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SF027 1
UNIT 9:Geometrical OpticsUNIT 9:Geometrical Optics
The study of light based on The study of light based on
the assumption that light the assumption that light
travels in straight lines and travels in straight lines and
is concerned with the laws is concerned with the laws
controlling the reflection controlling the reflection
and refraction of rays of and refraction of rays of
light.light.
SF027 2
Geometrical Optics
Reflection
Plane surface
Refraction
Spherical surface
Plane surface
- Law of reflection
- Ray diagram (plane mirror)
- Characteristic of image
- Term in used: principal axis. Centre of curvature( c ),radius of curvature( r ), focal point( F ),focal length( f ),pole of the spherical mirror
- Ray diagram (spherical mirror)
- Characteristic of image
- Sign convention( u, v , f )
- Snell’s Law
- Refractive index( n )
- Material of different n
- Convex and concave lenses
- Term in used: principal axis. focal point( F ),focal length( f ),optical centre of lens
- Ray diagram (lenses)
- Characteristic of image
- Sign convention( u, v , f )
- Thin lens formula:
Thin lenses
f
1
v
1
u
1=+
SF027 3
9.1 Reflection of Plane Mirror 9.1.1 Reflection of Plane Mirror
� Definition – is defined as the return of all or part of a beam of particles or waves when it encounters the boundary between twomedia.
� Laws of reflection state :
� The incident ray, the reflected ray and the normal all lie in the same plane.
� The angle of incidence, i is equal to the angle of reflection, r as shown in figure below.
i r
Plane mirrorPlane mirror
ri =
Simulation
SF027 4
A 'A
i
u v
i
r
i
� Image formation by a plane mirror. (ray diagrams)
� Point object
� Vertical (extended) object
Objecti
v
i
rr
u
Image
ihoh
distanceobject :uwhere
distance image :vheightobject :ohheight image :ih
Simulation
SF027 5
� The properties of image formed are
� virtual
� upright or erect
� laterally reverse
� the object distance, u is equal to the image distance, v� same size as the object where the linear magnification is
given by
� obey the laws of reflection.
� Example 1 :
Find the minimum vertical length of a plane mirror for an observer of
2.0 m height standing upright close to the mirror to see his whole
reflection. How should this minimum length mirror be placed on the
wall?
1h
hM
o
i ==height,Object
height, Image
SF027 6
Solution: By using the ray diagram as shown in figure below.
HE2
1AL =
EF2
1LB =
The minimum vertical length of the mirror is given by
LBALh +=
EF2
1HE
2
1h +=
( )EFHE2
1h +=
Height of observer
m01h .=
)head(H A)eyes(E
)feet(F
B
L
h
SF027 7
The mirror can be placed on the wall with the lower end of the mirror
is halved of the distance between the eyes and feet of the observer.
� Example 2 :
A rose in a vase is placed 0.250 m in front of a plane mirror. Nagar
looks into the mirror from 2.00 m in front of it. How far away from
Nagar is the image of the rose?
Solution: u=0.250 m
u v
m002 .
x
From the properties of the
image formed by the plane
mirror, thus
Therefore, the distance between Nagar and the image of the rose is
given by
SF027 8
9.2 Reflection at a Spherical surface9.2.1 Spherical mirror
� Definition – is defined as a reflecting surface that is part of a sphere.
� There are two types of spherical mirror. It is convexconvex (curving outwards) and concaveconcave (curving inwards) mirror.
� Figures below show the shape of concave and convex mirrors.
� Some terms 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 mirror forms a part.
(a) Concave (ConvergingConverging) mirror (b) Convex (DivergingDiverging) mirror
reflecting surface
imaginary sphere
CC CC
AA
BB
AA
BB
silver layer
r rPP PP
SF027 9
�� Radius of curvature, Radius of curvature, rr� is defined as the radius of the sphere of which a curved
mirror forms a part.
�� Pole or vertex (point P)Pole or vertex (point P)
� is defined as the point at the centre of the mirror.
