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PHYSICS CHAPTER 1The study of light based on the assumption that light light travels in straight linestravels in straight lines and is concerned with the
laws controlling the laws controlling the reflection and refractionreflection and refraction
of rays of lightlight.
CHAPTER 1: CHAPTER 1: Geometrical opticsGeometrical optics
(5 Hours)(5 Hours)
PHYSICS CHAPTER 1
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: StateState laws of reflection. laws of reflection. StateState the characteristics of image formed by a plane the characteristics of image formed by a plane
mirror.mirror. Sketch Sketch ray diagrams with minimum two rays.ray diagrams with minimum two rays.
Learning Outcome:
1.1 Reflection at a plane surface (1 hour)
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Figure 1.1Figure 1.1
1.1 Reflection at a plane surface1.1.1 Reflection of light is defined as the return of all or part of a beam of light when the return of all or part of a beam of light when
it encounters the boundary between two mediait encounters the boundary between two media. There are two types of reflection due to the plane surface
Specular (regular) reflectionSpecular (regular) reflection is the reflection of light from reflection of light from a smooth shiny surfacea smooth shiny surface as shown in Figure 1.1.
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Figure 1.2Figure 1.2
All the reflected rays are parallel to each another or move in the same direction.
Diffuse reflectionDiffuse reflection is the reflection of light from a rough reflection of light from a rough surfacesurface such as papers, flowers, people as shown in 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.
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Laws of reflectionLaws of reflection state : The incident ray, the reflected ray and the normal all lie incident ray, the reflected ray and the normal all lie
in the same planein the same plane. The angle of incidence, angle of incidence, ii equals the angle of reflection, equals the angle of reflection, rr
as shown in Figure 1.3.
i r
Plane surfacePlane surface
ri
Stimulation 1.1Figure 1.3Figure 1.3
Picture 1.1
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Image formation by a plane mirror as shown in Figures 1.4a and 1.4b.
Point object
1.1.2 Reflection at a plane mirror
Figure 1.4aFigure 1.4a
A 'Au v
i
i
r
i
distanceobject :uwhere
distance image :v
g
gangle glancing :g
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Vertical (extended) object
Stimulation 1.2
Figure 1.4bFigure 1.4b
Object
vu
ir
i
rImage
ihoh
where heightobject :ohheight image :ih
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The characteristics of the image formed by the plane mirror are virtual imagevirtual image
is seem to form by light coming from the image but seem to form by light coming from the image but light does not actually pass through the imagelight 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 imageupright or erect image laterally reverselaterally reverse
right-hand side of the object becomes the left-hand side of the image.
the object distance, object distance, uu equals the image distance, equals the image distance, vv the same sizesame size where the linear magnification, m is given by
obey the laws of reflectionobey the laws of reflection.
1height,Object
height, Image
o
i h
hm
Picture 1.2
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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?
Example 1 :
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A
B
L
Solution :Solution :a. The ray diagram to show the formation of the women’s image is
HE2
1AL
EF2
1LB
)feet(F
)eyes(E)head(H
h
y
m 60.1
m 10.0
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Solution :Solution :b. The minimum vertical length of the mirror is given by
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
LBAL h
EF2
1HE
2
1h
EFHE2
1h
Height of the women
m 80.060.12
1h
10.060.12
1y
m 75.0y
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u v
m 00.1
x
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 :Solution :
From the characteristic of the image formed by the plane mirror, thus
Therefore,
Example 2 :
m 350.0vuv
vx 00.1m 350.1x
m 350.0u
350.000.1 x
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Exercise 1.1 :1.
The two mirrors in Figure 1.5 meet at a right angle. The beam 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. :ANS. : 1.95 m 1.95 m ; 40; 40 to the mirror 2. to the mirror 2.
Figure 1.5Figure 1.5
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Exercise 1.1 :2.
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 1.6. Determine x.
ANS. :ANS. : 0.81 m0.81 m
Figure 1.6Figure 1.6
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Exercise 1.1 :3. Standing 2.00 m in front of a small vertical mirror, you see the
reflection of your belt buckle, which is 0.70 m below your eyes.
a. What is the vertical location of the mirror relative to the level of your eyes?
b. What is the angle do your eyes make with the horizontal when you look at the buckle?
c. If you now move backward until you are 6.0 m from the mirror, will you still see the buckle? Explain.
ANS. :ANS. : 35 cm below; 9.935 cm below; 9.9; U think; U think4. You are 1.80 m tall and stand 3.00 m from a plane mirror that
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. :ANS. : 11.13 m.13 m
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Sketch and useSketch and use ray diagrams to ray diagrams to determinedetermine the the
characteristics of image formed by spherical mirrors.characteristics of image formed by spherical mirrors. UseUse
for real object only.for real object only.
UseUse sign convention for focal length: sign convention for focal length:
+ + ff for concave mirror and – for concave mirror and – ff for convex mirror.for convex mirror. SketchSketch ray diagrams with minimum two rays. ray diagrams with minimum two rays. rr = 2 = 2ff only applies to spherical mirror. only applies to spherical mirror.
