 cm
a. By applying the mirror’s equation, thus
1
f
1
1
1
u
v
f 1.20
= +
=
1
+
1
( − 9.25)
f = +1.38 cm b. By using the magnification formula, thus
m=
v u
m=
9.25 1.20
= 7.71
(Concave mirror)
PHYSICS
CHAPTER 1
Example 4 : An upright image is formed 20.5 cm from the real object by using the spherical mirror. The image’s height is one fourth of object’s height. 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. c. Sketch and label a ray diagram to show the formation of the image. Solution : hi
= 0.25ho
Spherical
u O
mirror
20.5 cm
v I
PHYSICS
CHAPTER 1
=
0.25ho Solution : hi a. From the figure,
u+ v
= 20.5
(1)
By using the equation of linear magnification, thus
m=
hi ho
=
v
0.25ho
u
ho
v
= u v = 0.25u
(2)
By substituting eq. (2) into eq. (1), hence
u + 0.25u = 20.5 u = 16.4 cm The mirror should be placed 16.4 cm in front of the object object..
PHYSICS
CHAPTER 1
=
0.25ho Solution : hi b. By using the mirror’s equation, thus
1
f 1 f 1
1
1
u 1
v
u
( − 0.25u )
= + = + =
1
1
f 16.4
+
1
( − 0.25(16.4 ) )
f = −5.47 cm
r
f = 2 r = 2( 5.47 ) = 10.9 cm and
The type of spherical mirror is convex because the negative value of focal length.
PHYSICS
CHAPTER 1
=
0.25ho Solution : hi c. The ray diagram is shown below.
C
P
O
I front
F
back
PHYSICS
CHAPTER 1
Example 5 : A person of 1.60 m height stands 0.60 m from a surface of a hanging shiny globe in a garden. a. If the diameter of the globe is 18 cm, where is the image of the person relative to the surface of the globe? b. How large is the person’s image? c. State the characteristics of the person’s image. Solution :
ho
= 1.60 m; u = 0.60 m u ho
PHYSICS
CHAPTER 1
=
=
1.60 m; u 0.60 m Solution : ho a. Given d 0.18 m The radius of curvature of the globe’s surface (convex surface) is given by
=
r = −
0.18 2
= −0.09 m
By applying the mirror’s equation, hence
2
r 2
− 0.09
1
1
u
v
= + =
1 0.60
+
1 v
v = −0.042 m
(behind the globe’s surface)
PHYSICS
CHAPTER 1
Solution : ho
= 1.60 m; u = 0.60 m
b. By applying the magnification formula, thus
m=
hi
ho hi
1.60 hi
=
=
v u 0.042
0.60 = 0.112 m OR 11.2 cm
c. The characteristics of the person’s image are virtual upright diminished formed behind the reflecting surface.
PHYSICS
CHAPTER 1
Example 6 : A shaving or makeup mirror forms an image of a light bulb on a wall of a bathroom that is 3.50 m from the mirror. The height of the bulb is 8.0 mm and the height of its image is 40 cm. a. Sketch a labeled ray diagram d iagram to show the formation of the bulb’s image. b. Calculate i. the position of the bulb from the pole of the mirror, ii. the focal the v length homirror. 3.50ofm; 8.0 10 −3 m; hi 40 Solution : a. The ray diagram of the bulb is
=
=
×
=
O
I 8.0 mm C
40 cm 3 50
P F
u
× 10−2 m
PHYSICS
CHAPTER 1
Solution :
u = 3.50 m; ho
= 8.0 × 10−3 m; hi = 40 × 10 −2 m
b. i. By applying the magnification formula, thus
m=
hi ho
=
40 × 10 −2
v u 3.50
= 8.0 × 10 u u = 0.07 m OR 7.0 cm −3
The position of the bulb is 7.0 cm in front of the mirror. ii. By applying the mirror’s equation, thus
1
f 1
1
1
u
v
= + =
1
+
1
f 0.07 3.50 f = 0 0687 m OR 6 87 cm
PHYSICS
CHAPTER 1
Exercise 1.2 : 1.
