Example 7.3.1
An object 450 mm from a converging lens forms a real image
900 mm from the lens. What is the focal length of the lens.
Solution
5500
Example 7.3.2
An object is placed 10 cm from the a 15 cm focal length
converging lens. Find the image position and its characteristics.
Solution
Characteristics
- 3 x as large as the object
- upright
- virtual/in front of the lens)
5511
Example 7.3.3
An object 6 cm high is held 4 cm from a diverging meniscus lens of
focal length -24 cm. What are the nature, size, and location of the
image ?
Solution
virtual, upright 5522
5.16 cm, -.3.43 cm
Example 7.3.4
You are given a thin diverging lens. You find that a beam of parallel rays
spreads out after passing through the lens as though all the rays came from a
point 20.0 cm from the center of the lens. You want to use this lens to form
an erect virtual image that is 1/3 the height of the object.
(a) Where should the object be placed ?
53
Solution 7.3.4
a) f = – 20 cm = – 0.2
m
54
Example 7.3.5
A plano-convex is to be constructed out of glass so that it has a
focal length of 40 cm. What is the radius of curvature of the curve
surface.
Solution:
Given : R1 = ∞, R2 = ?, f = 40 cm, n = 1.50
55
Example 7.3.6
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
a. b.
56
57
7.4 HUYGEN’S PRINCIPLE
At the end of this subtopic, the students should be able to:
a) State Huygens's Principle (e.g. spherical and plane waves)
b) Sketch and explain the wave front of light after passing through
a single slit and obstacle using Huygens's principle.
58
7.4 a) STATE Huygens's Principle (e.g. spherical and plane waves)
HUYGEN’S PRINCIPLE
Huygens' principle is a geometrical method for
finding the shape of the new wave from the
known shape of wavefront.
Huygens's Principle state that :
Every point on a wavefront is a source of wavelets that
spread out in the forward direction at the same speed
as the wave itself. The new wavefront is a line tangent
to all of the wavelets.
59
The most common forms of wavefronts
are spherical and planar (plane).
(a) Spherical Wavefront (b) Plane wavefront
60
7.4 b) SKETCH and EXPLAIN the wave front of light after passing through a single slit
and obstacle using Huygens's principle.
Step of drawing the wave front
1) Draw a plane / plane spherical New wavefront
spherical wave front. wavefront wavefront at time t
2) Choose a few point on
the wave front ( ~ 5
points ).
Wavefront at New
3) Draw the wavelets for time t = 0 wavefront
at time t Wavefront at time
t=0
each point – (make
sure the wavelets is
forwarded semicircles
and all the wavelets
have the same radius).
4) Join all the tangent line l l
of the wavelets to build
the new wavefront.
61 WAVE FRONT OF LIGHT AFTER PASSING THROUGH A
SINGLE SLIT and OBSTACLE
a a a
A BC
Explanation:
According to Huygens’s principle, every point on a wave-
front can behave as a secondary source of light which can
propagate out light wave with the speed of light.
Figure above shows a new wave-front which is
constructed by drawing a line tangent to all the wavelets
from the secondary sources of the old wave-front.
After drawing several wave-front, the wave is seen to
bend round the edges of single slit or the obstacle.
TIfhthee sbizeenodf tihnegsliot fiswsmaavlle(as<<isλ)c, athlelendifafrsacdtioifnfrwailcl oticocunr.as shown in Figure C.
62
Same goes to obstacle, the edge of the wavefront bend after passing
through the obstacle.
Click here for simulation:
https://phet.colorado.edu/en/simulations/wave-interference
63
7.5 CONSTRUCTIVE & DESTRUCTIVE
INTERFERENCE
Learning Outcomes:
At the end of this subtopic, the students should be able to:
(a) Define coherence
(b) State the conditions for interference of light
(c) State the conditions of constructive interference and destructive
interference.
Remarks :
Emphasizes on the path difference and its equivalence to phase
difference.
64
INTERFERENCE PHENOMENON
Ü Interference is the process of overlapping of two or more
waves.