Learning Outcome:
1.2 Reflection at a spherical surface (1 hour)
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rvuf
2111
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CC
AA
BB
rPPCC
AA
BB
rPP
Figure 1.7aFigure 1.7a
1.1 Reflection at a spherical surface1.2.1 Spherical mirror is defined as a reflecting surface that is part of a spherea reflecting surface that is part of a sphere. There are two types of spherical mirror. It is convexconvex (curving
outwards) and concaveconcave (curving inwards) mirror. Figures 1.7a and 1.7b show the shape of concave and convex
mirrors.
reflecting surface
imaginary sphere
silver layer
Figure 1.7bFigure 1.7b
(a)Concave (ConvergingConverging) mirror
(b) Convex (DivergingDiverging) mirror
Picture 1.3
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Terms of spherical mirrorTerms of spherical mirror Centre of curvature (point C)Centre of curvature (point C)
is defined as the centre of the sphere of which a curved the centre of the sphere of which a curved mirror forms a partmirror forms a part.
Radius of curvature, Radius of curvature, rr is defined as the radius of the sphere of which a curved the radius of the sphere of which a curved
mirror forms a partmirror forms a part. Pole or vertex (point P)Pole or vertex (point P)
is defined as the point at the centre of the mirrorthe point at the centre of the mirror. Principal axisPrincipal axis
is defined as the straight line through the centre of the straight line through the centre of curvature C and pole P of the mirrorcurvature C and pole P of the mirror.
AB is called the apertureaperture of the mirror.
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Consider the ray diagram for a concave and convex mirrors as shown in Figures 1.8a and 1.8b.
Point FF represents the focal pointfocal point or focusfocus of the mirrors. Distance ff represents the focal lengthfocal length of the mirrors. The parallel incident raysparallel incident rays represent the object infinitely far object infinitely far
awayaway from the spherical mirror e.g. the sun.
CCPPCC PP
1.2.2 Focal point and focal length, f
Figure 1.8aFigure 1.8a
FFf
FFf
Incident Incident raysrays
Figure 1.8bFigure 1.8b
Incident Incident raysrays
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Focal point or focus, FFocal point or focus, F For concave mirror – is defined as a point where the incident a point where the incident
parallel rays converge after reflection on the mirrorparallel rays converge after reflection on the mirror. Its focal point is real (principal)real (principal).
For convex mirror – is defined as a point where the incident a point where the incident parallel rays seem to diverge from a point behind the mirror parallel rays seem to diverge from a point behind the mirror after reflectionafter reflection.
Its focal point is virtualvirtual.
Focal length, Focal length, ff is defined as the distance between the focal point (focus) F the distance between the focal point (focus) F
and pole P of the spherical mirrorand pole P of the spherical mirror. The paraxial raysparaxial rays is defined as the rays that are near to and the rays that are near to and
almost parallel to the principal axisalmost parallel to the principal axis.
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Consider a ray AB parallel to the principal axis of concave mirror as shown in Figure 1.9.
1.2.3 Relationship between focal length, f and
radius of curvature, r
Figure 1.9Figure 1.9
CC
PPFF DD
incident rayincident rayBBAA
fr
ii
i
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From the Figure 1.9,BCD
BFD By using an isosceles triangle CBF, thus the angle is given by
then
Because of AB is paraxial ray, thus point B is too close with pole P then
Therefore
ii CD
BDtan
FD
BDtan
Taken the angles are << Taken the angles are << small by considering the small by considering the ray AB is paraxial ray.ray AB is paraxial ray.
i2
rCPCD
fFPFD
This relationship also valid for convex mirror.This relationship also valid for convex mirror.
2
rf
CD
BD2
FD
BD
OR
FD2CD
f2r
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is defined as the simple graphical method to indicate the the simple graphical method to indicate the positions of the object and image in a system of mirrors or positions of the object and image in a system of mirrors or lenseslenses.
Figures 1.10a and 1.10b show the graphical method of locating an image formed by concave and convex mirror.
1.2.4 Ray diagrams for spherical mirrors
Figure 1.10aFigure 1.10a Figure 1.10bFigure 1.10b
(a) Concave mirror (b) Convex mirror
CC PPFF
11
33
33
11
I CC
FFPP
1122
22O O I
2233
11
22
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Ray 1Ray 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 2Ray 2 - Passes or directed towards focal point F reflected parallel to principal axis.
Ray 3Ray 3 - Passes or directed towards centre of curvature C, reflected back along the same path.
Images formed by a convex mirrorImages formed by a convex mirror Figure 1.11 shows the graphical method of locating an image
formed by a convex mirror.
At least any At least any two rays two rays for drawing for drawing the ray the ray diagram.diagram.
CC
FF
PP
O Iu v
frontfront backbackFigure 1.11Figure 1.11
Picture 1.4
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The characteristics of the image formed are virtualvirtual uprightupright diminished (smaller than the object)diminished (smaller than the object) formed at the back of the mirror (behind the mirror)formed at the back of the mirror (behind the mirror)
Object position any positionany position in front of the convex mirror. Convex mirror always being used as a driving mirrordriving mirror because it
has a wide field of viewwide field of view and providing an upright imageupright image.
Images formed by a concave mirrorImages formed by a concave mirror Concave mirror can be used as a shaving and makeup mirrorsshaving and makeup mirrors
because it provides an upright and virtual imagesupright and virtual images. Table 1.1 shows the ray diagrams of locating an image formed
by a concave mirror for various object distance, u.
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Object
distance, uRay diagram
Image characteristic
I
CC
FrontFront backback
FFPP
u > ru > r
u = ru = r
OI
O
Real Inverted Diminished Formed
between point C and F.