a. A conc concave ave mirr mirror or form forms s an inve invert rted ed imag image e four four time times s large larger r than the object. Calculate the focal length of the mirror, assuming the distance between object and image is 0.600 m. b. A convex convex mirror mirror forms forms a virtual virtual image half the size size of the object. Assuming the distance between image and object is 20.0 cm, determine the radius of curvature of the mirror. ANS. : 160 mm ; 267 mm 2. a. A 1.7 1.74 4 m tall tall shop shopp per in a dep depar arttmen ment st store ore is is 5.1 5.19 9 m from from a security mirror. The shopper notices that his image in the mirror appears to be only 16.3 cm tall. i. Is the shopper’s image upright or inverted? Explain. ii. Determine the radius of curvature of o f the mirror. b. A concave concave mirr mirror or of a focal focal leng length th 36 cm cm produc produces es an image whose distance from the mirror is one third of the object distance. Calculate the object and image distances. ANS. : u think, 1.07 m ; 144 cm, 48 cm
PHYSICS CHAPTER 1 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
1.3 Refraction Refraction at a pla plane ne and and spher spherical ical surfa surfaces ces (1 hour) At the end of this cchapter, hapter, students should be able to: this chapter, State and use the laws of refraction (Snell’s Law) for layers of materials with different densities. Apply
n1 u
+
n2 v
=
( n2 − n1 ) r
for spherical surface.
PHYSICS
CHAPTER 1
1.3 Refraction at a plane and spherical surfaces 1.3. 1. 3.1 1 Re Refr frac acti tion on at a pla plane ne su surf rfac ace e
Refraction 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. plane . For two given media, Snell’s law states
sin i sin r where
=
n2 n1
= constant
OR
n1 sin i = n2 sin r
n1 : refractive index of the medium 1 (Medium containing the incident ray) n2 : refractive index of the medium 2 (Medium containing the refracted ray) angle of refraction
PHYSICS
CHAPTER 1
Examples for refraction of light ray travels from one medium to another medium can be shown in Figures 1.13a and 1.13b.
(a) n1
< n2
(b) n1
(Medium 1 is less dense medium 2) Stimulation 1.3
> n2
(Medium 1 is denser than medium 2) Stimulation 1.4 Incident ray
Incident ray
i
i n1
n1 n2 r Refracted ray
Figure 1.13a The light ray is bent toward the normal,, thus r i normal
<
n2 r Figure 1.13b Refracted ray The light ray is bent away from the normal, normal, thus r i
>
PHYSICS
CHAPTER 1
Refractive index (index of refraction), n sin i for the two given is defined as the constant ratio media.. media sin r
The value of refractive index depends on the type of medium and the colour of the light. light . It is dimensionless and its value greater than 1. 1. Consider the light ray travels from f rom medium 1 into medium 2, the refractive index can be denoted by
1
n2
=
(Medium containing the incident ray)
velocity of light in medium 1 velocity of light in medium 2 (Medium containing the refracted ray)
=
v1 v2
PHYSICS
CHAPTER 1
Absolute refractive index, n (for the incident ray travels from vacuum or air into air into the medium medium)) is given by
n=
velocity of light in vacuum velocity of light in medium
=
c v
Table 1.3 shows the refractive indices for common substances. Substance Refractive index, n Table 1.3 Solids (If the density 2.42 Diamond of medium is 1.66 Flint glass 1.52 Crown glass greater hence greater hence 1.46 Fused quartz (glass) the refractive 1.31 Ice index is also Liquids greater ) Benzene 1.50 Ethyl alcohol 1.36 Water 1.33 Gases Carbon dioxide 1.00045 Air 1.000293
PHYSICS
CHAPTER 1
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 each medium is given by
= f λ 1 and v2 = f λ 2 c f λ 1 where v1 = =
v1 then
v1 v2
f λ 2
n1
and
v2
=
c n2
PHYSICS
CHAPTER 1
c n1 = λ 1 c λ 2 n2
n1λ 1 = n2λ 2 (Refractive index is inversely proportional to the wavelength)
If medium 1 is vacuum or air, then n1 = 1. Therefore the refractive index for any medium, n can be expressed as n