Ü To produce the stable interference, the overlapping of the
wave must obey 2 conditions:
i. The source must be coherent
ii. The source must have roughly the same amplitude.
( to obtain total cancellation at minimum and a good
contrast at maximum)
Ü What is the ‘coherence sources’ ???
Ü Coherent means the source of overlapping wave must have:
i. The same wavelength or frequency
ii. Constant phase difference
65
7.5 a) DEFINE coherence
Coherent sources are those which emit light waves of same
wavelength or frequency which are always in phase with each
other or have a constant phase difference.
66
2 types of coherent source:
(a) In phase coherent source (b) Anti phase coherent
source
Both S1 and S2 start with crest S1 start with trough
Or S2 start with crest
Both S1 and S2 start with trough
67 7.5 b) STATE the conditions for interference of light
Interference of Light waves
When 2 coherent light waves meet in a region, they undergo
superposition and result in constructive and destructive
interference.
Principle of Superposition:
“ When two or more waves overlap, the resultant
displacement at any instant of superposition is equal to the
vector sum of the individual displacements of the waves.”
68
The conditions for the interference of light
1. The source of light should be coherent (same
wavelength and constant phase difference.
2. Waves should be of the same frequency.
3. Direction of waves should also be the same.
4. The amplitudes of both the waves should also be
the same.
5. The slits of both the sources should be small
enough that it can be considered a point source
of light
69 Light Path 1
Path difference (PD)
• is difference in
distance (in terms of λ ,
unit m) traveled by two
waves from their
respective sources to
the same point in a
region.
• Is used to determine Light Path 2
the phase difference Path difference
(in terms of π, unit
radian) between the
two light rays when
they meet at a point.
70
7.5 c) STATE the conditions of constructive interference and destructive
interference.
TYPE OF INTERFERENCE
1. Constructive Interference :
It occurs when the crest of one wave falls on the crest of
another wave such that the amplitude is maximum.
Reinforcement of amplitude that will produce a bright
fringes (maximum)
The resultant wave amplitude is greater.
71 A=1
Constructive Interference +
A=1
§ The amplitude of the
resultant waves is the sum =
of the amplitudes of the A=2
individual waves which
reinforce each other.
§ If a crest of a wave meets a
crest of another wave at the
same point, the interference
constructively.
Two coherent sources are
in phase
72
Constructive interference : In- phase sources
Constructive interference for in-phase sources occurs if and
only if the path difference, ∆L between the waves has to
be any whole number of wavelength:
∆ = ! − " = 3 − 3 = 0
l Path difference, ∆ = 0, λ, 2λ, 3λ,…
l ∆ = mλ
l
where m =0,1,2,3,…
l l
l Phase difference, ΔΦ = m (2p)
ΔΦ = 0, 2π, 4π, ...
where m = 0, 1, 2, …
73
Constructive Interference : Anti- phase sources
Constructive interference for anti-phase sources occurs if and
only if the path difference, ∆L between the waves is an odd
multiple of λ/2 wavelength:
∆ = ! − " = 3 2 − 4 = 2
l/2 l l l Path difference, ∆ =#" , $# , %# , &"#,…
l " "
DL l ∆ = + λ
l where m =
l 0,P1h,2a,s3e,…difference
ΔΦ = (m + ½ ) (2p)
= π, 3π, 5π, …
where m = 0, 1, 2, 3, …
74
2. Destructive Interference
It occurs when the crest of one wave falls on the trough of
another wave such that the amplitude is minimum.
Total cancellation of amplitude that will produce a dark
fringes (minimum)
The overall amplitude is decreased.
75
Destructive Interference
§ The amplitude of the +
resultant waves is zero. =
§ If a crest of a wave meets Resultant A = 0
the crest, the interference
destructively.