Real Inverted Same size Formed at point
C.
CCFF
PP
FrontFront backback
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Object
distance, uRay diagram
Image characteristic
FFCC PP
FrontFront backback
f < u < rf < u < r
u = fu = f
O
Real Inverted Magnified Formed at a
distance greater than CP.
Real or virtual Formed at infinity.
IO
CC
FF
PP
FrontFront backback
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Linear (lateral) magnification of the spherical mirror, m is
defined as the ratio between image height, the ratio between image height, hhii and object and object
height, height, hhoo
Object
distance, uRay diagram
Image characteristic
u < fu < f
O
Virtual Upright Magnified Formed at the
back of the mirror
IFF
CC PP
FrontFront backback
u
v
h
hm
o
iwhere
pole thefrom distance image :vpole thefrom distanceobject :u
Table 1.1Table 1.1
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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.
1.2.5 Derivation of Spherical mirror equation
Figure 1.12Figure 1.12
O CC PPIv
u
BB
DD
From the figure, BOC BCIthen, eq. (1)(2) :
By using BOD, BCD and BID thus
(1)(1) (2)(2)
2 (3)(3)
ID
BD tan;
CD
BDtan ;
OD
BDtan
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By considering point B very close to the pole P, hence
then
therefore
vru IPID; CPCD ; OPOD tan; tan ; tan
v
BD
r
BD
u
BD ; ; Substituting this Substituting this
value in eq. (3)value in eq. (3)
fr 2
rvu
BD2
BD
BD
rvu
21
1 where
rvuf
21
11 Spherical mirror’s Spherical mirror’s
equationequation
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Table 1.2 shows the sign convention for spherical mirror’s equation .
Note: Real image is formed by the actual light rays that pass formed by the actual light rays that pass
through the imagethrough the image. Real image can be projected on the screenprojected on the screen.
Physical Quantity Positive sign (+) Negative sign (-)
Object distance, u
Image distance, v
Focal length, f
Real object Virtual object
Real image Virtual image
Concave mirror Convex mirror
(same side of the object) (opposite side of the object)
(in front of the mirror) (at the back of the mirror)
Table 1.2Table 1.2
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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 :Solution :
a. By applying the mirror’s equation, thus
b. By using the magnification formula, thus
Example 3 :
cm 38.1fvuf
111
u
vm
cm 9.25 cm; 20.1 vu
71.720.1
25.9m
25.9
1
20.1
11
f(Concave mirror)(Concave mirror)
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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 :Solution :
Example 4 :
oi 25.0 hh
O Icm 0.52
Spherical Spherical
mirrormirroru v
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Solution :Solution :a. From the figure,
By using the equation of linear magnification, thus
By substituting eq. (2) into eq. (1), hence
The mirror should be placed 16.4 cm in front of the object 16.4 cm in front of the object.
oi 25.0 hh
5.20 vu
u
v
h
hm
o
i
(1)(1)
u
v
h
h
o
o25.0
uv 25.0 (2)(2)
5.2025.0 uucm 4.16u
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Solution :Solution :b. By using the mirror’s equation, thus
The type of spherical mirror is convexconvex because the negative value of focal length.
oi 25.0 hh
cm 47.5f
vuf
111
uuf 25.0
111
4.1625.0
1
4.16
11
f
and2
rf
cm 9.1047.52 r
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Solution :Solution :c. The ray diagram is shown below.
oi 25.0 hh
FF
PP CC
O I
frontfront backback
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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 :Solution :
Example 5 :
m 0.60 m; 60.1o uh
u
oh
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Solution :Solution :a. Given The radius of curvature of the globe’s surface (convex surface) is given by
By applying the mirror’s equation, hence
m 18.0d
m 09.02
18.0r
vur
112
(behind the globe’s surface)(behind the globe’s surface)m 042.0v
m 0.60 m; 60.1o uh
v
1
60.0
1
09.0
2
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Solution :Solution :
b. By applying the magnification formula, thus
c. The characteristics of the person’s image are virtualvirtual uprightupright diminisheddiminished formed behind the reflecting surface.formed behind the reflecting surface.
u
v
h
hm
o
i
m 0.60 m; 60.1o uh
60.0
042.0
60.1i
h
m 112.0i h OR cm 2.11
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CC FFPP
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 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 length of the mirror.Solution :Solution :a. The ray diagram of the bulb is
Example 6 :
I
O
cm 40
mm 0.8
u
m 10 40 m; 10 0.8m; 3.50 2i
3o
hhv
m 50.3
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Solution :Solution :
b. i. By applying the magnification formula, thus
The position of the bulb is 7.0 cm in front of the mirror.The position of the bulb is 7.0 cm in front of the mirror. ii. By applying the mirror’s equation, thus
u
v
h
hm
o
i
u
50.3
100.8
10403
2
m 07.0u OR cm 0.7
m 10 40 m; 10 0.8m; 3.50 2i
3o
hhu
vuf
111
m 0687.0f50.3
1
07.0
11
fOR cm 87.6
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Exercise 1.2 :1. a. A concave mirror forms an inverted image four times
larger than the object. Calculate 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.
ANS. :ANS. : 160 mm 160 mm ; 267 mm; 267 mm2. a. A 1.74 m tall shopper in a department store is 5.19 m
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 the mirror.
b. A concave mirror of a focal length 36 cm produces an image whose distance from the mirror is one third of the object distance. Calculate the object and image distances.