�� Principal axisPrincipal axis
� is defined as the straight line through the centre of curvature C and pole P of the mirror.
� AB is called the aperture aperture of the mirror.
9.2.2 Focal point and focal length, f� Consider the ray diagram for concave and convex mirror as shown in
figures below.
CC
FFf
Incident Incident
raysrays
PP CC
FFf
Incident Incident
raysrays
PP
SF027 10
� From the figures,
� Point F represents the focal point or focus of the mirrors.
� Distance f represents the focal length of the mirrors.� The parallel incident rays represent the object infinitely far away
from the spherical mirror e.g. the sun.
�� Focal point or focus, FFocal point or focus, F
� for concave mirror – is defined as a point where the incident parallel rays converge after reflection on the mirror.
� Its focal point is real (principal).
� for convex mirror – is defined as a point where the incident parallel rays seem to diverge from a point behind the mirror after reflection.
� Its focal point is virtual.
�� Focal length, Focal length, ff� Definition – is defined as the distance between the focal point
(focus) F and pole P of the spherical mirror.
� The paraxial raysparaxial rays is defined as the rays that are near to and almost parallel to the principal axis.
SF027 11
9.2.3 Relationship between focal length, f and radius of curvature, r� Consider a ray AB parallel to the principal axis of concave mirror as
shown in figure below.
� From the figure,
�BCD
�BFD
� By using an isosceles triangle CBF, thus the angle θ is given by
CC
FF
incident rayincident ray
PPDD
BBAA
fr
iiθ
i
iCD
BDi ≈=tan
θθ ≈=FD
BDtan
Taken the angles are << Taken the angles are <<
small by considering the ray small by considering the ray
AB is paraxial ray.AB is paraxial ray.
i2=θ
SF027 12
then
� Because of AB is paraxial ray, thus point B is too close with pole P then
� Therefore
rCPCD =≈fFPFD =≈
This relationship also valid This relationship also valid
for convex mirror.for convex mirror.
2
rf =
=CD
BD2
FD
BD
or
FD2CD =
f2r =
SF027 13
9.2.4 Ray Diagrams for Spherical Mirrors
� Definition – is defined as the simple graphical method to indicate the positions of the object and image in a system of mirrors or lenses.
� Ray diagrams below showing the graphical method of locating an image formed by concave and convex mirror.
�� 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.
(a) Concave mirror (b) Convex mirror
CC PP
FF
11
33
33
11
I CC
FF
PP
11
22
22O O I
22
33
11
22
At least any At least any
two rays two rays
for drawing for drawing
the ray the ray
diagram.diagram.
SF027 14
9.2.5 Images formed by a convex mirror
� Ray diagrams below showing the graphical method of locating an image formed by a convex mirror.
� Properties of image formed are
� virtual
� upright
� diminished (smaller than the object)
� formed at the back of the mirror
� Object position → any position in front of the convex mirror.
CC
FF
PP
O I
u v
FrontFront backback
SF027 15
Image propertyRay diagramObject
distance, u
I
9.2.6 Images formed by a concave mirror
� Table below shows the ray diagrams of locating an image formed by a
concave mirror for various object distance, u.
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.CC
FF
PP
FrontFront backback
SF027 16
Image propertyRay diagramObject
distance, u
FFCC PP
FrontFront backback
f < u < rf < u < r
u = fu = f
O
� Real
� Inverted
� Magnified
� Formed at a distance greater than CP.
� Real
� Formed at infinity.