Two coherent sources are
in anti-phase
76
Destructive Interference : Anti- phase sources
Destructive interference for anti-phase sources occurs if and
only if the path difference, ∆L between the waves has to be any
whole number of wavelength:
∆ = ! − " = 3 − 3 = 0
Path difference, ∆ = 0, λ, 2λ, 3λ,…
l ∆ = mλ
l where m =0,1,2,3,…
l Phase difference, ΔΦ = m (2p)
ΔΦ = 0, 2π, 4π, ...
l where m = 0, 1, 2,…
l
l
77
Destructive Interference : In- phase sources
Destructive interference for in-phase sources occurs if and only
if the path difference, ∆L between the waves is an odd
multiple of λ/2 wavelength:
∆ = ! − " = 3 − 4 =
2 2
=#" $# %# &"#,…
Path difference, ∆ , " , " ,
l/2 l ∆ = + λ
l lO
where m =0,1,2,3,…
Phase difference
DL l ΔΦ = (m + ½ ) (2p)
l = π, 3π, 5π, …
l where m = 0, 1, 2, 3, …
l
78
SUMMARY
Condition (Path Difference) of Constructive & Destructive
Interference:
Two coherence Interference Path Difference
sources
In–phase Constructive ml
Destructive (m + ½) l
Anti–phase Constructive (m + ½) l
Destructive ml
79
7.6 INTERFERENCE OF TRANSMITTED LIGHT
THROUGH DOUBLE-SLITS
Learning Outcomes:
At the end of this subtopic, the students should be able to:
(a) USE: mλD
d
i) ym = for bright fringes ( maxima)
æ m + 1 ö λD for dark fringes (minima) where m = 0, ±1,±2,±3,...
çè 2 ø÷
ii) y m = d
(b) USE Δy = λD and EXPLAIN the effect of changing any of the
d
variables.
80
YOUNG DOUBLE-SLITS
EXPERIMENT
https://phet.colorado.edu/en/simulations/wave-interference
81
Introduction of Young Double-Slits Experiment:
Ü One of the experiment that shows us the results of interference
phenomenon.
FIGURE (a):
§ The monochromatic wave of wavelength, λ are shown entering
the slits S1 and S2 which are at a distance, d apart.
§ After passing the slits, the wave spread out in all direction.
82 FIGURE (a):
§ Since these two waves emerged from the same source , they could
be considered coming from two coherent sources.
§ Light waves from these two sources would interfere.
§ The interference pattern was then projected onto a screen and we
can see the pattern of alternating bright and dark fringes .
FIGURE (b): FIGURE (c):
3-Dimension of interference process Interference pattern on screen
83
Ym
d tan = !
D
For small angle, sin ≈ tan
!
sin =
Ü Consider Young’s Double Slit’s schematic diagram as shown in
the figure, Let P be a point on the screen.
D = the distance between the slits and the screen
d = the separation distance between the slits, S1 and S2
Ym = the distance from the centre of the central maximum to the
mth fringe ( bright or dark fringe)
84
7.6 a) Use equation for bright and dark fringes
Constructive interference
(Bright Fringes / Maximum)
ym ym = mλD
d
d Destructive interference
D (Dark Fringes / Minimum)
æ m + 1 ö λD
èç 2 ø÷
ym =
d
85
Constructive interference Destructive interference
( Bright Fringes / Maximum) (Dark Fringes / Minimum)
ym = mλD æ m + 1 ö λD
d èç 2 ÷ø
ym =
d
Meaning of each symbol :
ym = distance of fringes from central maximum
m = order number
λ = wavelength
d = distance between both slits
D = distance between the slit and the screen
86
HOW TO DETERMINE THE ORDER IN DOUBLE-SLITS
INTERFERENCE
ym = mλD æ m + 1 ö λD
d èç 2 ÷ø
ym =
d
Bright Bright Order, Dark Dark Order,
fringe order m fringe order m
0 0
Central 0th 1 1st 0th 1
1st 2 2nd 1st 2
1st 2nd 3 3rd 2nd 3
2nd 3rd … 4th 3rd
3rd … …
… … …
87
m=3
m=3
m=2
m=2
m=1
m=1
m=0
m=0
m=0
m=1
m=1
m=2
m=2
m=3
m=3
88
7.6 b) Use equation distance between adjacent fringes and
explain the effect of changing any of the variables.