ANS. :ANS. : u think, 1.07 m u think, 1.07 m ; 144 cm, 48 cm; 144 cm, 48 cm
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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: State and useState and use the laws of refraction (Snell’s Law) for the laws of refraction (Snell’s Law) for
layers of materials with different densities. layers of materials with different densities. ApplyApply
for spherical surface.for spherical surface.
Learning Outcome:
1.3 Refraction at a plane and spherical surfaces (1 hour)
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r
nn
v
n
u
n 1221
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1.3 Refraction at a plane and spherical surfaces1.3.1 Refraction at a plane surface RefractionRefraction is defined as the changing of direction of a light the changing of direction of a light
ray and its speed of propagation as it passes from one ray and its speed of propagation as it passes from one medium into anothermedium into another.
Laws of refractionLaws of refraction state : The incident ray, the refracted ray and the normal all lie incident ray, the refracted ray and the normal all lie
in the same planein the same plane. For two given media, Snell’s law Snell’s law states
constantsin
sin
1
2 n
n
r
irnin sinsin 21 OR
where 1 medium theofindex refractive:1nray)incident thecontaining Medium(
2 medium theofindex refractive:2nray) refracted thecontaining Medium(
refraction of angle :r
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The light ray is bent toward the bent toward the normalnormal, thus
The light ray is bent away from bent away from the normalthe normal, thus
Examples for refraction of light ray travels from one medium to another medium can be shown in Figures 1.13a and 1.13b.
21 nn (a)
ir
21 nn (b)
1n
2n
i
r
Incident ray
Refracted ray
1n
2n
i
r
Incident ray
Refracted ray
(Medium 1 is less dense (Medium 1 is less dense medium 2)medium 2)
(Medium 1 is denser than (Medium 1 is denser than medium 2)medium 2)
ir
Figure 1.13aFigure 1.13a Figure 1.13bFigure 1.13b
Stimulation 1.3 Stimulation 1.4
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Refractive index (index of refraction), Refractive index (index of refraction), nn is defined as the constant ratio for the two given constant ratio for the two given
mediamedia.
The value of refractive index depends on the type of mediumtype of medium and the colour of the lightcolour of the light.
It is dimensionlessdimensionless and its value greater than 1greater than 1. Consider the light ray travels from medium 1 into medium 2, the
refractive index can be denoted by
r
i
sin
sin
2
121 2 mediumin light ofvelocity
1 mediumin light ofvelocity
v
vn
(Medium containing (Medium containing the incident ray)the incident ray)
(Medium containing the (Medium containing the refracted ray)refracted ray)
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Absolute refractive index, n (for the incident ray travels from vacuum or air into the mediummedium) is given by
Table 1.3 shows the refractive indices for common substances.
v
cn
mediumin light ofvelocity
in vacuumlight ofvelocity
SubstanceSubstance Refractive index, Refractive index, nnSolidsSolids
DiamondFlint glassCrown glassFused quartz (glass)Ice
LiquidsLiquidsBenzeneEthyl alcoholWater
GasesGasesCarbon dioxideAir
2.421.661.521.461.31
1.501.361.33
1.000451.000293
Table 1.3Table 1.3
(If the density density of medium is of medium is greatergreater hence the refractive refractive index is also index is also greatergreater)
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Relationship between refractive index and the wavelength of Relationship between refractive index and the wavelength of
light light As light travels from one medium to another, its wavelength, wavelength,
changeschanges but its frequency, frequency, ff remains constant remains constant. The wavelength changes because of different materialdifferent material. The
frequency remains constant because the number of wave number of wave cycles arriving per unit time must equal the number leaving cycles arriving per unit time must equal the number leaving per unit timeper unit time so that the boundary surface cannot create or cannot create or destroy wavesdestroy waves.
By considering a light travels from medium 1 (n1) into medium 2
(n2), the velocity of light in each medium is given by
then
11 fv 22 fv and
2
1
2
1
f
f
v
v where
11 n
cv
22 n
cv and
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If medium 1 is vacuum or air, then n1 = 1. Therefore the
refractive index for any medium, n can be expressed as
2
1
2
1
nc
nc
2211 nn
(Refractive index is inversely (Refractive index is inversely proportional to the wavelength)proportional to the wavelength)
where
0n
in vacuumlight ofh wavelengt:0mediumin light ofh wavelengt:
Picture 1.5 Picture 1.6
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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 :Solution :
Example 7 :
33.1; 1.00 wa nn
wheredepthapparent :AB
m 3.00 depth actual :AC
A
i
Air (na)
C
r
B
Water (nw)
ir
m 00.3
D
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Solution :Solution :From the diagram,
ABD
ACD
By considering only small angles of r and i , thus
33.1; 1.00 wa nn
AB
ADtan r
AC
ADtan i
andrr sintan ii sintan
AC
AB
ABADACAD
sin
sin
tan
tan
r
i
r
ithen
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Note : Note : (Important)(Important)
Solution :Solution :From the Snell’s law,
33.1; 1.00 wa nn
w
a
AC
AB
n
n
w
a
1
2
sin
sin
n
n
n
n
r
i
33.1
00.1
3.00
AB
m 26.2AB
depthapparent
depth real
1
2 n
nn
Other equation for absolute refractive index in term of depth is given by
PHYSICS CHAPTER 1
53
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 3.00 108 m s-1.)Solution :Solution :
Example 8 :
33.1; 1.31 wi nn
Ice (Ice (nnii))
Water (Water (nnww))
BottomBottom
m 00.4
m 32.0i h
32.000.4w hm 68.3w h
PHYSICS CHAPTER 1
54
Solution :Solution :The speed of light in ice and water are
Since the light propagates in ice and water at constant speed thus
Therefore the time required is given byt
sv
v
st
ii v
cn
18i s m 1029.2 v
33.1; 1.31 wi nn
i
81000.331.1
v
ww v
cn
18w s m 1026.2 v
w
81000.333.1
v
wi ttt
88w
w
i
i
1026.2
68.3
1029.2
32.0
v
h
v
ht
s 1077.1 8t
PHYSICS CHAPTER 1
55
Figure 1.14 shows a spherical surface with radius, r forms an
interface between two media with refractive indices n1 and n2.