IO
CC
FF
PP
FrontFront backback
SF027 17
Image propertyRay diagramObject
distance, u
� Linear (lateral) magnification of the spherical mirror, M is defined as the ratio
between image height, hi and object height, ho
Negative sign indicates that the object and image are on opposite sides of the
principal axis (refer to the real image), If ho is positive, hi is negative.
u < fu < f
O
� Virtual
� Upright
� Magnified
� Formed at the back of the mirror
IFF
CC PP
FrontFront backback
i
o
h vM
h u= =
where
pole from distance image :vpole from distanceobject :u
Simulation
SF027 18
O CC PPIv
u
BB
θθ
α φ βDD
� By considering point B very close to the pole P, hence
then
� From the figure,
�BOC
�BCI
then, eq. (1)-(2) :
By using �BOD, �BCD and �BID thus
9.2.7 Derivation of Spherical mirror equation
� Figure below shows an object O at a distance u and on the principal
axis of a concave mirror. A ray from the object O is incident at a point
B which is close to the pole P of the mirror.
θαφ += (1)(1)
θφβ += (2)(2)
φαβφ −=−φβα 2=+ (3)(3)
ID
BD
CD
BD
OD
BD=== βφα tan; tan ; tan
vIPIDrCPCDuOPOD =≈=≈=≈ ; ;
v
BD
r
BD
u
BD=== βφα ; ;
ββφφαα ≈≈≈ tan; tan ; tan
Substituting this Substituting this
value in eq. (3)value in eq. (3)
SF027 19
therefore
� Table below shows the sign convention for equation of spherical mirror .
f2r =
=+r
BD2
v
BD
u
BD
r
2
v
1
u
1=+ where
v
1
u
1
f
1 += Equation (formula) Equation (formula)
of spherical mirrorof spherical mirror
Negative sign (-)Positive sign (+)Physical Quantity
Object distance, u
Image distance, v
Focal length, f
Linear
magnification, M
Real object Virtual object
Real image Virtual image
Concave mirror Convex mirror
Upright (erect)
imageInverted image
(same side of the object) (opposite side of the object)
(in front of the mirror) (at the back of the mirror)
SF027 20
� Example 7 :
An object is placed 10 cm in front of a concave mirror whose focal length is 15 cm. Determine
a. the position of the image.
b. the linear magnification and state the properties of the image.
Solution: u=+10 cm, f=+15 cm
a. By applying the equation of spherical mirror, thus
The image is 30 cm from the mirror on the opposite side of the
object (or 30 cm at the back of the mirror).
b. The linear magnification is given by
cm30v −=
3M =
SF027 21
� Example 8 :
An upright image is formed 30 cm from the real object by using the spherical mirror. The height of image is twice the height of object.
a. Where should the mirror be placed relative to the object?
b. Calculate the radius of curvature of the mirror and describe the type
of mirror required.
Solution: hi=2ho
a. From the figure above,
By using the equation of linear magnification, thus
O Icm30
Spherical Spherical
mirrormirror
u v
SF027 22
By substituting eq. (2) into eq. (1), hence
b. By using the equation of spherical mirror,
and therefore
The type of spherical mirror is concaveconcave because the positive value
of focal length.
SF027 23
� Example 9 :
A mirror on the passenger side of your car is convex and has a radius of curvature 20.0 cm. Another car is seen in this side mirror and is 11.0 m behind the mirror. If this car is 1.5 m tall, calculate the height of the car image . (Similar to No. 34.66, pg. 1333, University Physics with Modern Physics,11th edition, Young & Freedman.)
Solution: ho=1.5x102 cm, r=-20.0 cm, u=+11.0x102 cm
By applying the equation of spherical mirror,
From equation of linear magnification,
cm351hi .=
and
SF027 24
I
O
CC FF
PP
cm035 .
mm05 .
m203 .
u
� Example 10 :
A concave mirror forms an image on a wall 3.20 m from the mirror of the filament of a headlight lamp. If the height of the filament is 5.0 mm and the height of its image is 35.0 cm, calculate
a. the position of the filament from the pole of the mirror.
b. the radius of curvature of the mirror.
Solution: hi=-35.0 cm, v=3.20x102 cm, ho=0.5 cm
a. By applying the equation of linear magnification,
cm574u .=cm019r .=
SF027 25
b. By applying the equation of spherical mirror, thus
� Example 11 : (exercise)
a. A concave mirror forms an inverted image four times larger than the object. Find the focal length of the mirror, assuming the distance between object and image is 0.600 m.
b. A convex mirror forms a virtual image half the size of the object. Assuming the distance between image and object is 20.0 cm, determine the radius of curvature of the mirror.