• Equation:
∆y Δy = λD
d
∆y where :
screen ∆y = distance between adjacent bright fringes /
adjacent dark fringes
λ = wavelength
D = distance between slits and screen
d = separation distance between the slits
89
THE EFFECT OF CHANGING, ,D AND d TO THE
INTERFERENCE PATTERN, ∆y
Ü Interference pattern, ∆y is depends on λ, D and d Δy = λD
d
Ü If D and d constant,
Ä λ increase : ∆y increase : interference pattern is become wider.
Ä λ decrease : ∆y decrease : interference pattern is become
closer.
Ü If λ and d constant,
Ä D increase : ∆y increase : interference pattern is become wider.
Ä D decrease : ∆y decrease : interference pattern is become
closer.
Ü If λ and D constant,
Ä d increase : ∆y decrease : interference pattern is become closer
Ä d decrease : ∆y increase : interference pattern is become wider.
90
SUMMARY
Bright Fringe / Maximum Dark Fringe / Minimum
d sin θ = mλ d sin θ = æ m + 1 ö λ
èç 2 ø÷
ym = mλD æ m + 1 ö λD
d çè 2 ÷ø
ym =
d
Adjacent / Consecutive Bright @ Dark Fringes
Δy = λD
d
91
EXAMPLE 7.6.1 Solution:
An interference pattern is ym = mλD
formed on a screen when light d
of wavelength 550 nm is
incident on two parallel 4.5× 10-2 = 2 ( 550× 10-9 )D
slits 50 μm apart. 50× 10-6
The second order bright m=2 3D
m=2 2B
m=1 2D D = 2.05 m
fringes is 4.5 cm y2 = 4.5 cm m=1 1B
from the central m=0 1D
maximum. m=0 CB
How far from the slits is the
screen? m=0 1D
m=1 1B
m=1 2D
m=2 2B
m=2 3D
92
EXAMPLE 7.6.2 Solution:
Red light of wavelength 600 nm from a
point source passes through two
parallel and narrow slits which æ m+ 1 ö λD
m=2 3D çè 2 ø÷
are 1.5 mm apart. m=2 2B ym =
Determine the distance d
m=1 2D
between the central bright m=1 1B æ 3+ 1 ö ( 600 × 10-9 )(1.2)
fringe and the forth dark fringes çè 2 ø÷ 1.5× 10-3
formed on a screen parallel m=0 1D =
to the plane of the slits, m=0 CB
if the distance between m=0 1D ym = 1.68×10-3 m
the slits and the m=1 1B
screen is 1.2 m. y3 m=1 2D
m=2 2B
m=2 3D
m=3 3B
m=3 4D
m=4 4B
m=4 5D
93
EXAMPLE 7.6.3 Solution:
In a lab experiment, m=2 3D (a) = mλD
monochromatic light passes m=2 2B ym d
through two narrow slits that are m=1 2D
0.050 mm apart. The interference m=1 1B 2.4× 10-2 = 2 ( λ)(1.0)
pattern is observed on a white m=0 1D 0.050× 10-3
wall 1.0 m from the slits, and the m=0 CB
λ = 6×10-7 m
second bright is 2.4 cm from the m=0 1D (b)
center of the central maximum. m=1 1B
m=1 2D ym = y5 - y3
m=2 2B
a) What is the wavelength of m=2 3D æç5 + 1 ö÷λD æç3 + 1 ö÷λD
the light? m=3 3B è 2 ø è 2 ø
b) What is the distance between 4D ym = d - d
m=3 4B
third order and fifth order ym m=4 ym = 0.066 - 0.042
dark fringes?