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 where n1 < n2. Point C is the centre of curvature of the spherical surface and
BC is normal.
1.3.2 Refraction at a spherical surface
Figure 1.14Figure 1.14
P
B
O ICD
1n
vr
u
2ni
PHYSICS CHAPTER 1
56
From the figure,
BOC
BIC
From the Snell’s law
By using BOD, BCD and BID thus
By considering point B very close to the pole P, hence
then Snell’s law can be written as
i (1)
(2)
sinsin 21 nin
ID
BD tan;
CD
BDtan ;
OD
BDtan
vru IPID; CPCD ; OPOD
21 nin
tan; tan ; tan ; sin ; sin ii
(3)
PHYSICS CHAPTER 1
57
By substituting eq. (1) and (2) into eq. (3), thus
then
)()( 21 nn )( 1221 nnnn
rnn
vn
un
BD)(
BDBD1221
r
nn
v
n
u
n )( 1221
where pole from distance image :vpole from distanceobject :u
1 medium ofindex refractive :1nray)incident thecontaining Medium(
2 medium ofindex refractive :2nray) refracted thecontaining Medium(
Equation of spherical Equation of spherical refracting surfacerefracting surface
PHYSICS CHAPTER 1
58
Note : If the refracting surface is flat (plane)flat (plane) :
then
The equation (formula) of linear magnification for refraction by the spherical surface is given by
021 v
n
u
n
r
un
vn
h
hm
2
1
o
i
PHYSICS CHAPTER 1
59
Table 1.4 shows the sign convention for refraction or thin refraction or thin lenseslenses:
Physical Quantity Positive sign (+) Negative sign (-)
Object distance, u
Image distance, v
Focal length, f
Real object Virtual object
Real image Virtual image
Converging lens Diverging lens
(same side of the object)
(opposite side of the object)
(in front of the refracting surface)
(at the back of the refracting surface)
Radius of
curvature, r
Centre of curvature is located in more dense medium
Centre of curvature is located in less dense medium
Table 1.4Table 1.4
(convex surface) (concave surface)
PHYSICS CHAPTER 1
60
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.
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)
Example 9 :
O ICP
cm 0.10
3.00 cm
glassair
Figure 1.15Figure 1.15
PHYSICS CHAPTER 1
61
Solution :Solution :a. By using the equation of spherical refracting surface, thus
The image is 20.7 cm at the back of the convex surface.The image is 20.7 cm at the back of the convex surface.b. The linear magnification of the image is given by
r
nn
v
n
u
n agga
cm 3.00 cm; 0.10; 1.52g run
un
vnm
g
a
cm 7.20v
00.3
00.152.152.1
0.10
00.1
v
un
vnm
2
1
0.1052.1
7.2000.1m
36.1m
PHYSICS CHAPTER 1
62
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.
Determine
a. the position of the image formed by 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)
Example 10 :
Figure 1.16Figure 1.16
OP
cm 0.20
Glass sphere
air
cm 0.5
PHYSICS CHAPTER 1
63
Solution :Solution :a. By using the equation of spherical refracting surface, thus
The image is 21.5 cm at the back of the first surface P.The image is 21.5 cm at the back of the first surface P.
OR
r
nn
v
n
u
n agga
cm .05 cm; 0.20; 1.63g run
cm 5.21v
0.5
00.163.163.1
0.20
00.1
v
O C 1I
cm020u . cm 5.21v
P
gnan
r
PHYSICS CHAPTER 1
64
Solution :Solution :b.
From the figure above, the image I1 formed by the first surface P
is in the glass and 11.5 cm from the second surface Q. I1 acts
as a virtual objectvirtual object for the second surface and
O C
2I cm 1.52
P
gnan
First surface
1I
an
Q
cm 1.51
Second surface
cm; 5.11 1.00; ; 1.63 a2g1 unnnn
cm 5.00rCentre of curvature is located in Centre of curvature is located in more dense mediummore dense medium
PHYSICS CHAPTER 1
65
Solution :Solution :b. By using
The image is real and 3.74 cm at the back of the second The image is real and 3.74 cm at the back of the second
surface Q.surface Q.
r
nn
v
n
u
n gaag
cm 74.3v
0.5
63.100.100.1
5.11
63.1
v
PHYSICS CHAPTER 1
66
Exercise 1.3 :1. A student wishes 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 1.17.
Calculate the depth of the pool.