where
=
λ 0 λ
λ 0 : wavelength of light in vacuum λ : wavelength of light in medium
Picture 1.5
Picture 1.6
PHYSICS
CHAPTER 1
Example 7 : A fifty cent coin is at the bottom of a swimming pool of depth 3.00 m. The refractive index of air and water are 1.00 and 1.33 respectively. Determine the apparent depth of the coin. Solution : na 1.00 ; nw 1.33
=
=
Air (na)
A
Water (nw)
3.00 m
B
r r
i
D
i
C where
AB : apparent depth AC : actual depth = 3.00 m
PHYSICS
CHAPTER 1
=
Solution : na 1.00 ; nw From the diagram, ABD
ACD
= 1.33
tan r =
AD
tan i =
AD
AB
AC
By considering only small angles of r and i , thus
tan r ≈ sin r and tan i ≈ sin i
AD tan i sin i AC AB = = = tan r sin r AD AC AB
then
PHYSICS
CHAPTER 1
=
1.00 ; nw Solution : na From the Snell’s law,
sin i sin r AB
AC AB
=
= =
n2 n1 na
= 1.33
=
na nw
nw 1.00
3.00 1.33 AB = 2.26 m
Note : (Important) Other equation for absolute refractive index in term of depth is given by
n=
n2 n1
=
real depth apparent depth
PHYSICS
CHAPTER 1
Example 8 : A pond with a total depth (ice + water) of 4.00 m is covered by a transparent layer of ice of thickness 0.32 m. Determine the time required for light to travel vertically from the surface of the ice to the bottom of the pond. The refractive index of ice and water are 1.31 and 1.33 respectively. (Given the speed of light in vacuum is 3.00 Solution : ni 1.31 ; nw 1.33
=
=
Ice (ni)
Water (nw)
hi
hw hw
×
108 m s-1.)
= 0.32 m
= 4.00 − 0.32 = 3.68 m
4.00 m
PHYSICS
CHAPTER 1
=
=
Solution : ni 1.31 ; nw 1.33 The speed of light in ice and water are
ni
nw
= =
c
vi c vw
1.31 =
3.00 × 108
vi vi = 2.29 × 108 m s −1 8 × 3.00 10 1.33 = vw 8 −1 vw = 2.26 × 10 m s
Since the light propagates in ice and water at constant speed thus
v=
s
t =
t
s
v
Therefore the time required is given by
t = t i
+ t w
t =
hi vi
+
hw vw
0.32 3.68 = + 8 8 2.29 ×10 2.26 ×10
t = 1 77 ×10 −8 s
PHYSICS
CHAPTER 1
1.3. 1. 3.2 2 Re Refr frac actio tion n at at a sp sphe heri rica call sur surfa face ce
Figure 1.14 shows a spherical surface with radius, r forms an interface between two media with refractive indices n1 and n2.
i
n1
B
θ
α
n2
φ
β
P D
O
C
I
r u
v
Figure 1.14 The surface forms an image I of a point object O.
The incident ray OB making an angle i with the normal and is refracted to ray BI making an angle θ where n1 < n2.
Point C is the centre of curvature of the spherical surface and
PHYSICS
CHAPTER 1
From the figure, BOC BIC
i = α + φ
(1)
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
; tan φ =
thus
BD 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 ; tanβ ≈ β OD ≈ OP = u ; CD ≈ CP = r ; ID ≈ IP = v
then Snell’s law can be written as
n1i = n2θ
(3)
PHYSICS
CHAPTER 1
By substituting eq. (1) and (2) into eq. (3), thus
then
n1 u
n1 (α + φ ) = n2 (φ − β ) n1α + n2 β = (n2 − n1 )φ
BD BD BD n1 + n2 = (n2 − n1 ) u v r +
where
n2 v
=
(n2 − n1 )
r
Equation of spherical refracting surface
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)
PHYSICS
CHAPTER 1
Note : If the refracting surface is flat (plane) :
r = ∞
then
n1 u
+
n2 v
=0
The equation (formula) of linear magnification for refraction by the spherical surface is given by
m=
hi ho
=
n1v n2u
PHYSICS
CHAPTER 1
Table 1.4 shows the sign convention for refraction or thin lenses:: lenses
Physical Quantity Object distance u ,
Positive sign (+)
Negative sign (-)
Real object
Virtual object
(in front of the refracting surface)
Real image
Image distance, v
(opposite side of the object)
Focal length, f
Converging lens
Radius of curvature, r
(at the back of the refracting surface)
Virtual image
(same side of the object)
Diverging lens
Centre of curvature is located in more dense medium
Centre of curvature is located in less dense medium
(convex surface)
(concave surface)
Table 1.4
PHYSICS
CHAPTER 1
Example 9 : A cylindrical glass rod in air has a refractive index of 1.52. One end is ground to a hemispherical surface with radius, r =3.00 cm as shown in Figure 1.15. air
glass
P
c m 0 0 0 C 3.