No. 14, pg. 1169,Physics for scientists and engineers with modern physics,
Serway & Jewett,6th edition.
Ans. : 160 mm, -267 mm
and
SF027 26
9.3 Refraction on plane surface� Definition – is defined as the changing of direction of a light ray and its
speed of propagation as it passes from one medium into another.
� Laws of refraction state :
� The incident ray, the refracted ray and the normal all lie in the same plane.
� For two given media,
incidence of angle :iwhere
refraction of angle :r1 medium theofindex refractive:1nray)incident thecontaining Medium(
constantsin
sin==
1
2
n
n
r
i
rnin 21 sinsin =
Or
2 medium theofindex refractive:2nray) refracted thecontaining Medium(
SnellSnell’’s laws law
SF027 27
� Examples for refraction of light ray travels from one medium to another medium can be shown in figures below.
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 (Medium 1 is less
dense than medium 2)dense than medium 2)(Medium 1 is denser (Medium 1 is denser
than medium 2)than medium 2)
ir <
The light ray is bent toward the
normal, thus
The light ray is bent away from the
normal, thus
Simulation-1 Simulation-2
SF027 28
� Refractive index (index of refractionindex of refraction)
� Definition – is defined as the constant ratio for the two given media.
� The value of refractive index depends on the type of medium and the colour of the light.
� It is dimensionless and its value greater than 1.
� Consider the light ray travels from medium 1 into medium 2, the refractive index can be denoted by
� Absolute refractive index, n (for the incident ray is travelling in
vacuumvacuum or airor air and is then refracted into the medium concernedmedium concerned) is given by
r
i
sin
sin
2
121
v
vn ==
2 mediumin light ofvelocity
1 mediumin light ofvelocity
(Medium containing (Medium containing
the incident ray)the incident ray)
(Medium containing (Medium containing
the refracted ray)the refracted ray)
v
cn ==
mediumin light ofvelocity
in vacuumlight ofvelocity
SF027 29
� Table below shows the indices of refraction for yellow sodium light having a wavelength of 589 nm in vacuum.
(If the density of medium is greater hence the refractive index (If the density of medium is greater hence the refractive index is also greater)is also greater)
SF027 30
� The relationship between refractive index and the wavelength of light.
� As light travels from one medium to another, its wavelength, wavelength, λλλλλλλλchangeschanges but its frequency, frequency, ff remains constantremains constant.
� The wavelength changes because of different materialdifferent 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 wavescannot create or destroy waves.
� By considering a light travels from medium 1 (n1) into medium 2
(n2), the velocity of light in each medium is given by
then11 fv λ=
22 fv λ=and
2
1
2
1
f
f
v
v
λλ
= where
1
1n
cv =
2
2n
cv =and
2
1
2
1
n
c
n
c
λλ
=
2211 nn λλ =
(Refractive index is inversely (Refractive index is inversely
proportional to the wavelength)proportional to the wavelength)
SF027 31
� If medium 1 is vacuum or air, then n1 = 1. Hence the refractive
index for any medium, n can be expressed as
� Example 3 :
A fifty cent coin is at the bottom of a swimming pool of depth 2.00 m.
The refractive index of air and water are 1.00 and 1.33 respectively.
What is the apparent depth of the coin?
Solution: na=1.00, nw=1.33
where
λλ0n = in vacuumlight ofh wavelengt:0λ
mediumin light ofh wavelengt:λ
wheredepthapparent :AB
m 2.00 depth actual : =AC
A
i
Air (na)
C
r
B
Water (nw)
i
r
m002 .