(Given nwater = 1.33 and nair = 1.00)
ANS. :ANS. : 5.16 m5.16 m
Figure 1.17Figure 1.17
PHYSICS CHAPTER 1
67
Exercise 1.3 :
2. 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. Determine the position of the image.
(Given the refractive index of glass =1.52 and the refractive index of air =1.00)
ANS. :ANS. : 20.83 cm in front of the 220.83 cm in front of the 2ndnd refracting surface. refracting surface.
3. 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. Determine the image position of the source.
(Given the refractive index of glass =1.52 and the refractive index of air =1.00)
ANS. :ANS. : 25.2 cm at the back of the 225.2 cm at the back of the 2ndnd refracting surface. refracting surface.
PHYSICS CHAPTER 1
68
At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Sketch and useSketch and use ray diagrams to ray diagrams to determinedetermine the the
characteristics of image formed by diverging and characteristics of image formed by diverging and converging lenses. converging lenses.
UseUse equation stated in 1.3 to equation stated in 1.3 to derivederive thin lens formula, thin lens formula,
for real object only.for real object only.
UseUse lensmaker’s equation: lensmaker’s equation:
UseUse the thin lens formula for a combination of the thin lens formula for a combination of converging lenses.converging lenses.
Learning Outcome:1.4 Thin lenses (2 hours)
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
ww
w.k
mp
h.m
atri
k.ed
u.m
y/p
hys
ics
fvu
111
21
111
1
rrn
f
PHYSICS CHAPTER 1
69
1.4 Thin lenses is defined as a transparent material with two spherical a transparent material with two spherical
refracting surfaces whose thickness is thin compared to refracting surfaces whose thickness is thin compared to the radii of curvature of the two refracting surfacesthe radii of curvature of the two refracting surfaces.
There are two types of thin lenses. It is convergingconverging and divergingdiverging lenses.
Figures 1.18a and 1.18b show the various types of thin lenses, both converging and diverging.
(a) Converging (Convex) lensesConverging (Convex) lenses
BiconvexBiconvex Plano-convexPlano-convex Convex meniscusConvex meniscus
Figure 1.18aFigure 1.18a
rr11
(+ve)(+ve)rr22
(+ve)(+ve)rr11
(+ve)(+ve)rr22
(())rr11
(+ve)(+ve)rr22
((ve)ve)
PHYSICS CHAPTER 1
70
1.4.1 Terms of thin lenses Figures 1.19 show the shape of converging (convex) and
diverging (concave) lenses.
(b) Diverging (Concave) lensesDiverging (Concave) lenses
BiconcaveBiconcave Plano-concavePlano-concave Concave meniscusConcave meniscusFigure 1.18bFigure 1.18b
(a) Converging lens (b) Diverging lens
CC11 CC22
rr11
rr22OO CC11 CC22
rr11
rr22
OO
Figure 1.19Figure 1.19
rr11
((ve)ve)rr22
((ve)ve)rr11
((ve)ve)rr22
(())rr11
(+ve)(+ve)rr22
((ve)ve)
PHYSICS CHAPTER 1
71
Centre of curvature (point CCentre of curvature (point C11 and C and C22)) is defined as the centre of the sphere of which the surface the centre of the sphere of which the surface
of the lens is a partof the lens is a part. Radius of curvature (rRadius of curvature (r11 and r and r22))
is defined as the radius of the sphere of which the surface the radius of the sphere of which the surface of the lens is a partof the lens is a part.
Principal (Optical) axisPrincipal (Optical) axis is defined as the line joining the two centres of curvature the line joining the two centres of curvature
of a lensof a lens. Optical centre (point O)Optical centre (point O)
is defined as the point at which any rays entering the lens the point at which any rays entering the lens pass without deviationpass without deviation.
PHYSICS CHAPTER 1
72
Consider the ray diagrams for converging and diverging lenses as shown in Figures 1.20a and 1.20b.
From the figures, Points F1 and F2 represent the focus of the lenses. Distance f represents the focal length of the lenses.
1.4.2 Focal point and focal length, f
FF11 FF22OO
ffff
Figure 1.20aFigure 1.20a Figure 1.20bFigure 1.20b
FF11 FF22OO
PHYSICS CHAPTER 1
73
Focus (point FFocus (point F11 and F and F22)) For converging (convex)converging (convex) lens – is defined as the point on the the point on the
principal axis where rays which are parallel and close to the principal axis where rays which are parallel and close to the principal axis converges after passing through the lensprincipal axis converges after passing through the lens.
Its focus is real (principal). For diverging (concave)diverging (concave) lens – is defined as the point on the the point on the
principal axis where rays which are parallel to the principal principal axis where rays which are parallel to the principal axis seem to diverge from after passing through the lensaxis seem to diverge from after passing through the lens.
Its focus is virtual.
Focal length ( Focal length ( ff )) is defined as the distance between the focus F and the the distance between the focus F and the
optical centre O of the lensoptical centre O of the lens.
PHYSICS CHAPTER 1
74
Figures 1.21a and 1.21b show the graphical method of locating an image formed by a converging (convex) and diverging (concave) lenses.
1.4.3 Ray diagram for thin lenses
Figure 1.21aFigure 1.21a
FF11
FF22
(a) Converging (convex) lens
11
11
22
22
OO
33
33
II
u v
PHYSICS CHAPTER 1
75
Ray 1Ray 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 2Ray 2 - Passes through the optical centre of the lens is undeviated.