O
I
10 .0 cm
Figure 1.15 Calculate, a. the position of the image for a small object on the axis of the rod, 10.0 cm to the left of the pole as shown in figure. b. the linear magnification. (Given the refractive index of air , na= 1.00)
PHYSICS
CHAPTER 1
=
=
=+
ng
(n − n )
Solution : ng 1.52 ; u 10.0 cm; r 3.00 cm a. By using the equation of spherical refracting surface, thus
na
+
u v 1.00 1.52 10.0
+
v
= =
g
a
r (1.52 − 1.00)
+ 3.00
v = +20.7 cm
The image is 20.7 cm at the back of the convex surface. b. The linear magnification of the image is given by
m=
n1v n2u
m=
na v ng u
(1.00 ) ( 20.7 ) m= (1.52 ) (10.0 )
m = 1 36
PHYSICS
CHAPTER 1
Example 10 : Figure 1.16 shows an object O placed at a distance 20.0 cm from the surface P of a glass sphere of radius 5.0 cm and refractive index of 1.63. Glass sphere air
O
P 5.0 cm
20 .0 cm
Figure 1.16 Determine a. the position of the image formed by b y the surface P of the glass sphere, b. the position of the final image formed by the glass sphere. (Given the refractive index of air , na= 1.00)
PHYSICS
CHAPTER 1
=
=
=+
Solution : ng 1.63 ; u 20.0 cm; r 5.0 cm a. By using the equation of spherical refracting surface, thus
na
+
ng
u v 1.00 1.63 20.0
+
v
= =
(n − n ) g
a
r (1.63 − 1.00)
+ 5.0
v = +21.5 cm
The image is 21.5 cm at the back of the first surface P. OR
ng
na O
C
P u = 20.0 cm
r
v = 21.5 cm
I1
PHYSICS
CHAPTER 1
Solution : b.
ng
na O
P
C
na
Q I2
First surface
21.5 cm 11.5 cm
I1
Second surface From the figure above, the image I1 formed by the first surface P is in the the glass and 11.5 cm from the second surface Q. I1 acts as a virtual object for the second surface and
= ng = 1.63 ; n2 = na = 1.00; u = −11.5 cm; r = +5.00 cm
n1
Centre of curvature is located in
PHYSICS
CHAPTER 1
Solution : b. By using
ng u 1.63
− 11.5
+
+
na
v 1.00 v
= =
(n − n ) a
g
r (1.00 − 1.63)
+ 5.0
v = +3.74 cm
The image is is real and 3.74 3.74 cm at the the back of the second surface Q.
PHYSICS
CHAPTER 1
Exercise 1.3 : 1.
A stu stude dent nt wish wishes es to dete determ rmin ine e the the dept depth h of of a swim swimmi ming ng pool pool x = 5.50 m) and then filled with water by measuring the width ( x noting that the bottom edge of the pool is just visible at an angle of 14.0 ° above the horizontal as shown in Figure 1.17.
Figure 1.17 Calculate the depth of the pool. (Given nwater = 1.33 and n air = 1.00) ANS. : 5.16 m
PHYSICS
CHAPTER 1
Exercise 1.3 : 2.
A sma smallll stri strip p of of pap paper er is past pasted ed on one one sid side e of of a glas glass s sph spher ere e of radius 5 cm. The paper is then view from the opposite surface of the sphere. Determine the position of the image. (Given the refractive index of glass =1.52 and the refractive index of air =1.00)
ANS. : 20.83 cm in front of the 2 nd refracting surface. 3.