D
SF027 32
From the diagram,
�ABD
�ACD
By considering only small angles of r and i , hence
From the Snell’s law,
and
AB
ADr =tan
w
a
n
n
AC
AB=
m501AB .=
rr sintan ≈
AC
ADi =tan
depthapparent
depth real=n
ii sintan ≈
AC
AB
AB
AD
AC
AD
r
i
r
i=
==sin
sin
tan
tan
w
a
1
2
n
n
n
n
r
i==
sin
sin
then
Note : Note : (Important)(Important)
Other equation for absolute
refractive index in term of
depth is given by
SF027 33
� Example 4 :
A light beam travels at 1.94 x 108 m s-1 in quartz. The wavelength of
the light in quartz is 355 nm.
a. Find the index of refraction of quartz at this wavelength.
b. If this same light travels through air, what is its wavelength there?
(Given the speed of light in vacuum, c = 3.00 x 108 m s-1)
No. 33.3, pg. 1278, University Physics with Modern Physics,11th edition, Young &
Freedman.
Solution: v=1.94 x 108 m s-1, λ=355 x 10-9 ma. By applying the equation of absolute refractive index, hence
b. By using the equation below, thus
551n .=
nm550m10x505 7
0 @. −=λ
SF027 34
� Example 5 : (exercise)
We wish to determine the depth of a swimming pool filled with water by measuring the width (x = 5.50 m) and then noting that the bottom edge of the pool is just visible at an angle of 14.0° above the horizontal as shown in figure below. (Gc.835.60)
Calculate the depth of the pool. (Given nwater = 1.33 and nair = 1.00)
Ans. : 5.16 m
� Example 6 : (exercise)
A person whose eyes are 1.54 m above the floor stands 2.30 m in front of a vertical plane mirror whose bottom edge is 40 cm above the floor as shown in figure below. (Gc.832.10)
Find x.
Ans. : 0.81 m
SF027 35
9.4 Thin Lenses� Definition – is defined as a transparent material with two spherical
refracting surfaces whose thickness is thin compared to the radii of curvature of the two refracting surfaces.
� There are two types of thin lens. It is convergingconverging and diverging diverging lens.
� Figures below show the various types of thin lenses, both converging and diverging.
(a) Converging (Convex) lensesConverging (Convex) lenses
(b) Diverging (Concave) lensesDiverging (Concave) lenses
BiconvexBiconvex PlanoPlano--convexconvex Convex meniscusConvex meniscus
BiconcaveBiconcave PlanoPlano--concaveconcave Concave meniscusConcave meniscus
SF027 36
9.4.1 Terms of lens
� Figures below show the shape of converging (convex) and diverging (concave) lenses.
�� Centre of curvature (point CCentre of curvature (point C11 and Cand C22))
� is defined as the centre of the sphere of which the surface of the lens is a part.
�� Radius of curvature (rRadius of curvature (r11 and rand r22))
� is defined as the radius of the sphere of which the surface of the lens is a part.
�� Principal (Optical) axisPrincipal (Optical) axis
� is defined as the line joining the two centre's of curvature of a lens.
�� Optical centre (point O)Optical centre (point O)
� is defined as the point at which any rays entering the lens pass without deviation.
(a) Converging lens (b) Diverging lens
CC11 CC22
rr11
rr22OO CC11 CC22
rr11
rr22
OO
SF027 37
FF11 FF22OO
ff
FF11 FF22OO
ff
9.4.2 Focus (Focal point) and focal length
� Consider the ray diagrams for converging and diverging lens as shown in figures below.
� From the figures,
� Point F1 and F2 represent the focus of the lens.
� Distance f represents the focal length of the lens.�� Focus (point FFocus (point F11 and Fand F22))
� For converging (convex)converging (convex) lens – is defined as the point on the principal axis where rays which are parallel and close to the principal axis converges after passing through the lens.
� Its focus is real (principal).
� For diverging (concave)diverging (concave) lens – is defined as the point on the principal axis where rays which are parallel to the principal axis seem to diverge from after passing through the lens.
� Its focus is virtual.
