Matriculation Physics ( Physical Optics ) 633r3
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PHYSICS
CHAPTER 2 The study of interference, diffraction and polarization of light. light Light is treated as waves rather than as rays.
CHAPTER 2: Physical optics (9 Hours)
1
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PHYSICS CHAPTER 2 Learning Outcome: 2.1 Huygen’s principle (1 hour) At the end of this chapter, students should be able to: Explain Huygen’s principle governing the propagation of wave fronts. Include spherical and plane wavefronts. Explain diffraction patterns by using Huygen’s principle.
2
PHYSICS CHAPTER 2 2.1 Huygen’s principle 2.1.1 Wavefronts
is defined as a line or surface, in the path of a wave motion, on which the disturbances at every point have the same phase. phase Figure 2.1 shows the wavefront of the sinusoidal waves. wavefront A D
Figure 2.1
B
E
C
F
v
λ
Line ing all point of adjacent wave, e.g. A, B and C or D,E and F are in phase Wavefront always perpendicular to the direction of wave 3 propagation.
PHYSICS
CHAPTER 2
Type of wavefronts Circular wavefronts as shown in Figure 2.2 are produced by a point source generates two-dimensional waves. circular wavefront
λ ray
point source
Figure 2.2 4
PHYSICS
CHAPTER 2
Spherical wavefronts as shown in Figure 2.3 are produced by a point source generates three-dimensional waves. spherical wavefronts
point source
rays Figure 2.3
5
PHYSICS
CHAPTER 2
Plane wavefronts as shown in Figures 2.4a and 2.4b are produced by a point source generates three-dimensional waves at large distance from the source.
plane wavefront rays
plane wavefront
Figure 2.4a : (3-D)
rays
Figure 2.4b : (2-D)
6
PHYSICS
CHAPTER 2
Ray is defined as a line represents the direction of travel of a wave. wave It is at right angle to the wavefronts as shown in Figure 2.5. ray wavefront
λ Figure 2.5
Beam of light is a collection of rays or a column of light. light parallel beam, e.g. a laser beam (shown in Figure 2.6a) Source of light from infinity Figure 2.6a
7
PHYSICS
CHAPTER 2
divergent beam, e.g. a lamp near you (shown in Figure 2.6b)
Figure 2.6b
convergent beam as shown in Figure 2.6c.
Figure 2.6c
8
PHYSICS
CHAPTER 2
2.1.2 Huygen’s principle
states that every point on a wavefront can be considered as a source of secondary wavelets that spread out in the forward direction at the speed of the wave. The new wavefront is the envelope of all the secondary wavelets i.e. the tangent to all of them. them secondary wavefront
wavelets
Figure 2.7
9
PHYSICS
CHAPTER 2
Application of Huygen’s principle a. Construction of new wavefront for a plane wave
P1
A
A’ Q
Q2
P3
Q3
s B
If the wave speed is v, hence in time t the distance travels by the wavelet is s = vt.
From Huygens’ Principle, points P1, P2, P3 and P4 on the wavefront AB are the sources of secondary wavelets.
From the points, draw curves of radius s.
Then draw a straight line A’B’ which is tangent to the curves at points Q1,Q2,Q3 and Q4
Hence, line A’B’ is the new wavefront after t second.
1
P2
P4
Q4 B’
Figure 2.8
10
PHYSICS
CHAPTER 2
b. Construction of new wavefront for a circular wave A’
A
Q1
P1
source
B B’
s
Q2
Explanation as in the construction of new wavefront for a plane wavefront.
But the wavefront A’B’ is a curve touching points Q1,Q2,Q3 and Q4.
The curve A’B’ is the new (circular) wavefront after t second.
P2 P3
P4
Q3 Q4
ray Figure 2.9
11
PHYSICS
CHAPTER 2
c. Diffraction of wave at a single slit
Figure 2.10
Huygens’ principle can be used to explain the diffraction of wave.
Each of the point in Figure 2.10, acts as a secondary source of wavelets (red circular arc)
The tangent to the wavelets from points 2, 3 and 4 is a plane wavefront.
But at the edges, points 1 and 5 are the last points that produce wavelets.
Huygens’ principle suggest that in conforming to the curved shape of the wavelets near the edges, the new wavefront bends or diffracts around the edges - applied to all kinds of waves. Stimulation 2.1 12
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PHYSICS CHAPTER 2 Learning Outcome: 2.2 Constructive interference and destructive interference (1 hour) At the end of this chapter, students should be able to: Define coherence. State the conditions to observe interference of light.
State the conditions of constructive and destructive interference.
13
PHYSICS
CHAPTER 2
2.2 Constructive interference and destructive interference 2.2.1 Interference of light
Light wave is an electromagnet waves (emw). It consists of varying electric field E and varying magnetic field B which are perpendicular to each other as shown in Figure 2.11.
Figure 2.11 Electric field:
E = E0 sin (ωt-kx)
Magnetic field:
B = B0 sin (ωt-kx)
14
PHYSICS
CHAPTER 2
Interference is defined as the effect of interaction between two or more waves which overlaps or superposed at a point and at a particular time from the sources. sources For light the Interference is occurred when two light waves meet at a point, a bright or a dark region will be produced in accordance to the Principle of superposition. Principle of superposition states the resultant displacement at any point is the vector sum of the displacements due to the two light waves. waves Constructive interference is defined as a reinforcement of amplitudes of light waves that will produce a bright fringe (maximum). (maximum) Destructive interference is defined as a total cancellation of amplitudes of light waves that will produce a dark fringe (minimum). (minimum)
15
PHYSICS
CHAPTER 2
2.2.2 Conditions for permanent interference
Permanent interference between two sources of light only take place if they are coherent sources. It means the sources must have the same wavelength or frequency (monochromatic). the sources must have a constant phase difference between them. The light waves that are interfering must have the same or approximately of amplitude to obtain total cancellation at minimum or to obtain a good contrast at maximum. The distance between the coherent sources should be as small as possible of the light wavelength ( ≤ λ ).
16
PHYSICS
CHAPTER 2
2.2.3 Path difference, ∆ L
is defined as the difference in distance from each source to a particular point. point
P x1 S1
x2
Figure 2.12
S2
screen
∆L
Path difference, ∆ L = |S2P − S1P| = |x2 –x1|
17
PHYSICS
CHAPTER 2
Interference of two coherent sources in phase Path difference for constructive interference S1 and S2 are two coherent sources in phase
S1
x1 P (maximum)
S2
x2
+
=
Figure 2.13
18
PHYSICS
CHAPTER 2
A bright fringe is observed at P thus Δφ = 2mπ where m = 0,±1,±2,...
At P, then
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 )
Δφ = (ωt − kx2 ) − (ωt − kx1 ) Δφ = k ( x1 − x2 ) since k = 2π and ( x1 − x2 ) = ∆L λ 2π order Δφ = ∆L λ 2π ∆L = mλ therefore 2mπ = ∆L Note : λ When
m=0
m=±1 m=±2
where
m = 0,±1,±2,.....
λ : wavelength Central bright fringe (zeroth order bright) 1st bright fringe (1st order bright) 2nd bright fringe (2nd order bright)
19
PHYSICS
CHAPTER 2
Path difference for destructive interference S1 and S2 are two coherent sources in phase
x1
Q (minimum)
S1 x2 S2
+
=
Figure 2.14
20
PHYSICS
CHAPTER 2
A dark fringe is observed at Q thus Δφ = 2m + 1 π where m = 0,±1,±2,...
At P,
(
then
)
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 )
Δφ = (ωt − kx2 ) − (ωt − kx1 ) Δφ = k ( x1 − x2 ) 2π Δφ = ∆L λ 2π ∆L therefore ( 2m + 1)π = λ Note : When
m=0 m=±1 m=±2
1 ∆L = m + λ 2 where m = 0,±1,±2,.....