A poi point nt sour source ce of light light is plac placed ed at a dis dista tanc nce e of of 25. 25.0 0 cm cm fro from m the centre of a glass sphere of radius 10 cm. Determine the image position of the source. (Given the refractive index of glass =1.52 and the refractive index of air =1.00)
ANS. : 25.2 cm at the back of the 2 nd refracting surface.
PHYSICS CHAPTER 1 Learning Outcome: s c i s y h p / y m . .. u d e . . ki r t a m . . h p m . .k w w w
1.4 1. 4 Th Thin in le lens nses es (2 ho hour urs) s) At the end of this t his chapter, students should be able to: this Sketch and use ray diagrams to determine the characteristics of image formed by diverging and converging lenses. Use equation stated in 1.3 to derive thin lens formula,
1
u
1
1
v
f
+ =
for real object only.
Use lensmaker’s equation:
1 1 = ( n − 1) + f r 1 r 2 1
Use the thin lens formula for a combination of
PHYSICS
CHAPTER 1
1.4 Thin lenses
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. surfaces . There are two types of thin lenses. It is converging and diverging lenses. Figures 1.18a and 1.18b show the various types of thin lenses, both converging and diverging. (a) Converging (Convex) lenses
r 1 (+ve)
r 2 r 1 (+ve) (+ve) Biconvex
r 2 (
Plano-convex Figure 1.18a
r 1 ) (+ve)
r 2 ( ve)
Convex meniscus
PHYSICS
CHAPTER 1
(b) Diverging (Concave) lenses
r 1 ( ve)
r 2 ( ve)
Biconcave
r 1 ( ve)
r 2 (
Plano-concave Figure 1.18b
r 2 ( ve)
r 1 ) (+ve)
Concave meniscus
1.4. 1. 4.1 1 Ter erm ms of of thi thin n len lense ses s
Figures 1.19 show the shape of converging (convex) and diverging (concave) lenses.
(a) Converging lens
(b) Diverging lens
r 1
r 1 C1
O
r 2
C2
C1
Figure 1.19
O
r 2
C2
PHYSICS
CHAPTER 1
Centre of curvature (point C 1 and C2)
is defined as the centre of the sphere of which the surface of the lens is a part. 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. part . Principal (Optical) axis is defined as the line joining the two centres of curvature of a lens. lens. Optical centre (point O) is defined as the point at which any rays entering the lens pass without deviation. deviation .
PHYSICS
CHAPTER 1
1.4. 1. 4.2 2 Fo Foca call poi point nt and and fo foca call len lengt gth, h, f
Consider the ray diagrams for converging and diverging lenses as shown in Figures 1.20a and 1.20b.
O
F1
f
F2
f
O
F1
f
F2
f
Figure 1.20a Figure 1.20b From the figures, Points F and F represent the focus of the lenses. 1 2
Distance f represents the focal length of the lenses.
PHYSICS
CHAPTER 1
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. 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. lens. Its focus is virtual.
Focal length ( f ) is defined as the distance between the focus F and the optical centre O of the lens. lens.
PHYSICS
CHAPTER 1
1.4. 1. 4.3 3 Ra Ray y dia diagr gram am fo forr thi thin n len lense ses s
Figures 1.21a and 1.21b show the graphical method of locating an image formed by a converging (convex) and diverging (concave) lenses. (a) Converging (convex) lens 1 2 3
O
F2
I
F1 2 3
u
v Figure 1.21a
1
PHYSICS
CHAPTER 1
(b) Diverging (concave) lens 1 2
O
3
F2
u
At least any two rays for drawing the ray diagram.
1
3
I
F1
v 2
Figure 1.21b 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 F 1 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
PHYSICS
CHAPTER 1
Images formed by a diverging lens Figure 1.22 shows the graphical method of locating an image formed by a diverging lens.
O
F2 Front
I
F1 back
Figure 1.22
The characteristics of the image formed are virtual upright diminished (smaller than the object) formed in front of the lens. lens . Object position → any position in front of the diverging lens.