SF027 38
FF11
FF22
�� Focal length ( Focal length ( f f ))� Definition – is defined as the distance between the focus F and the
optical centre O of the lens.
9.4.3 Ray Diagrams for Lenses
� Ray diagrams below showing the graphical method of locating an image formed by converging (convex) and diverging (concave) lenses.
(a) Converging (convex) lens
11
11
22
22
OO
33
33
II
u v
SF027 39
(b) Diverging (concave) lens
�� 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 not deviated.
�� Ray 3Ray 3 - Passes through the focus point F1 of a converging lens or appears to converge towards the focus F1 of a diverging lens, after refraction by the lens the ray parallel to the principal axis.
OO FF22FF11
11
11
22
22
33
33
II
vu
At least any At least any
two rays two rays
for drawing for drawing
the ray the ray
diagram.diagram.
SF027 40
9.4.4 Images formed by a diverging (concave) lens
� Ray diagrams below showing the graphical method of locating an image formed by a diverging lens.
� Properties of image formed are
� virtual
� upright
� diminished (smaller than the object)
� formed in front of the lens.
� Object position → any position in front of the diverging lens.
FrontFront backback
OO FF22FF11II
SF027 41
Image propertyRay diagramObject
distance, u
FF11FF22
OO 2F2F22
2F2F11
9.4.5 Images formed by a converging lens
� Table below shows the ray diagrams of locating an image formed by a
converging lens for various object distance, u.
FrontFront backback
u > 2fu > 2f
u = 2fu = 2f
� Real
� Inverted
� Diminished
� Formed between point F2 and 2F2.
(at the back of the lens)
� Real
� Inverted
� Same size
� Formed at point 2F2. (at the back of the lens)
FF11FF22OO 2F2F222F2F11
I
FrontFront backback
I
SF027 42
Image propertyRay diagramObject
distance, u
FF11FF22
OO
2F2F222F2F11
f < u < 2ff < u < 2f
u = fu = f
� Real
� Inverted
� Magnified
� Formed at a distance greater
than 2f at the back of the lens.
� Real
� Formed at infinity.
FF11FF22OO 2F2F222F2F11
FrontFront backback
I
FrontFront backback
SF027 43
Image propertyRay diagramObject
distance, u
� Linear (lateral) magnification of the thin lenses, M is defined as the ratio
between image height, hi and object height, ho
Negative sign indicates that when u and v are both positive, the image is
inverted and ho and hi have opposite signs.
u < fu < f
� Virtual
� Upright
� Magnified
� Formed in front of the lens.
u
v
h
hM
o
i −==where
centre optical from distance image :vcentre optical from distanceobject :u
Simulation
FF11 FF22OO 2F2F222F2F11
FrontFront backback
I
SF027 44
Thin Lenses Formula and Lens maker’s
Equation
u
O
vo
v
Io
n1
n2
I
first surfacesecond surface
A ray from an object O in medium of refractive index n1 passes
through the first surface of thin lens with refractive index n2. An
image is formed (Io).
Io acts as the virtual object for the second surface of the lens and
finally form a final image
SF027 45
� By using the equation of spherical refracting surface, the refraction by first surface and second surface are given by
�� Convex surface (Convex surface (r = +rr = +r11))
�� Concave surface (Concave surface (r = r = --rr22))
adding up eq. (1) and eq. (2):
1
12
o
21
r
)nn(
v
n
u
n −=+
2
211
o
2
r
)nn(
v
n
v
n
−−
=+−
(1)(1)
2
21
1
121
o
2
o
21
r
)nn(
r
nn
v
n
v
n
v
n
u
n
−−
+−
=+−
++
(2)(2)
2
12
1
1211
r
nn
r
nn
v
n
u
n −+
−=+
SF027 46
+−=
+21
121r
1
r
1)nn(
v
1
u
1n
+
−=
211
2
r
1
r
11
n
n
f
1
+
−=+
211
12
r
1
r
1
n
nn
v
1
u
1
Lens makerLens maker’’s s
equationequation
where
length focal :fsurface refractingfirst of curvature of radius :1r
medium theofindex refractive :1nmaterial lens theofindex refractive :2n
surface refracting second of curvature of radius :2r
�If u at infinity then v = f, hence eq. (3) becomes
(3)(3)
SF027 47
� By equating eq. (3) with lens maker’s equation, hence
therefore in general,
� Note :
� If the medium is airair (n1= nair=1) thus the lens maker’s equation will be
� For thin lens formula and lens maker’s equation, Use the sign sign
conventionconvention for refractionrefraction.