1st dark fringe (zeroth order dark) 2nd dark fringe (1st order dark) 3rd dark fringe (2nd order dark)
21
PHYSICS
CHAPTER 2
Interference pattern for two coherent sources in phase Fringe 2nd bright 2nd dark 1st bright 1st dark Central bright 1st dark 1st bright 2nd dark 2nd bright
S1 S2
∆φ
m
∆L
4π 3π 2π π 0 π 2π 3π 4π
2 1 1 0 0 0 −1 −1 −2
2λ 3 2
λ
1 2
λ
λ 0 1 2
λ
3 2
λ
λ
2λ
screen Figure 2.15 22
PHYSICS
CHAPTER 2
Interference of two coherent sources in antiphase Path difference for constructive interference S1 and S2 are two coherent sources in antiphase
P (maximum)
x1 S1
x2
S2
+
=
Figure 2.16
23
PHYSICS
CHAPTER 2
A bright fringe is observed at P thus Δφ = 2mπ where m = ±1,±2,...
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 − π ) then Δφ = (ωt − kx − π ) − (ωt − kx ) 2 1 Δφ = k ( x1 − x2 ) − π 2π Δφ = ∆L − π λ 1 2π ∆L = m + λ ∆L − π therefore 2mπ = 2 λ Note : where m = 0,±1,±2,..... When
At P,
m=0 m=±1 m=±2
1st bright fringe (zeroth order bright) 2nd bright fringe (1st order bright) 3rd bright fringe (2nd order bright)
24
PHYSICS
CHAPTER 2
Path difference for destructive interference S1 and S2 are two coherent sources in antiphase
S1
x1 Q (minimum)
S2
x2
+
=
Figure 2.17
25
PHYSICS
CHAPTER 2
A dark fringe is observed at Q thus Δφ = 2m + 1 π where m = 0,±1,±2,...
(
)
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 + π ) then Δφ = (ωt − kx2 + π ) − (ωt − kx1 ) Δφ = k ( x1 − x2 ) + π 2π Δφ = ∆L + π λ 2π ∆L = mλ ∆L + π therefore ( 2m + 1)π = λ Note : where m = 0,±1,±2,..... When
At P,
m=0 m=±1 m=±2
Central dark fringe (zeroth order dark) 1st dark fringe (1st order dark) 2nd dark fringe (2nd order dark)
26
PHYSICS
CHAPTER 2
Interference pattern for two coherent sources in antiphase Fringe 2nd dark 2nd bright 1st dark 1st bright Central dark 1st bright 1st dark 2nd bright 2nd dark
S1 S2
∆φ
m
∆L
5π 4π 3π 2π π 2π 3π 4π 5π
2 1 1 0 0 0 −1 −1 −2
2λ 3 2
λ
1 2
λ
λ 0 1 2
λ
3 2
λ
λ
2λ
screen Figure 2.18 27
PHYSICS
CHAPTER 2
Table 2.1 shows the summary of chapter 2.2.3. Two Coherent Bright fringe Dark fringe sources
In phase
∆L = mλ
m = 0,±1,±2,... Δφ = 2mπ m = 0,1,2,... Antiphase
1 ∆L = m + λ 2 m = 0,±1,±2,... Δφ = 2mπ m = 1,2,... Table 2.1
1 ∆L = m + λ 2 m = 0,±1,±2,... Δφ = (2m + 1)π m = 0,1,2,... ∆L = mλ
m = 0,±1,±2,... Δφ = (2m + 1)π m = 0,1,2,... 28
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PHYSICS CHAPTER 2 Learning Outcome: 2.3 Interference of transmitted light through doubleslits (2 hours) At the end of this chapter, students should be able to: Derive with the aid of a diagram and use
mλD ym = d
xm
for bright fringes (maxima)
( m + 12 ) λD = d
for dark fringes (minima),
where m = 0, ±1, ±2, ±3, … .
Use expression
λD ∆y = d
and
explain the effect of changing any of the variables.
29
PHYSICS
CHAPTER 2
2.3 Interference of transmitted light through double-slits 2.3.1 Methods of obtaining two coherent sources Division of wavefront
S
S1 S2
monochromatic light source single slit
double slits
A slit S is placed at equal distance from slits S1 and S2 as shown in figure.
Light waves from S that arrived at S1 and S2 are in phase.
Therefore, both slits S1 and S2 are two new coherent sources, e.g. in Young’s double slit experiment
Figure 2.19 30
PHYSICS
CHAPTER 2
Division of amplitude incident ray
1 2 partial reflection
air film
t
Both reflected waves 1 and 2 are coherent and will result in interference when they superpose.
e.g. Newton’s ring, air wedge fringes and thin film interference.
partial transmission
air
The incident wavefront is divided into two waves by partial reflection and partial transmission.
Figure 2.20
31
PHYSICS
CHAPTER 2
2.3.2 Young’s double-slit experiment Figure 2.21 shows the schematic diagram of Young’s double-slit Intensity experiment.
Max
m=2
Min Max S monochromatic light beam single slit Picture 2.1
S1
Min
S2
Min
Max Max
double slits Figure 2.21
m=1 m=0 m = −1
Min Max screen
m = −2 interference pattern 32
PHYSICS
CHAPTER 2
Explanation of Young’s double-slit experiment by using Huygens’ principle Wavefront from light source falls on a narrow slit S and diffraction occurs. Every point on the wavefront that falls on S acts as sources of secondary wavelets that will produce a new wavefront that propagate to slits S1 and S2 .
S1 and S2 are produced two new sources of coherent waves in phase because they originate from the same wavefront and their distance from S are equal. An interference pattern consisting of bright and dark fringes is formed on the screen as shown in Figure 2.21. The bright fringes are occurred when the light from slits S1 and S2 superposes constructively. The dark fringes are occurred when the light from slits S1 and S2 superposes destructively. 33
PHYSICS
CHAPTER 2
Derivation of Young’s double-slit equations Equation for separation between central bright fringe and mth bright fringe (m+1)th bright M
∆y
P
θ
ym
S1
d
θ
Q
O
N
S2
mth bright
y m +1 Central bright
mλ
D double-slit
screen Figure 2.22
34
PHYSICS
CHAPTER 2
Suppose P in Figure 2.22 is the mth order bright fringe, thus
S2 P − S1P = mλ
Let OP = ym = distance from P to O .
In practice d is very small (<1mm) and D>>d, then S1N meets PQ at right angle. Hence NP = S P then S N = S P− NP = mλ . 1 2 2 angle PQO = angle S2S1N = θ From the figure, S2 N mλ sin θ = = ΔS2S1N
S2S1 d PO ym tanθ = = ΔPQO QO D Since θ is small, thus sinθ = tanθ
mλ ym = d D
35
PHYSICS
CHAPTER 2
Therefore, the separation between central bright and mth bright fringes, ym is given by
mλD ym = d
where
(2.1)
m : order = 0,±1,±2,... λ : wavelength D : distance between double - slits and the screen d : separation between double - slits
Note: For bright fringes Central bright fringe (Zeroth order maximum) m=0
m = ±1 m = ±2 m = ±3
1st bright fringe (1st order maximum) 2nd bright fringe (2nd order maximum) 3rd bright fringe (3rd order maximum) 36
PHYSICS
CHAPTER 2
Equation for separation between central bright fringe and mth dark fringe R
mth order dark
∆y
(m−1)th order dark
θ
xm
S1
d
θ
Q S2
N
1 m + λ 2
double-slit
O
Central bright
D screen Figure 2.23 37
PHYSICS
CHAPTER 2
Suppose R in Figure 2.23 is the mth order dark fringe, thus
1 S2 R − S1R = m + λ 2 Let OR = xm = distance from R to O .
In practice d is very small (<1mm) and meets RQ at right angle. Hence,
D>>d, then S1N
1 λ NR = S R then S2 N = S2 R − NR = m + 1 2 angle RQO = angle S S N = θ 2 1 S2 N 1λ From the figure, sin θ = = m + ΔS2S1N S2S1 2 d RO xm tanθ = = ΔRQO QO D 1 λ xm Since θ is small, thus sinθ = tanθ m + = 2 d D
38
PHYSICS
CHAPTER 2
Therefore, the separation between central bright and mth order dark fringes, xm is given by
1 λD xm = m + 2 d
where
(2.2)
m : order = 0,±1,±2,...