PHYSICS
CHAPTER 1
Images formed by a converging lens Table 1.5 shows the ray diagrams of locating an image formed by a converging lens for various object distance, u. Object distance, u
Ray diagram
Image characteristic
Real Inverted Diminished Formed between point F2 and 2F2.
u > 2 f
I O 2F1
F1 Front
F2 back
2F2
(at the back of the lens)
PHYSICS
CHAPTER 1
Object distance, u
Ray diagram
Image characteristic
Real Inverted Same size Formed at point 2F2. (at the back of the lens)
u = 2 f
O
2F2 F1
F2
Front
back
2F1
I
Real Inverted Magnified Formed at a distance greater than 2 f at the back of the lens.
I f < u < 2 f 2F1 O
F1
F2
Front
back
2F2
Pdistance, HObject YSICuS
Ray diagram
Image characteristic CHAPTE R1 Real or virtual Formed at infinity.
u = f
O 2F1
F1
F2
Front
back
2F2
Virtual Upright Magnified Formed in front of the lens.
u < f
I
2F1
F1 O
F2
Front
back
2F2
Stimulation 1.5
PHYSICS
CHAPTER 1
1.4.4 1.4. 4 Th Thin in len lens s form formul ula, a, len lens s make maker’ r’s s and and line linear ar magnification equations Thin lens formula and lens maker’s equation Considering the ray diagram of refraction for two spherical surfaces as shown in Figure 1.23.
u1
r 1
u2
v1
r 2
v2
D
A
n1
= t − v1
n2 C1
O
Figure 1.23
I1
n1 C2 P2
P1
B
t
E
I2
PHYSICS
CHAPTER 1
By using the equation of spherical refracting surface, the refraction by first surface AB and second surface DE are given by
Surface AB (r = +r 1)
n1
u1
+
n2 v1
=
Surface DE (r = +r 2)
n2
+
n1
( t − v1 ) v2
=
( n2 − n1 )
+ r 1
(1)
( n2 − n1 )
+ r 2
Assuming the lens is very thin thus t = 0,
n2
− v1
+
n1
v2 n2
v1
=
( n2 − n1 )
r 2
n2 − n1 = − v2 r 2 n1
(2)
PHYSICS
CHAPTER 1
By substituting eq. (2) into eq. (1), ( 1), thus
n1 n2 − n1 (n2 − n1 ) = + − u1 v2 r 2 r 1 n1 n1 (n2 − n1 ) (n2 − n1 ) + = + n1
u1 1
u1
+
v2 1
v2
r 1
r 2 1
n2 1 + = − 1 n r r 1 1 2
(3)
If u1 = ∞ and v2 = f thus eq. (3) becomes
n2 1 1 = − 1 + f n1 r 1 r 2 1
where f : focal length
Lens maker’s equation
r 1 : radius of curvature for 1stndrefracting surface r 2 : radius of curvature for 2 refracting surface n1 : refractive index of the medium refractive index of the lens material
PHYSICS
CHAPTER 1
By equating eq. (3) and the lens maker’s equation, thus
1
u1
+
1
v2
=
1
f
therefore in general,
1 f
1
1
u
v
= +
Thin lens formula
Note :
If the medium is air air ((n1= nair =1) thus the lens maker’s equation can be written as
1 1 = ( n − 1) + f r 1 r 2 1
where
n : refractive index of the lens material
For thin lenses and lens maker’s equations, use the sign convention for refraction refraction..
PHYSICS
CHAPTER 1
Linear magnification, m
is defined as the ratio between image height, hi and object height, ho.
m=
where
1
Since
f
written as
hi ho
=
v u
v : image distance from optical centre u : object distance from optical centre
1
1
u 1
v 1
= +
the linear magnification equation can be
1
= + ×v f u v v v = +1 f
u
m=
v −1 f
PHYSICS
CHAPTER 1
Example 11 : A person of height 1.75 m is standing 2.50 m in from of a camera. The camera uses a thin biconvex lens of radii of curvature 7.69 mm. The lens made from the crown glass of refractive index 1.52. a. Calculate the focal length of the lens. b. Sketch a labelled ray diagram to show the formation of the image. c. Determine the position of the image and its height. d. State the characteristics of the image.