m=4 5D ym = 0.024 m
94
EXAMPLE 7.6.4 Solution:
Calculate the distance Given equation :
between two slits that
produces the first minimum sin qm = ym æ m + 1 ö lD
for 410 nm violet light at an D èç 2 ÷ø
angle of 45o? ym =
d
1st Dark Fringe
m=0 Then, æ 1 ö
èç 2 ÷ø
m + lD
sin qm = dD
So, æ m + 1 ö l
çè 2 ÷ø
sin qm =
d
( )sin æ 1 ö
450 çè 0 + 2 ø÷ 410 ´10-9
d
=
d = 290 nm
95
EXAMPLE 7.6.5 Solution:
A laser beam of wavelength 630 m=2 3D a) the separation between
nm is incident on two slits 0.3 m=2 2B two consecutive bright
mm apart. The screen is 3 m Dy m=1 2D fringes
away from the slits.
a) Find the separation between m=1 1B Dy = lD
two consecutive bright m=0 1D d
fringes m=0 CB
b) What is the separation ( )630´10-9 (3)
between two adjacent dark m=0 1D ( )Dy = 0.3´10-3
fringes? m=1 1B Dy
m=1 2D
m=2 2B Dy = 6.3´10-3 m
m=2 3D
b) the separation between
two adjacent dark
fringes
Dy = 6.3´10-3 m
96
EXERCISE 7.6
1. Monochromatic light illuminates a double-slit system with a slit separation d = 0.30
mm. The second-order maximum occurs at y= 4.0 mm on a screen 1.0 m from the slits.
Find
a) the wavelength (600 nm)
b) the distance y on the screen between the central maximum and the third order-
minimum (7 mm)
2. With two slits spaced 0.2 mm apart, and a screen at a distance of 1.0 m, the third
bright fringe is found to be displaced at 7.5 mm from the central fringe. Find the
wavelength of the light used. (λ = 500 nm)
3. White light passes through two slits 0.50 mm apart an interference pattern is
observed on a screen 2.5 m away. The first-order fringe resembles a rainbow with
violet and red light at either end. The violet light falls about 2.0 mm and the red 3.5
mm from the center of the central white fringe(figure below). Estimate the
wavelengths of the violet and red lights.
( λviolet= 400 nm, λred= 700 nm)
97
4. A screen containing two slits 0.100mm apart is 1.20 m from the viewing screen. Light of wavelength λ =
500 nm falls on the slits from a distant source.
(a) Approximately how far apart will the bright interference fringes be on the screen?
(b) What happens to the interference pattern if the incident light (500 nm) if replaced
by light of wavelength 700 nm.
(c) What happens instead if the slits are moved farther apart?
(∆y = 6 mm, ∆y = 8.4 mm, DIY, DIY)
5. Two slits 0.05 mm apart are illuminated by green light of wavelength 520 nm. An
interference pattern is formed on a viewing screen 2.0 m away. What is the
distance from the center of the screen to the first bright fringe ? What is the
distance to the third dark fringe ?
(2.08 cm, 5.20 cm)
98
7.7 INTERFERENCE OF REFLECTED LIGHT IN
THIN FILMS
Learning Outcomes:
At the end of this subtopic, the students should be able to:
(a) IDENTIFY the occurrence of phase change upon reflection !
"
(From lower to higher refractive index, phase change = rad or path difference = ).
(b) DESCRIBE with the aid of a diagram the interference of light in
thin films at Normal incidence. (Limited to 3 media)
(c) USE the following equations for reflected light with no phase
difference (Non-reflective coating)
i) Constructive interference (or reflective coating) : 2nt = mλ
ii) Destructive interference 2nt = æ m + 1 ö λ
(or Non-reflective coating) : çè 2 ÷ø
99
Continue Learning Outcomes:
(d) USE the following equations for reflected light of phase
difference
π rad (Reflective coating): 1
2
i) Constructive interference : 2nt = æ m + ö λ
çè ø÷
ii) Destructive interference : 2nt = mλ
where m = 0, ±1, ±2,±3,...
(e) EXPLAIN the application of thin films (eg: solar panel, glass tint).