Note: For dark fringes 1st dark fringe (Zeroth order minimum) m=0
m = ±1 m = ±2 m = ±3
2nd dark fringe (1st order minimum) 3rd dark fringe (2nd order minimum) 4th dark fringe (3rd order minimum)
39
PHYSICS
CHAPTER 2
Equation for separation between successive (consecutive) bright or dark fringes, ∆ y (Figure 2.22) is given by
λD mλD and y m +1 = ( m + 1) ∆y = ym +1 − ym where ym = d d
λD mλD ∆y = ( m + 1) − d d λD ∆y = d where
(2.3)
∆y : separation between consecutive bright or dark fringes λ : wavelength D : distance between double - slits and the screen d : separation between double - slits 40
PHYSICS
CHAPTER 2
Appearance of Young’s double-slit experiment From the equation (2.3), (2.3)
λD ∆y = d
∆ y depends on :
the wavelength of light, λ
the distance apart, d of the double slits,
distance between slits and the screen, D Explanation for the above factors:
if λ is short and thus Δy decreases for fixed D and d. The interference fringes are closer to each other and vice-versa. if the distance apart d of the slits diminished, Δy increased for fixed D and λ and vice-versa. if D increases Δy also increases for fixed λ and vice-versa. 41
PHYSICS
CHAPTER 2
if a source slit S (Figure 2.21) is widened the fringes gradually disappear. The slit S then equivalent to large number of narrow slits, each producing its own fringe system at different places. The bright and dark fringes of different systems therefore overlap, giving rise to a different illumination. if one of the slit, S1 or S2 is covered up, the fringes disappear. if the source slit S is moved nearer the double slits, Δy is unaffected but their intensity increases. if the experiment is carried out in a different medium, for example water, the fringe separation Δy decreased or increased depending on the wavelength, λ of the medium. if white light is used the central bright fringe is white, and the fringes on either side are coloured. Violet is the colour nearer to the central fringe and red is farther away as shown in Figure 2.24. 42
PHYSICS
CHAPTER 2
Figure 2.24 Table 2.2 shows the range of wavelength for colours of visible light. Colour Range of λ/ nm
Stimulation 2.2 Stimulation 2.3
Violet
400 – 450
Blue
450 – 520
Green
520 – 560
Yellow
560 – 600
Orange
600 – 625
Red
625 - 700 Table 2.2
43
PHYSICS
CHAPTER 2
Example 1 : A double-slits pattern is view on a screen 1.00 m from the slits. If the third order minima are 25.0 cm apart, determine a. the ratio of wavelength and separation between the slits, b. the distance between the first order minimum and fourth order maximum on the screen. Solution : a.
d
D = 1.00 m; ∆x3 = 0.25 m; m = 3
3rd order minimum
S1
∆x3
S2
D
x3 zeroth order maximum
x3 3rd order minimum
From the figure,
∆x3 0.25 x3 = = 2 2 x3 = 0.125 m
44
PHYSICS
CHAPTER 2
Solution : D = 1.00 m; ∆x3 = 0.25 m; m = 3 a. By using the equation of separation between central bright and
1 λD x3 = 3 + 2 d 1 λ (1.00 ) 0.125 = 3 + 2 d λ = 3.57 ×10 − 2 d
mth order dark 1 λDthus fringes,
xm = m + 2 d
b. The separation between central max and the 1st order min. is
1 λD x1 = 1 + 2 d λD x1 = 1.5 d
45
PHYSICS
CHAPTER 2
Solution : b. and the separation between central max and the 4th order max. (m = 4) is given by
mλD ym = d
λD y4 = 4 d
Therefore the distance between the first order minimum and fourth order maximum on the screen is
∆d = y4 − x1 λD λD ∆d = 4 − 1. 5 d d λD ∆d = 2.5 d ∆d = 2.5( 3.57 ×10 −2 )(1.00 ) ∆d = 8.93 ×10 −2 m
46
PHYSICS
CHAPTER 2
Example 2 : a. How would you expect the interference pattern of a double-slit experiment to change if white light is used instead of monochromatic light? b. Describe the changes that would be observed in a double-slit interference pattern if the entire experiment were submerged in water. (Physics, 3rd edition, J. S. Walker, Q4&Q6, p.963)
Solution : a. The locations of bright and dark fringes depends on the wavelength of light. light Therefore, if white light is used in a double-slit experiment, each bright fringe will show some separation into colours, colours giving a “rainbow” effect. effect
47
PHYSICS
CHAPTER 2
Solution : b. Submerging the double-slit experiment in water would reduce the wavelength of the light from λ to λ /n, where n = 1.33 is the refraction index of water. Therefore, the bright or dark fringe separation would be reduced, reduced according to the λD equation below:
∆y =
d
It follows that the interference pattern fringes get closer to each other. other
48
PHYSICS
CHAPTER 2
Example 3 : In a Young’s double-slit experiment, when a monochromatic light of wavelength 600 nm shines on the double slits, the fringe separation of the interference pattern produced is 7.0 mm. When another monochromatic light source is used, the fringe separation is 5.0 mm. Calculate the wavelength of the second light. Solution : λ1
= 600 ×10 −9 m; ∆y1 = 7.0 ×10 −3 m; ∆y2 = 5.0 ×10 −3 m
1st case:
d
∆y1
S1 S2
∆y1 Central of interference pattern
D By applying the fringe separation equation, thus
λ1 D ∆y1 = d
7.0 × 10 −3
( 600 ×10 ) D = −9
d
(1) 49
PHYSICS
CHAPTER 2
Solution : λ1 2nd case:
= 500 ×10 −9 m; ∆y1 = 7.0 ×10 −3 m; ∆y2 = 5.0 ×10 −3 m
d
S2
λ2 D ∆y2 = d (2)÷ (1):
∆y2
S1
∆y2 Central of interference pattern
D
λ2 D 5.0 ×10 = d −3 5.0 ×10 λ2 = −3 7.0 ×10 600 ×10 −9 −3
(2)
λ2 = 429 ×10 −9 m OR 429 nm 50
PHYSICS
CHAPTER 2
Example 4 :
A
S1 1.2 mm S2
2.5 m Figure 2.25
B
Figure 2.25 shows two coherent sources (S1 and S2) of light in phase. The separation of S1 and S2 is 1.2 mm and the screen is 2.5 m from the sources. a. The frequency of the light is 5.77 × 1014 Hz. Calculate i. the wavelength of the light used ii. the separation between two consecutive bright fringes if the experiment is carried out in air. b. If the experiment is carried out in water of refractive index 1.33, calculate the separation of two consecutive dark fringes. (The speed of light in vacuum, c = 3.00 × 108 m s−1)
51
PHYSICS
CHAPTER 2 −3
Solution : d = 1.2 × 10 m; D = 2.5 m 14 a. i. Given f = 5.77 × 10 Hz By applying the wave speed equation, thus
c = λf 8 14 3.00 ×10 = λ 5.77 ×10 λ = 520 ×10 −9 m OR 520 nm
(
)
ii. By using the equation of fringe separation, thus
λD ∆y = d −9 520 ×10 ( 2.5) ∆y = −3 1.2 × 10 ∆y = 1.08 × 10 −3 m
(
)
52
PHYSICS
CHAPTER 2 −3
Solution : d = 1.2 × 10 m; D = 2.5 m b. Given n = 1.33 The wavelength of light in water is given by
λ n= λw
520 ×10 −9 1.33 = λw −7 λw = 3.91×10 m
Therefore the dark fringes separation is
λD ∆x = d 3.91×10 −7 ( 2.5) ∆x = 1.2 ×10 −3
(
)
∆x = 8.15 × 10 −4 m 53
PHYSICS
CHAPTER 2
Exercise 2.1 : 1.