= 1.75 m; u = 2.50−3m; n = 1.52; r 1 = r 2 = +7.69 × 10 m
Solution : ho
a. By applying the lens maker’s equation in air, thus
1 1 = ( n − 1) + f r 1 r 2 1
PHYSICS
CHAPTER 1
= 1.75 m; u = 2.50−3m; n = 1.52; r 1 = r 2 = +7.69 × 10 m 1 1 1 = (1.52 − 1) −3 + − 3 f 7.69 × 10 7.69 × 10 f = +7.39 × 10 −3 m
Solution : ho a.
b. The ray diagram for the case is
I O 2F1
F1
F2
Front
back
2F2
PHYSICS
CHAPTER 1
Solution : c. The position of the image formed is
1
f
1
= + u
1
1
+ 7.39 × 10
v
1
1
= + 2.50 v v = 7.41× 10 −3 m
−3
(at the back of the lens) By using the linear magnification equation, thus
m=
hi ho
=
v
hi
u
1.75 hi
=
7.41× 10
−3
2.50 = 5.19 × 10 −3 m OR 5.19 mm
d. The characteristics of the image are
real
inverted
diminished
formed at the back of the lens
PHYSICS
CHAPTER 1
Example 12 : A thin plano-convex lens is made of glass of refractive index 1.66. When an object is set up 10 cm from the lens, a virtual image ten times its size is formed. Determine a. the focal length of the lens, b. the radius of curvature of the convex surface.
=
=
=
Solution : n 1.66; u 10 cm; m 10 a. By applying the linear magnification equation for thin lens, thus
m=
v u
v = 10u
= 10
By using the thin lens formula, for mula, thus
1
f 1 f
1
1
u 1
v
u
( − 10u )
= + = +
Virtual image
1
PHYSICS Solution : a.
CHAPTER 1 n = 1.66; u = 10 cm; m = 10 1 1 1
=
f 10
+
( − 10) (10)
f = +11.1 cm b. Since the thin lens is plano-convex thus r 2 Therefore
1 1 = ( n − 1) + f r 1 r 2 1 1 1 = (1.66 − 1) + 11.1 r 1 ∞ 1
r 1 = +7.33 cm
=∞
PHYSICS
CHAPTER 1
Example 13 : The radii of curvature of the faces of a thin concave meniscus lens of material of refractive index 3/2 are 20 cm and 10 cm. What is the focal length of lens a. in air, b. when completely immersed in water of refractive index 4/3? Solution :
n2
= 3/ 2
r 1 = +20 cm
r 2
= −10 cm
a. By applying the lens maker’s equation in air,
1 1 and n = n = 3 / 2 = ( n − 1) + 2 f r 1 r 2 1
PHYSICS
CHAPTER 1
= 3/ 2 1 3 1 1 = − 1 + f 2 + 20 ( − 10 ) f = −40 cm n1 = 4 / 3
Solution : n2 a.
b. Given
By using the general lens maker’s equation, therefore
n2 1 1 = − 1 + f n1 r 1 r 2 1 1 ( 32 ) 1 = 4 − 1 + f ( 3 ) + 20 ( − 10 ) 1
f = −160 cm
PHYSICS
CHAPTER 1
1.4 .4.5 .5 Com omb bin inat atio ion n of of len lense ses s
Picture 1.7 Picture 1.8
Many optical instruments, such as microscopes and telescopes,, use two converging lenses together to produce telescopes an image. In both instruments, the 1 st lens (closest to the object )is called the objective and the 2nd lens (closest to the eye) eye) is referred to as the eyepiece or ocular . The image formed by the 1st lens is treated as the object for the 2nd lens and the final image is the image formed by the 2nd lens.. lens The position of the final image in a two lenses system can be determined by applying the thin lens formula to each lens separately.. separately The overall magnification of a two lenses system is the product of the magnifications of the separate lenses . where
m = m1m2
m : overall magnification m1 : magnification due to the 1st lens m2 : magnification due to the 2 nd lens
PHYSICS
CHAPTER 1
Example 14 : The objective and eyepiece of the compound microscope are both converging lenses and have focal lengths of 15.0 mm and 25.5 mm respectively. A distance of 61.0 mm separates se parates the lenses. The microscope is being used to examine a sample placed 24.1 mm in front of the objective. a. Determine i. the position of the final image, ii. the overall magnification of the microscope. b. State the characteristics of the final image.