Ray 3Ray 3 - Passes through the focus F1 of a converging lens or appears to converge towards the focus F1 of a
diverging lens, after refraction by the lens the ray parallel to the principal axis.
(b) Diverging (concave) lens
OO FF22 FF11
11
11
22
22
33
33
II
vu
Figure 1.21bFigure 1.21b
At least At least any two any two rays for rays for drawing drawing the ray the ray diagram.diagram.
PHYSICS CHAPTER 1
76
Images formed by a diverging lensImages formed by a diverging lens Figure 1.22 shows the graphical method of locating an image
formed by a diverging lens.
The characteristics of the image formed are virtualvirtual uprightupright diminished (smaller than the object)diminished (smaller than the object) formed in front of the lensformed in front of the lens.
Object position any positionany position in front of the diverging lens.
FrontFront backback
OO FF22 FF11II
Figure 1.22Figure 1.22
PHYSICS CHAPTER 1
77
Object
distance, uRay diagram
Image characteristic
FF11FF22 2F2F222F2F11
Images formed by a converging lensImages 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.
FrontFront backback
u > u > 22ff
Real Inverted Diminished Formed between
point F2 and 2F2.
(at the back of the lens)
OOI
PHYSICS CHAPTER 1
78
Object
distance, uRay diagram
Image characteristic
OO
FF11FF22
2F2F22
2F2F11u = u = 22ff
Real Inverted Same size Formed at point
2F2. (at the back of the lens)FrontFront backback
I
f < u < f < u < 22ff
Real Inverted Magnified Formed at a
distance greater than 2f at the back of the lens.
OO FF11FF22 2F2F222F2F11
FrontFront backback
I
PHYSICS CHAPTER 1
79
Object
distance, uRay diagram
Image characteristic
OOFF11
FF22 2F2F222F2F11u = fu = f
Real or virtual Formed at infinity.
FrontFront backback
u < fu < f
Virtual Upright Magnified Formed in front
of the lens.
OOFF11 FF22 2F2F222F2F11
FrontFront backback
I
Table 1.5Table 1.5
Stimulation 1.5
PHYSICS CHAPTER 1
80
Thin lens formula and lens maker’s equationThin lens formula and lens maker’s equation Considering the ray diagram of refraction for two spherical
surfaces as shown in Figure 1.23.
1.4.4 Thin lens formula, lens maker’s and linear magnification equations
Figure 1.23Figure 1.23
OOCC11
CC22II11
II22PP11 PP22
EEBB
AA DD
1u 1v2v
1r 2r
t
1n
12 vtu
1n2n
PHYSICS CHAPTER 1
81
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 == +r +r11))
Surface DE ( (r r == +r +r22))
Assuming the lens is very thin thus very thin thus tt = 0= 0,
1
12
1
2
1
1 )(
r
nn
v
n
u
n
2
12
2
1
1
2 )(
r
nn
v
n
vt
n
(1)(1)
2
12
2
1
1
2
r
nn
v
n
v
n2
12
2
1
1
2 )(
r
nn
v
n
v
n
(2)(2)
PHYSICS CHAPTER 1
82
By substituting eq. (2) into eq. (1), thus
If u1 = and v2 = f thus eq. (3) becomes
1
12
2
12
2
1
1
1 )(
r
nn
r
nn
v
n
u
n
2
12
1
12
2
1
1
1 )()(
r
nn
r
nn
v
n
u
n
(3)(3)
211
2
21
111
11
rrn
n
vu
211
2 111
1
rrn
n
fLens maker’s Lens maker’s equationequation
where length focal :fsurface refracting 1for curvature of radius : st
1r
medium theofindex refractive :1nmaterial lens theofindex refractive :2n
surface refracting 2for curvature of radius : nd2r
PHYSICS CHAPTER 1
83
By equating eq. (3) and the lens maker’s equation, thus
therefore in general,
Note : If the medium is airair (n1= nair=1) thus the lens maker’s
equation can be written as
For thin lenses and lens maker’s equations, use the sign sign conventionconvention for refractionrefraction.
fvu
111
21
vuf
111 Thin lens formulaThin lens formula
where material lens theofindex refractive :n
21
111
1
rrn
f
PHYSICS CHAPTER 1
84
Linear magnification, Linear magnification, mm is defined as thethe ratio between image height, ratio between image height, hhii and object and object
height, height, hhoo.
Since the linear magnification equation can be
written as
u
v
h
hm
o
i
where centre optical from distance image :vcentre optical from distanceobject :u
vuf
111
vvuf
111
1u
v
f
v1
f
vm
PHYSICS CHAPTER 1
85
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.
Solution :Solution :
a. By applying the lens maker’s equation in air, thus
Example 11 :
;52.1 m; 50.2 m; 1.75o nuhm 1069.7 3
21rr
21
111
1
rrn
f
PHYSICS CHAPTER 1
86
Solution :Solution :
a.