Young’s double-slit experiment is performed with 589-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits. (Physics for scientists and engineers,6th edition,Serway&Jewett, Q37.5, p.1198)
ANS. : 1.54 mm 2. A Young’s interference experiment is performed with monochromatic light. The separation between the slits is 0.500 mm, and the interference pattern on a screen 3.30 m away shows the first side maximum 3.40 mm from the centre of the pattern. What is the wavelength? (Physics for scientists and engineers,6th edition,Serway&Jewett, Q37.2, p.1197)
ANS. : 515 nm 54
PHYSICS
CHAPTER 2
Exercise 2.1 : 3.
A coherent light that contains two wavelength, 660 nm (red) and 470 nm (blue) es through two narrow slits separated by 0.3 mm and the interference pattern is observed on a screen 5.00 m from the slits. Determine the distance between the first order bright fringes for each wavelength. (University physics,11th edition, Young&Freedman, Q35.14, p.1362)
ANS. : 3.17 mm 4. A monochromatic light of wavelength 560 nm es through a Young’s double-slit system of unknown slit separation. After that, the slits is illuminated by a monochromatic light of unknown wavelength. It was observed that the 4th order minimum of the known wavelength light overlapped with the 5th order maximum of the unknown wavelength light on a screen. Calculate the wavelength of the unknown wavelength light. ANS. : 504 nm 55
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PHYSICS CHAPTER 2 Learning Outcome: 2.4 Interference of reflected light in thin films (1 hour) At the end of this chapter, students should be able to: Explain with the aid of a diagram the interference of light in thin films for normal incidence. For non-reflective coating: Constructive interference : 2nt = mλ
Destructive interference : 2nt = (m + ½ )λ For reflective coating: Constructive interference : 2nt = (m + ½ )λ
Destructive interference : 2nt = mλ where m = 0, ±1, ±2, ±3, …
56
PHYSICS
CHAPTER 2
2.4 Interference of reflected light in thin films
Interference due to reflected waves is observed in many everyday circumstances such as bright colours reflected from oil film on water and soap bubble. The reflected waves can change their phase in two ways: The phase changes in proportion to the distance of the waves travel. travel The phase changes as a result of the reflection process itself. Optical path is defined as the product between a distance travelled by light and the refractive index of the medium OR
L = nl
where
L : optical path n : refractive index of a medium l : distance travelled by light in the medium 57
PHYSICS
CHAPTER 2
2.4.1 Phase changes due to reflection
A light wave travelling in a medium of lower refractive index (n1) when reflected from a medium’s surface of higher refractive index (n2) undergoes a π radian phase change as shown in Figure 2.26a.
π radian phase change
incident wave
n1 reflected wave
incident pulse
n2 transmitted wave
transmitted pulse
Note:
∆φ = π rad λ ∆L = 2
n1
n2
Figure 2.26a
reflected pulse Figure 2.26b: string analogous 58
PHYSICS
CHAPTER 2
A light wave travelling in a medium of higher refractive index (n2) when reflected from a medium’s surface of lower refractive index (n1) undergoes no phase change as shown in Figure 2.27a. no phase change incident wave
n2 reflected wave
incident pulse
n1 transmitted wave
transmitted pulse
Note:
∆φ = 0 ∆L = 0
n2
n1
Figure 2.27a Stimulation 2.4
reflected pulse Figure 2.27b: string analogous 59
PHYSICS
CHAPTER 2
2.4.2 Interference from thin films on a denser medium
Figure 2.28 shows the light waves reflected from the upper and lower surfaces of a thin film (refractive index, n) on a denser medium. π rad phase π rad phase change change E F A 1
n1 = 1.0
D
B
n = 1.5
2
t : thickness of thin film
C
n2 = 3.5 Figure 2.28
60
PHYSICS
CHAPTER 2
When an incident ray falls on a thin film surface almost normal to the surface (point B) division of amplitude occurs, part of ray are reflected (ray 1→ ray ABE), part of ray are refracted and reflected (ray 2→ ray ABCDF), point D very close to B (BC and CD become straight line). At B,
the reflected ray (ray 1) undergoes π radian phase change. change because the ray 1 reflected from a surface of higher refractive index (denser medium). medium) At C,
the reflected ray (ray 2) undergoes π
radian phase change. change Therefore both rays 1 and 2 are two coherent sources in phase because the phase difference, ∆ φ is
∆φ = interference π − π = 0 pattern. and meet at a point produces 61
PHYSICS
CHAPTER 2
The optical path difference between rays 1 and 2 is given by
∆L = ABCDF− ABE ∆L = BC + CD ∆L = nt + nt ∆L = 2nt
Constructive interference: where
2nt = mλ
λ : wavelength of light in vacuum m = 0,±1,±2,...
Destructive interference:
1 where 2nt = m + λ 2 m = 0,±1,±2,...
Example of thin film on a denser medium: Non-reflective (anti-reflective) coating Oil film on water and etc… Picture 2.2
Picture 2.3
Picture 2.4 62
PHYSICS
CHAPTER 2
2.4.3 Interference from thin films on a less dense medium
Figure 2.29 shows the light waves reflected from the upper and lower surfaces of a thin film (refractive index, n) in a less dense medium. π rad phase no phase change change E F A 1
nair = 1.0 n = 1.33
2
D
B
t : thickness of thin film
C
nair = 1.0 Figure 2.29
63
PHYSICS
CHAPTER 2
When an incident ray falls on a thin film surface almost normal to the surface (point B) division of amplitude occurs, part of ray are reflected (ray 1→ ray ABE), part of ray are refracted and reflected (ray 2→ ray ABCDF), point D very close to B (BC and CD become straight line). At B, the reflected ray (ray 1) undergoes π radian phase change. change because the ray 1 reflected from a surface of higher refractive index (denser medium). medium) At C, the reflected ray (ray 2) undergoes no phase change. change Therefore both rays 1 and 2 are two coherent sources antiphase because the phase∆difference, is φ = π − 0∆= φπ rad
and meet at a point produces interference pattern.
64
PHYSICS
CHAPTER 2
The optical path difference between rays 1 and 2 is given by
∆L = ABCDF− ABE λ ∆L = BC + CD + λ 2 λ ∆L = nt + nt + ∆L = 2nt + 2 2
Constructive interference:
λ 2nt + = mλ 2
1 where 2nt = m + λ m = 0,±1,±2,... 2
Destructive interference:
λ 1 2nt + = m + λ 2 2
where
2nt = mλ m = 0,±1,±2,...