= +15.0 mm; f 2 = +25.5 mm; d = 61.0 mm; u1 = 24.1 mm
Solution : f 1
d
f 1 O
u1
F1
f 1 F1
f 2 F2
f 2 F2
PHYSICS
CHAPTER 1
= +15.0 mm; f 2 = +25.5 mm; d = 61.0 mm; u1 = 24.1 mm
Solution : f 1
a. i. By applying the thin lens l ens formula for the 1 st lens (objective),
1
f 1
=
1
u1
+
1
1
v1
+ 15.0 d
f 1 O
u1
F1
=
f 2 F2
F1
+
1
24.1 v1 v1 = +39.7 mm (real)
f 1
f 2 F2
I1
v1
u 2 = d − v1 u 2 = 61.0 − 39.7 u2
1
= 21.3 mm
u2
PHYSICS
CHAPTER 1
= +15.0 mm; f 2 = +25.5 mm; d = 61.0 mm; u1 = 24.1 mm
Solution : f 1
a. i. and the position of the final image formed by the 2 nd lens
1 (eyepiece) is
f 2
=
1
u2
+
1
1
v2
+ 25.5
=
1
+
1
21.3 v2 v2 = −129 mm
(in front of the 2 nd lens)
d
f 1 O
u1
F1
f 1
f 2
F1
F2
v1
I2
v
= 129 mm
f 2 F2
I1
u2
PHYSICS
CHAPTER 1
= +15.0 mm; f 2 = +25.5 mm; d = 61.0 mm; u1 = 24.1 mm
Solution : f 1
a. ii. The overall (total) magnification of the microscope is given by
m = m1m2 m= m=
v1 u1
×
39.7 24.1
where
m1
v2
=
v1 u1
and
m2
=
u2
×
( − 129) 21.3
m = 9.98
b. The characteristics of the final image are
virtual
inverted
magnified
formed in front of the 1 st and 2nd lenses lenses..
v2 u2
PHYSICS
CHAPTER 1
Exercise 1.4 : 1.
a. A glas glass s of refr refrac acti tive ve inde index x 1.50 1.50 plan planoo-co conc ncav ave e lens lens has has a focal length of 21.5 cm. Calculate the radius of the concave surface. b. A rod of length length 15.0 cm is placed horizontally horizontally along the principal axis of a converging lens of focal length 10.0 cm. If the closest end of the rod is 20.0 cm from from the lens calculate the length of the image formed.
ANS. : 10.8 cm; 6.00 cm 2. An obje object ct is plac placed ed 16.0 16.0 cm to the the lef leftt of of a lens lens.. The The lens lens forms an image which is 36.0 cm to the right of the lens. a. Calculate the focal length of the lens and state state the type type of the lens. b. If the object is 8.00 mm tall, calculate the height of the image. c. Sketch a labelled ray diagram for for the case above. ANS. : 11.1 cm; 1.8 cm
PHYSICS 3.
CHAPTER 1
When When a sma small ll ligh lightt bul bulb b is is pla place ced d on on the the left left side side of a converging lens, a sharp image is formed on a screen placed 30.0 cm on the right side of the lens. When the lens is moved 5.0 cm to the right, the screen has to be moved 5.0 cm to the left so that a sharp image is again formed on the screen. What is the focal length of the lens? ANS. : 10.0 cm 4. A conv conver ergi ging ng lens lens of of foca focall leng length th 8.0 8.00 0 cm cm is 20.0 20.0 cm cm to the the left left of a converging lens of focal length 6.00 cm. A coin is placed 10.0 cm to the left of the 1 st lens. Calculate a. the distance of the final image from the 1 st lens, b. the total magnification of the system. ANS. : 24.6 cm; 0.924 5. A conv conver ergi ging ng lens lens wit with h a foc focal al len lengt gth h of of 4.0 4.0 cm cm is to the the lef leftt of a second identical lens. When a feather is placed 12 cm to the left of the first lens, the final image is the same size and orientation as the feather itself. Calculate the separation between the lenses. ANS. : 12.0 cm
PHYSICS
CHAPTER 1
Next Chapter… CHAPTER 2 : Physical optics