b. The ray diagram for the case is
;52.1 m; 50.2 m; 1.75o nuhm 1069.7 3
21rr
33 1069.7
1
1069.7
1152.1
1
fm 1039.7 3f
FF11FF22 2F2F222F2F11
FrontFront backback
OOI
PHYSICS CHAPTER 1
87
Solution :Solution :
c. The position of the image formed is
By using the linear magnification equation, thus
d. The characteristics of the image are realreal invertedinverted diminisheddiminished formed at the back of the lensformed at the back of the lens
vuf
111
m 1041.7 3vv
1
50.2
1
1039.7
13
(at the back of the lens)(at the back of the lens)
u
v
h
hm
o
i
50.2
1041.7
75.1
3i
h
m 1019.5 3i
h OR mm 19.5
PHYSICS CHAPTER 1
88
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 :Solution :
a. By applying the linear magnification equation for thin lens, thus
By using the thin lens formula, thus
Example 12 :
10 cm; 10 1.66; mun
10u
vm uv 10
vuf
111
uuf 10
111
Virtual imageVirtual image
PHYSICS CHAPTER 1
89
Solution :Solution :
a.
b. Since the thin lens is plano-convex thus
Therefore
10 cm; 10 1.66; mun
2r
21
111
1
rrn
f
11
166.11.11
1
1r
cm 33.71 r
1010
1
10
11
f
cm 1.11f
PHYSICS CHAPTER 1
90
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 :Solution :
a. By applying the lens maker’s equation in air,
Example 13 :
2/32 n
21
111
1
rrn
f
cm 201 r cm 102 r
2/32 nnand
PHYSICS CHAPTER 1
91
Solution :Solution :
a.
b. Given
By using the general lens maker’s equation, therefore
3/41 n
211
2 111
1
rrn
n
f
cm 160f
cm 40f
2/32 n
10
1
20
11
2
31
f
10
1
20
11
1
34
23
f
PHYSICS CHAPTER 1
92
Many optical instruments, such as microscopes and microscopes and telescopestelescopes, use two converging lensestwo converging lenses together to produce an image.
In both instruments, the 1st lens (closest to the objectclosest to the object )is called the objectiveobjective and the 2nd lens (closest to the eyeclosest to the eye) is referred to as the eyepiece eyepiece or ocular ocular.
The image formed image formed by the 1 1stst lens lens is treatedtreated as the object for object for the 2the 2ndnd lens lens and the final imagefinal image is the image formed by the 22ndnd lenslens.
The position of the final imageposition of the final image in a two lenses system can be determined by applying the thin lens formula to each lens thin lens formula to each lens separatelyseparately.
The overall magnification of a two lenses systemoverall magnification of a two lenses system is the product of the magnifications of the separate lensesproduct of the magnifications of the separate lenses.
1.4.5 Combination of lenses
21mmm where
ionmagnificat overall :mlens 1 the todueion magnificat : st
1mlens 2 the todueion magnificat : nd
2m
Picture 1.7
Picture 1.8
PHYSICS CHAPTER 1
93
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 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.
Solution :Solution :
Example 14 :
mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u
d
1u
1f1f 2f2fFF11 FF11 FF22 FF22O
objective (1objective (1stst)) eyepiece(2eyepiece(2ndnd) )
PHYSICS CHAPTER 1
94
d
1u
1f1f 2f2fFF11 FF11 FF22 FF22O
Solution :Solution :
a. i. By applying the thin lens formula for the 1st lens (objective),
mm 7.391 v111
111
vuf
mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u
1
1
1.24
1
0.15
1
v
(real)(real)
1I
1v 2u
12 vdu 7.390.612 u
mm 3.212 u
PHYSICS CHAPTER 1
95
Solution :Solution :
a. i. and the position of the final image formed by the 2nd lens
(eyepiece) is
mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u
222
111
vuf
2
1
3.21
1
5.25
1
v
mm 1292 v
(in front of the 2(in front of the 2ndnd lens) lens)
2I
mm 1292 v
d
1u
1f1f 2f2fFF11 FF11 FF22 FF22O
1I
1v 2u
PHYSICS CHAPTER 1
96
Solution :Solution :
a. ii. The overall (total) magnification of the microscope is given by
b. The characteristics of the final image are virtualvirtual invertedinverted magnifiedmagnified formed in front of the 1formed in front of the 1stst and 2 and 2ndnd lenses lenses.
mm; 61.0 mm; 5.25 mm; 0.15 21 dffmm 24.11 u
21mmm where1
11 u
vm
2
22 u
vm and
2
2
1
1
u
v
u
vm
3.21
129
1.24
7.39 m 98.9m
PHYSICS CHAPTER 1
97
Exercise 1.4 :1. a. A glass of refractive index 1.50 plano-concave lens
has a focal length of 21.5 cm. Calculate the radius of the concave surface.
b. A rod of length 15.0 cm is placed 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 the lens
calculate the length of the image formed.
ANS. :ANS. : 10.8 cm10.8 cm; 6.00 cm; 6.00 cm
2. An object is placed 16.0 cm to the left of a lens. The 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 the 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 the case above.
ANS. :ANS. : 11.1 cm; 1.8 cm11.1 cm; 1.8 cm
PHYSICS CHAPTER 1
98
3. When a small light bulb is placed on the left 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. :ANS. : 10.0 cm10.0 cm4. A converging lens of focal length 8.00 cm is 20.0 cm to the left
of a converging lens of focal length 6.00 cm. A coin is placed 10.0 cm to the left of the 1st lens. Calculate a. the distance of the final image from the 1st lens,b. the total magnification of the system.
ANS. :ANS. : 24.6 cm; 0.92424.6 cm; 0.9245. A converging lens with a focal length of 4.0 cm is to the left 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. :ANS. : 12.0 cm12.0 cm
99
PHYSICS CHAPTER 1
Next Chapter…CHAPTER 2 :Physical optics