Example of a thin film on a less dense medium: Soap bubbles Reflective coating and etc… Picture 2.5
Picture 2.6
65
PHYSICS
CHAPTER 2
Example 5 : A non-reflective coating of magnesium fluoride of refractive index 1.38 covers the camera lens of refractive index 1.52. The coating prevents reflection of yellow-green light of wavelength in vacuum 565 nm. Determine the minimum non zero thickness of the magnesium fluoride. (Physics,7th edition, Cutnell&Johnson, Q48, p.886)
Solution :
λ = 565 ×10 −9 m π rad phase change
n1 = 1.0 n = 1.38
n2 = 1.52
1
2
π rad phase change
2 coherent sources in phase
t min 66
PHYSICS
CHAPTER 2
Solution : λ = 565 × 10 −9 m By using the condition of destructive interference for non-reflective coating, thus
1 2nt = m + λ 2
For minimum thickness of MgF2,
2nt min 2(1.38) t min
m=0
1 = 0 + λ 2 1 = ( 565 ×10 −9 ) 2
t min = 1.02 ×10 −7 m OR
102 nm 67
PHYSICS
CHAPTER 2
Example 6 : White light is incident on a soap film of refractive index 1.30 in air. The reflected light looks bluish because the red light of wavelength 670 nm is absent in the reflection. a. State the condition for destructive interference. b. What is the minimum thickness of the soap film? (Physics,3rd edition, J.S.Walker, Q26, p.966)
Solution :
λ = 670 ×10 −9 m π rad phase change
n1 = 1.0 n = 1.30
n1 = 1.0
1
2
no phase change 2 coherent sources antiphase
t min 68
PHYSICS
CHAPTER 2
Solution : λ = 670 × 10 −9 m a. The condition of destructive interference is given by
2nt = mλ b. For minimum thickness of soap film,
2nt min = (1) λ
(
2(1.30 ) t min = 670 ×10 −9
m =1
)
−7
t min = 2.58 × 10 m OR
258 nm
69
www.kmph.matrik.edu.my/physic s
PHYSICS CHAPTER 2 Learning Outcome: 2.5 Interference of reflected light in air wedge and Newton’s rings (1 hour) At the end of this chapter, students should be able to: Explain with the aid of a diagram the interference in air wedge. Explain with the aid of a diagram the formation of Newton’s rings. Use 2t = (m + ½)λ for bright fringes (maxima) 2t = mλ for dark fringes (minima), where m = 0, 1, 2, 3, …
70
PHYSICS CHAPTER 2.5 Interference of reflected light2in air wedge and Newton’s ring 2.5.1 Air wedge
S
travelling microscope
α X glass plate
l
monochromatic light source
L Q
O P B
t
T
Y
L Figure 2.31
m= 0 0 1 1 2 2 3 3 4 4 5 glass slide thin foil
1st dark fringe Figure 2.30: Apparatus setup
∆x
Figure 2.32
air
71
PHYSICS
CHAPTER 2 Ray S falls almost normal to the surface of a glass slide. At point O, Ray S is partially reflected (ray OL) partially refracted (OB) and then reflected at B (ray PQ) The two refracted rays (OL and PQ) are coherent since both have originated from the same source O. OL and PQ produces interference pattern if it is brought together as shown in Figure 2.32. Since the incidence is nearly normal (point P very close to O), the path difference between the rays at O (ray OL and ray OBPQ) is given by, path difference, ∆
L = OB + BP = nt + nt = 2nt
where n is refractive index of air = 1.0
At X, t = 0 and thus the path difference = 0 and a bright fringe is expected, but a dark fringe is observed at X. This is due to the phase change of π radian for ray PQ (reflected on a denser medium at B). 72
PHYSICS
CHAPTER 2 Hence, ray PQ is in antiphase with ray OL and when brought together (by the retina or lens) to interfere, a dark fringe is obtained. Constructive interference (bright fringe):
1 2t = mλ + λ 2
1 2t = m + λ 2 where
(2.4)
m = 0,1,2,...
Destructive interference (dark fringe):
2t = mλ Note:
where
A phase change of difference of ½ λ
(2.5)
m = 0,1,2,...
π radian is equivalent to a path 73
PHYSICS
CHAPTER 2 From equation (2.4), When
m = 0; t = 14 λ m = 1; t = 34 λ m = 2; t = 54 λ
1st bright fringe (Zeroth order maximum) 2nd bright fringe (1st order maximum) 3rd bright fringe (2nd order maximum)
i.e. bright fringes are formed when the thickness of air film, t = 14 λ , 34 λ , 54 λ ,.......
From equation (2.5), When
m = 0; t = 0 m = 1; t = 12 λ m = 2; t = λ
1st dark fringe (Zeroth order minimum) 2nd dark fringe (1st order minimum) 3rd dark fringe (2nd order minimum)
i.e. dark fringes are formed when the thickness of air film, t = 0, 12 λ , λ , 32 λ ,.......
74
PHYSICS
CHAPTER 2 Equation for separation between the 1st dark fringe and the mth order dark fringe, l
From Figure 2.31,
T t tan α = = L l
t (2.6) l= tan α mλ Rearrange eq. (2.5): t = substitute into eq. (2.6) 2 mλ l= (2.7) 2 tan α where
m : order = 0,1,2,... λ : wavelength of light in vacuum
α : angle of inclination of glass slide 75
PHYSICS
CHAPTER 2 Equation for separation between the 1st dark fringe and the mth order bright fringe, l
Rearrange eq. (2.4):
( m + 12 ) λ substitute into eq. (2.6) t=
( m + 12 ) λ l= 2 tan α
where
2
(2.8)
m : order = 0,1,2,...
Equation for separation between adjacent dark fringes or bright fringes, ∆ x
Put m = 1 into eq. (2.7),
λ ∆x = 2 tan α
(2.9) 76
PHYSICS
CHAPTER 2 C
2.5.2 Newton’s ring Figure 2.34
travelling microscope
monochromatic light source
A
t
glass plate
L
R R −t
Q
S
Y O X
d
P B
plano-convex lens glass block
Figure 2.33: Apparatus setup
Figure 2.35: Newton’s ring
77
PHYSICS
CHAPTER 2 Ray S falls almost normal to the surface of a plano-convex. At point O, Ray S is partially reflected (ray OL) partially refracted (OB) and then reflected at B (ray PQ) The two refracted rays (OL and PQ) are coherent since both have originated from the same source O. OL and PQ produces interference pattern if it is brought together as shown in Figure 2.35. The pattern is a series of circular interference fringes called Newton’s ring. This because of a curved piece of glass with a spherical cross section. Since the incidence is nearly normal (point P very close to O), the path difference between the rays at O (ray OL and ray OBPQ) is given by, path difference, ∆
L = OB + BP = nt + nt = 2nt
where n is refractive index of air = 1.0 78
PHYSICS
CHAPTER 2 At X, t = 0 and thus the path difference = 0 and a bright spot is expected, but a dark spot is observed at X. This is due to the phase change of π radian for ray PQ (reflected on a denser medium at B). Hence, ray PQ is in antiphase with ray OL and when brought together (by the retina or lens) to interfere, a dark spot is obtained. Constructive interference (bright ring):
1 2t = mλ + λ 2
1 2t = m + λ 2
(2.8)
where m = 0,1,2,... Destructive interference (dark ring):
2t = mλ where
m = 0,1,2,...
(2.9) 79
PHYSICS
CHAPTER 2 Relationship between diameter of ring, d and thickness of air gap, t
From Figure 2.34, C
By using the Phytogorean theorem, thus the distance AY is
AY 22 = AC2 − YC 2
R R −t A
Y
d 2
d 2 2 = R − ( R − t) 2 d2 = 2 Rt + t 2 4
Since t is very thin thus t2 ≈ 0
d2 = 2 Rt 4
(2.10)
80
PHYSICS
CHAPTER 2 Equation for diameter of dark ring
mλ substitute into eq. (2.10) Rearrange eq. (2.9): t = 2 d2 mλ = 2 R 4 2 When
m = 0; t = 0 m = 1; t = 12 λ m = 2; t = λ m = 3; t = 32 λ
d 2 = 4 Rmλ where
(2.11)
m : order = 0,1,2,...
Central dark spot (zeroth order minimum), d = 0 1st dark ring (1st order minimum) 2nd dark ring (2nd order minimum) 3rd dark ring (3rd order minimum)
81
PHYSICS
CHAPTER 2 Equation for diameter of bright ring
( m + 12 ) λ substitute into eq. Rearrange eq. (2.9): t = d ( m + )λ = 2R 4 2 2
1 2
2
d = 4 R( m + 12 ) λ 2
where When
(2.10)
(2.12)
m : order = 0,1,2,...
m = 0; t = 14 λ m = 1; t = 34 λ m = 2; t = 54 λ m = 3; t = 74 λ
1st bright ring (zeroth order maximum) 2nd bright ring (1st order maximum) 3rd bright ring (2nd order maximum) 4th bright ring (3rd order maximum) 82
PHYSICS
CHAPTER 2 From Figure 2.35,
The rings become more closely spaced as one moves farther from the centre of the Newton’s ring.