� The radius of curvature of flat refracting surface is infinity, r = r = ∞∞∞∞∞∞∞∞.
f
1
v
1
u
1=+
v
1
u
1
f
1+= Thin lens formulaThin lens formula
where material lens theofindex refractive :n
( )
+−=
21 r
1
r
11n
f
1
Very ImportantVery Important
SF027 48
� Example 16 :
A biconvex lens is made of glass with refractive index 1.52 having the radii of curvature of 20 cm respectively. Calculate the focal length of the lens in
a. water,
b. carbon disulfide.
(Given nw = 1.33 and nc=1.63)
Solution: r1=+20 cm, r2=+20 cm, ng=n2=1.52
a. Given the refractive index of water, nw = n1By using the lens maker’s equation, thus
b. Given the refractive index of carbon disulfide, nc = n1By using the lens maker’s equation, thus
+
−=
21w
g
r
1
r
11
n
n
f
1
cm70f +=
+
−=
21c
g
r
1
r
11
n
n
f
1
cm18148f .−=
SF027 49
� Example 17 :
A converging lens with a focal length of 90.0 cm forms an image of a 3.20 cm tall real object that is to the left of the lens. The image is 4.50 cm tall and inverted. Find
a. the object position from the lens.
b. the image position from the lens. Is the image real or virtual?
No. 34.26, pg. 1331, University Physics with Modern Physics,11th edition,
Young & Freedman.
Solution: f=+90.0 cm, ho=3.20 cm, hi=-4.50 cma. By using the linear magnification equation, hence
By applying the thin lens formula,
u
v
h
hM
o
i −==
u411v .=
v
1
u
1
f
1+=
v
1
u
1
090
1+=
.
(1)(1)
(2)(2)
SF027 50
By substituting eq. (1) into eq. (2),hence
The object is placed 154 cm in front of the lens.
b. By substituting u = 154 cm into eq. (1),therefore
The image forms 217 cm at the back of the lens (at the
opposite side of the object placed) and the image is real.
� Example 18 :
An object is placed 90.0 cm from a glass lens (n=1.56) with one concave surface of radius 22.0 cm and one convex surface of radius 18.5 cm. Determine
a. the image position.
b. the linear magnification. (Gc.862.28)
Solution: u=+90.0 cm, n=1.56, r1=-22.0 cm, r2=+18.5 cma. By applying the lens maker’s equation in air,
cm217v =
cm154u =
( )
+−=
21 r
1
r
11n
f
1
cm208f +=
SF027 51
By applying the thin lens formula, thus
The image forms 159 cm in front of the lens (at the same side
of the object placed)
b. By applying equation of linear magnification for thin lens, thus
� Example 19 : (exercise)
A glass (n=1.50) plano-concave lens has a focal length of 21.5 cm. Calculate the radius of the concave surface. (Gc.862.26)
Ans. : -10.8 cm
� Example 20 : (exercise)
An object is 16.0 cm to the left of a lens. The lens forms an image 36.0 cm to the right of the lens.
a. Calculate the focal length of the lens and state the type of the
lens.
b. If the object is 8.00 mm tall, find the height of the image.
c. Sketch the ray diagram for the case above. (UP. 1332.34.34)
Ans. : +11.1 cm, -1.8 cm
cm159v −=
771M .=
v
1
u
1
f
1+=
u
vM −=
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