The reason is that the convex surface of the lens moves away from the lower glass block at a progressively faster rate therefore the thickness of air film increases rapidly. rapidly
Newton’s ring can be used to test the accuracy with which a lens has been ground. ground
The rings are not circular if the surface is not spherical (or the glass block is not flat). flat)
83
PHYSICS
CHAPTER 2
Example 7 : An air wedge is formed by placing a human hair between two glass slides of length 44 mm on one end, and allowing them to touch on the other end. When this wedge is illuminated by a red light of wavelength 771 nm, it is observed to have 265 bright fringes. Determine a. the diameter of hair, b. the angle of air wedge, c. the thickness of the air film for 99th dark fringe to be observed, d. the separation between two consecutive bright fringes. −9 −3 Solution : λ = 771 × 10 m; L = 44 × 10 m
d
α
L
84 265th bright fringe
PHYSICS
CHAPTER 2
Solution : λ = 771 × 10 −9 m; L = 44 × 10 −3 m a. Assuming the diameter of the hair, d = the thickness of air film, t at 265th bright fringe Therefore the diameter of the hair is given by
1 2t = m + λ and m = 264 2 1 2d = 264 + 771×10 −9 2 d = 1.02 × 10 −4 m
(
)
b. The angle of air wedge is
d tan α = L
1.02 ×10 −4 tan α = −3 44 ×10 α = 0.13 85
PHYSICS
CHAPTER 2
Solution : λ = 771 × 10 −9 m; L = 44 × 10 −3 m c. By applying the equation for dark fringe (air wedge), thus 2t = mλ and m = 98
(
2t = 98 771×10 −9
)
t = 3.78 ×10 −5 m d. The separation between two consecutive bright fringes is
λ ∆x = 2 tan α −9 771×10 ∆x = 2 tan 0.13 ∆x = 1.70 ×10 −4 m
86
PHYSICS
CHAPTER 2
Example 8 : a. Explain why the central spot in Newton’s ring is dark. b. In a Newton’s ring experiment, the radius of the qth bright ring is 0.32 cm and the radius of the (q+19)th dark ring is 0.67 cm. Determine the radius of curvature of the plano-convex used in the experiment if the wavelength of light used is 589 nm. Solution : a. A ray of light reflected from the lower surface of the convex surface has no phase change. Meanwhile, a ray of light reflected from the top surface of glass block undergoes a π radian phase change. Thus the two reflected rays are two coherent sources in antiphase. At the centre of the interference pattern, the thickness of the air film is zero, hence the path difference for these two rays goes to zero. These resulting a destructive interference at the central of the Newton’s ring. 87
PHYSICS
CHAPTER 2
= 0.32 × 10 −2 m; r( q +19 ) = 0.67 × 10 −2 m λ = 589 ×10 −9 m b. For qth bright ring, m = q − 1 Solution : rq
d 2 = 4 R( m + 12 ) λ
( 2r ) ( 2r )
2
q
and
= 4 R[ ( q − 1) + 12 ]λ
d = 2rq
= 4 R( q − 0.5) λ For (q+19) dark ring, m = q + 19 d 2 = 4 Rmλ and d = 2r( q +19 ) 2
q
(1)
th
( 2r( ( 2r(
) ))
q +19 ) q +19
2
= 4 R( q + 19 ) λ
2
= 4 R( q + 19 ) λ
(2)
88
PHYSICS
CHAPTER 2
= 0.32 × 10 −2 m; r( q +19 ) = 0.67 × 10 −2 m λ = 589 ×10 −9 m 2 r( q +19 ) b. (2) ÷ (1) : = q + 19 r q − 0.5 q −2 2 0.67 ×10 q + 19 = q = 6.27 −2 0.32 ×10 q − 0.5 Solution : rq
By substituting q = 6.27 into eq. (1) thus
( 2( 0.32 ×10 ) ) −2
(
= 4 R( 6.27 − 0.5) 589 ×10 −9 R = 3.01 m 2
89
)
PHYSICS
CHAPTER 2
Exercise 2.2 : 1. A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the film and reflects into your eyes. Although the sunlight is white, since it contains all colours, the film has a yellow hue, because destructive interference has occurred eliminating the colour of blue (λ =469 nm) from the reflected light. If the refractive indices for gasoline and water are 1.40 and 1.33 respectively, Calculate the minimum thickness of the film. ANS. : 168 nm 2. White light is incident normally on a thin soap film (n =1.33) suspended in air. a. What are the two minimum thickness that will constructively reflect yellow light of wavelength 590 nm? b. What are the two minimum thickness that will destructively reflect yellow light of wavelength 590 nm? (Physics,3rd edition, J.S.Walker, Q34, p.966)
ANS. : 110 nm, 330 nm ; 220 nm, 440 nm ;
90
PHYSICS
CHAPTER 2
3. Two plane glass plates which are in at one edge are separated by a piece of metal foil 12.5 cm from that edge. Interference fringes parallel to the line of are observed in reflected light of wavelength 546 nm and are found to be 1.50 mm apart. Determine the thickness of the foil. ANS. : 2.27× 10−5 m 4. Newton’s rings are formed by reflection between an biconvex lens of focal length 100 cm made of glass of refractive index 1.50 and in with a glass block of refractive index 1.60. Calculate the diameter and thickness of air film for fifth bright ring using light of wavelength 6000 Å . Given 1 angstrom (Å) = 10−10 m ANS. : 3.28 mm; 1.35 µ m 5. Newton’s rings are formed with light of wavelength 589 nm between the plano-convex lens of radius of curvature 100 cm and a glass block, in perfect . a. Determine the radius of the 20th dark ring from the centre. b. How will this ring move and what will its radius become if the lens and the block are slowly separated to a distance apart 5.00 × 10−4 cm? 91 ANS. : 3.43 mm; inwards, 1.26 cm
www.kmph.matrik.edu.my/physic s
PHYSICS CHAPTER 2 Learning Outcome: 2.6 Diffraction by a single slit (1 hour) At the end of this chapter, students should be able to: Explain with the aid of a diagram the diffraction of a single slit. Derive and use formula
nλD yn = a
yn
for dark fringes (minima)
( n + 12 ) λD = a
for bright fringes (maxima),
where n = ±1, ±2, ±3, ... Explain with the aid of a diagram the effect of changing wavelength on the resolution of single slit from two coherent sources. 92
PHYSICS
CHAPTER 2
2.6 Diffraction by a single slit 2.6.1 Diffraction of light
is defined as the bending of light waves as they travel around obstacles or through an aperture or slit comparable to the wavelength of the light waves. waves Figures 2.36a, 2.36b and 2.36c show the bending of plane wavefront.
λ
Figure 2.36a: obstacle
λ
Figure 2.36b: slit, a > λ
λ
Figure 2.36c: slit, a ≈ λ
93
PHYSICS
CHAPTER 2
2.6.2 Diffraction by a single silt
Figure 2.37 shows an apparatus setup of diffraction by a single slit. 3rd minimum 2nd maximum 2nd minimum 1st maximum 1st minimum Central maximum 1st minimum 1st maximum 2nd minimum 2nd maximum 3rd minimum
θ1 θ2 S
θ1
θ2
Animation 2.1 Picture 2.7 Picture 2.8
single slit
intensity Figure 2.37
where
screen
θ : angle of diffraction
94
PHYSICS
CHAPTER 2
Explanation of single slit diffraction experiment Wavefront from light source falls on a narrow slit S and diffraction occurs. Every point on the wavefront that falls on S acts as sources of secondary wavelets and superposed each another to form an interference pattern on the screen as shown in Figure 2.37. The central fringe is bright (maximum) and widen compare to other bright fringes. The central fringe has the highest intensity compare to the other bright fringes. The intensity of bright fringes reduce as the distance increase from the central bright fringe.
Other rays with angle θ2 and θ1 will produce minimum and maximum on both sides of the central maximum.
95
PHYSICS
CHAPTER 2
Derivation of single slit diffraction equations Equation for separation between central maximum (bright) and nth minimum (dark) fringes P nth minimum
yn
p st r i 1 θ1 nd strip θ1 2 aE st
a
where
a 2
A
a 2
C B
2
Central Q maximum
sin θ1
a sin θ1 2
D
a : slit width D : distance between single slit and screen Figure 2.38
screen 96
PHYSICS
CHAPTER 2
A single slit is split into two equal parts, AC and CB. A,C and B are new sources of secondary wavelets. (Huygen’s principle) When the wavelets from A, C and B superpose, interference will occur at P. As AB is very small, small thus AE is perpendicular to and AP = EP, EP the outgoing rays are considered parallel, parallel and therefore the path difference at P between ray AP a and is :
∆L = CE =
2
sin θ1
Consider two narrow strips as shown in Figure 2.38, for the two strips superposed destructively thus both strip of light must in antiphase to each another which is equivalence to a path difference of ½λ . If the 1st minimum (1st order minimum) is at P, hence :
a λ ∆L = sin θ1 = 2 2
a sin θ1 = λ
97
PHYSICS
CHAPTER 2
a
For the 2nd minimum and 3rd minimum, minimum AB is split into 4 equal parts, 6 equal parts and so on as shown in Figures 2.39 and 2.40. ip p r t p st s tri i r t st s 1 nd s rip t a a 1 p 2 i r rd s rip t nd s 3 th st rip 6 4 2 p ri 4 th st trip rd st 3 5 th s rip a th st 6 4
2λ 2nd minimum (2nd order minimum)
a λ sin θ 2 = 4 2 a sin θ 2 = 2λ
Figure 2.39
θ3
θ2
3λ 3rd minimum (3rd order minimum)
a λ sin θ 3 = 6 2 a sin θ 3 = 3λ
Figure 2.40 98
PHYSICS
CHAPTER 2
In general, for minima (dark fringes)
a sin θ n = nλ
where n : order = ±1,±2,±3,.. If the distance of single slit to the screen is D, and D>>a yn then:
sin θ n = tan θ n =
D
Therefore the distance of nth minimum from central maximum is:
When
yn a = nλ D
n = ±1 n = ±2 n = ±3
nλ D yn = a
1st minimum fringe (1st order minimum) 2nd minimum fringe (2nd order minimum) 3rd minimum fringe (3rd order minimum) 99
PHYSICS
CHAPTER 2
Equation for separation between central maximum (bright) and nth maximum (bright) fringes R nth maximum
yn
p stri ip 1 d str n 2 rd strip 3 θ1 st
a
a 3 a 3 a 3
θ1 A C aE sin θ1 D 3 B a sin θ
Central Q maximum
1
D screen Figure 2.41
100
PHYSICS
CHAPTER 2
A single slit is split into three equal parts, AC,CD and DB. A,C,D and B are new sources of secondary wavelets. (Huygen’s principle) When the wavelets from A,C,D and B superpose, interference will occur at R. As AB is very small, thus AE is perpendicular to and AP = EP, the outgoing rays are considered parallel, and therefore the path difference at P between ray AP a and is :
∆L = CE =
3
sin θ1
Consider three narrow strips as shown in Figure 2.41, the first two strips (pair) superposed destructively at which the path difference is ½λ and leave the third strip. strip The 3rd strip produces the maximum (bright) fringe at R.
If the 1st maximum (1st order maximum) is at R, hence :
a λ ∆L = sin θ1 = 3 2
3λ a sin θ1 = 2
101
PHYSICS
CHAPTER 2
a
For the 2nd maximum and 3rd maximum, maximum AB is split into 5 equal parts, 7 equal parts and so on as shown in Figures 2.42 and 2.43. ip r t st s trip p ri 1 nd s trip st st p 2 rd s trip 1 nd stri a a 3 th s trip p i 2 r 7 4 th s tripp 5 rd st p 3 th stri 5 th s tri 6 th s 4 th strip a 7 5
5λ 2
2nd maximum (2nd order maximum)
θ3
θ2
a λ sin θ 2 = 5 2 5λ a sin θ 2 = 2
Figure 2.42
3rd maximum (3rd order minimum)
7λ 2 a λ sin θ 3 = 7 2 7λ a sin θ 3 = 2
Figure 2.43 102
PHYSICS
CHAPTER 2
In general, for maxima (bright fringes)
1 where a sin θ n = n + λ n = ±1,±2,±3,.. 2 If the distance of single slit to the screen is D, and D>>a then:
yn sin θ n = tan θ n = D
Therefore the distance of nth maximum from central maximum is:
1 yn a = n + λ 2 D
When
n = ±1 n = ±2 n = ±3
yn
( n + 12 ) λD = a
1st maximum fringe (1st order maximum) 2nd maximum fringe (2nd order maximum) 3rd maximum fringe (3rd order maximum) 103
PHYSICS
CHAPTER 2
Equation for central maximum (bright) fringe
D 1st minimum
θ1 θ1
a single slit
y1 y1
Q Central maximum 1st minimum
Figure 2.44 A C D E B
Figure 2.45
screen
104
PHYSICS
CHAPTER 2
Figure 2.45 shows five sources of Huygen’s wavelets and the screen is to be so far from the slit (D>>a) thus the rays from each source are nearly parallel. parallel
All the wavelets from each source travel the same distance to the point Q (Figure 2.44) and arriving there in phase. phase
Therefore, the constructive interference is occurred at the central of the single slit diffraction pattern. pattern
The angular width of central maximum, θ
w
is given by
θ w = 2θ1 and θ1 : 1st minimum diffraction angle −1 λ θ1 = sin a
−1 λ θ w = 2 sin a 105
PHYSICS
CHAPTER 2
The width of central maximum, w is given by
w = 2 y1
and
y1 : separation of 1st minimum and central maximum λD y1 = a
λD w = 2 a
Note:
To calculate the maximum number of orders observed, observed take the diffraction angle,θ = 90°.
From both equations for minima and maxima, we obtain
sin θ n ∝ λ
and
yn ∝ λ
By using this two relations, the changes of single slit diffraction pattern can be explained. 106
PHYSICS
CHAPTER 2
Example 9 : A sodium light of wavelength 580 nm shines through a slit and produces a diffraction pattern on a screen 0.60 m away. The width of the central maximum fringe on the screen is 5.0 cm. Determine a. the width of the slit, b. the angular width of the central maximum fringe, c. the number of minimum that can be observed on the screen.
−9 −2 λ = 580 × 10 m; D = 0 . 60 m, w = 5 . 0 × 10 m Solution :
w
a a. Since
w = 2 y1
λD w = 2 a
and
1st minimum Central maximum 1st minimum
λD y1 = a
5.0 × 10
−2
( 580 × 10 )( 0.60) =2 −9
a
a = 1.39 × 10−5 m
107
PHYSICS
CHAPTER 2
Solution :λ = 580 × 10 m; D = 0.60 m, w = 5.0 × 10 b. The angular width of the central maximum fringe is given by −9
−2
λ θ w = 2θ1 and θ1 = sin a −9 580 × 10 −1 λ −1 θ w = 2 sin θ w = 2 sin −5 1.39 × 10 a θ w = 4.78 −1
c. By applying the equation for minimum fringe,
a sin θ = nλ
For the maximum no. of order for minimum fringe, θ
(1.39 ×10 ) sin 90 −5
nmax
(
= nmax 580 ×10 −9 = 23.97 ≈ 23
)
= 90
Therefore the number of minimum that can be observed is
23 × 2 = 46 fringes
108
m
PHYSICS
CHAPTER 2
Example 10 : a. State the similarities and differences of double-slit interference and single slit diffraction patterns. b. How many bright fringes will be produced on the screen if a green light of wavelength 553 nm is incident on a slit of width 8.00 µ m? Solution : a. The similarities are Double-slit interference pattern
Single slit diffraction pattern
Both patterns consist of alternating dark and bright fringes. The central for both patterns is bright fringe.
109
PHYSICS
CHAPTER 2
a. The differences are Double-slit interference pattern
The width of each fringe is the same. The intensity of each bright fringe is constant.
Single slit diffraction pattern The central fringe is wider than the other fringes. The intensity of bright fringes reduce as a distance increase from the central bright.
b. Given λ = 553 × 10 −9 m; a = 8.00 × 10 −6 m By applying the equation for bright (maximum) fringe,
1 a sin θ = n + λ 2
For the maximum no. of order for bright fringe, θ = 90 Central bright −6 −9 8 . 00 × 10 sin 90 = n + 0 . 5 553 × 10 fringe max
(
)
(
nmax = 13.97 ≈ 13
)(
)
Therefore the number of bright that can be observed is
(13 × 2)+1 = 27 fringes
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PHYSICS
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Exercise 2.3 : 1. Monochromatic light of wavelength 689 nm falls on a slit. If the angle between first bright fringes on either side of the central maximum is 38°, calculate the slit width. (Physics for scientist & engineers ,3rd edition, Giancoli, Q4, p.913)
ANS. : 3.2 µ m 2. Light of wavelength 633 nm from a distant source is incident on a single slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. Determine the distance between the two dark fringes on either side of the central bright fringe. (University physics,11th edition, Young&Freedman, Q36.4, p.1396)
ANS. : 5.91 mm 3. A screen is placed 1.00 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.60 cm wide. What is the distance between the two second order minima? (Physics,3rd edition, J.S.Walker, Q45, p.967)
ANS. : 3.20 cm
111
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PHYSICS CHAPTER 2 Learning Outcome: 2.7 Diffraction grating (2 hours) At the end of this chapter, students should be able to: Explain with the aid of a diagram the formation of diffraction. Apply formula,
d sinθ n = nλ
where
1 d= N
Describe with the aid of diagram the formation of spectrum by using white light.
112
PHYSICS
CHAPTER 2
2.7 Diffraction grating
is defined as a large number of equally spaced parallel slits. slits Diffraction grating can be made by ruling very fine parallel lines on glass or metal by a very precise machine. machine The untouched spaces between the lines serve as the slits as shown in Figure 2.46.
d
lines
Light es through the slit because it is transparent.
The spaces between the lines are the slits, for example : if there are four lines then we have 3 slits.
slit
Figure 2.46
113
PHYSICS
CHAPTER 2
If there N lines per unit length, then slit separation, d is given by: 1
d=
N
e.g. if a diffraction grating has 5000 lines per cm, then
1 1 d= = N 5000 cm
d = 2 × 10−4 cm
The light that es through the slits are coherent . The Interference pattern is narrower and sharper than doubleslits. There are two type of diffraction grating which are transmission grating (usual diffraction grating) reflection grating e.g. CD and DVD Diffraction grating is used in spectrometer to determine the wavelength of light and to study spectra. spectra 114
PHYSICS
CHAPTER 2
2.7.1 Explanation of diffraction by using Huygen’s principle for diffraction grating Figure 2.47 shows an incident lights fall on the transmission diffraction grating. C first order wavefront A source of secondary wavelets E second order wavefront
incident lights
third order wavefront D
grating
F zeroth order wavefront
Figure 2.47
B
115
PHYSICS
CHAPTER 2
Using Huygens’ principle, each maximum is located by taking the tangent of the wavelets from the slits. If the wavelets from each of the slits are drawn and a tangent AB is drawn, a plane wavefront parallel to the diffraction grating is obtained. This represents the zeroth-order maximum (n = 0). If the wavelets are grouped such that the first wavelet from one slit is combined with the second wavelet from the next slit, the third wavelet from the third slit and so on, the tangent CD will represent the first-order maximum (n =1). For the second-order maximum, the wavelets are grouped are such that the second wavelet of one slit is combined with the fourth wavelet of the next slit, the sixth wavelet from the third slit and so on. (tangent EF) EF Similarly, the third-, fourth-,…. order maximum may be obtained.
116
PHYSICS
CHAPTER 2
2.7.2 Equation of diffraction grating
Figure 2.48 illustrates light travels to a distant viewing screen from five slits of the grating. first order maximum (n = 1)
incoming plane wavefront of light
central or zeroth order maximum (n = 0) first order maximum (n = −1)
diffraction grating
d sin θ
Figure 2.48
d Figure 2.49
θ θ 117
PHYSICS
CHAPTER 2
The maximum (bright) fringes are sometimes called the principal maxima or principal fringes since they are placed where the light intensity is a maximum. Since the screen is far so that the rays nearly parallel while the light travels toward the screen as shown in Figure 2.49. In reaching the place on the screen while the 1st order maximum is located, light from one slit travels a distance of one wavelength farther than light from adjacent slit. Therefore the path difference for maximum fringe (constructive interference) is given by
d sinθ n = nλ n : order = 0,±1,±2,±3,.. θ n : n th order of diffraction angle
where
When
n=0 n = ±1 n = ±2
Central maximum fringe (0th order maximum) 1st maximum fringe (1st order maximum) 118 2nd maximum fringe (2nd order maximum)
PHYSICS
CHAPTER 2
The maximum fringes produce by a grating are much narrower and sharper than those from a double-slit as the intensity graph in Figures 2.50a and 2.50b.
Figure 2.50a
n = −2
−1
0
1
2
n = −2
−1
0
1
2
Figure 2.50b
119
PHYSICS
CHAPTER 2
Figures 2.51 shows the diffraction grating pattern.
n=2
Parallel beam of monochromatic light
θ2 θ1
second-order maximum
first-order n = 1 maximum
n=0
zero-order maximum
n = −1first-order
maximum
grating
n = −2 second-order maximum
Figure 2.51 120
PHYSICS
CHAPTER 2
If the white light is falls on the grating, a rainbow colours would be observed to either side of the central fringe on the screen which is white as shown in Figure 2.51. This because the white light contains wavelengths between violet and red. red
n = −2
−1
1
0
Rainbow Rainbow
white Rainbow
2 Rainbow
White light Figure 2.51
121
PHYSICS
CHAPTER 2
Note:
To calculate the maximum number of orders for bright fringes observed, observed take the diffraction angle,θ = 90°. Therefore
d sin 90 = nmaxλ d nmax = λ
where nmax : maximum number of orders that can be observed. From the equation for maxima, we obtain
sin θ n ∝ λ
and
1 sin θ n ∝ d
By using this two relations, the changes of diffraction grating pattern can be explained. 122
PHYSICS
CHAPTER 2
Example 11 : A monochromatic light of unknown wavelength falls normally on a diffraction grating. The diffraction grating has 3000 lines per cm. If the angular separation between the first order maxima is 35°. Calculate a. the wavelength of the light, b. the angular separation between the second-order and thirdorder maxima. Solution : N
= 3000 cm −1 ; 2θ1 = 35 ; n = 1
35
θ1
1st order max. max
1st order max. max
123
PHYSICS
CHAPTER 2
Solution : N = 3000 cm ; 2θ1 = 35 ; n = 1 a. The diffraction angle for 1st order maximum is −1
2θ1 = 35 θ1 = 17.5
And the slit separation, d is given by
1 d= −1 3000 cm d = 3.33 × 10 −4 cm OR 3.33 × 10−6 m
1 d= N
Therefore the wavelength of the light is
d sin θ n = nλ
(3.33 ×10 ) sin θ (3.33 ×10 ) sin 17.5 −6
−6
=λ =λ λ = 1.00 × 10 −6 m 1
124
PHYSICS
CHAPTER 2
Solution : b.
n=3 θ3 θ2
∆θ 23
3rd order maximum
2nd order n = 2 maximum
n=0
0th order maximum
2nd order n = −2 maximum 3rd order n = −3 maximum By using the equation of diffraction grating for maxima,
d sin θ n = nλ 125
PHYSICS
CHAPTER 2
Solution : b. For 2nd order maximum,
n=2 d sin θ 2 = 2λ (3.33 ×10−6 ) sin θ 2 = 2(1.00 ×10−6 ) θ 2 = 36.9 For 3 order maximum, n = 3 d sin θ 3 = 3λ (3.33 ×10−6 ) sin θ3 = 3(1.00 ×10−6 ) θ 3 = 64.3 rd
Therefore the angular separation,
∆θ 23 = θ 3 − θ 2 ∆θ 23 = 64.3 − 36.9 ∆θ 23 = 27.4 126
PHYSICS
CHAPTER 2
Example 12 : The second-order maximum produced by a diffraction grating with 560 lines per centimeter is at an angle of 3.1°. a. What is the wavelength of the light that illuminates the grating? b. Determine the number of maximum can be observed on a screen. c. State and giving reason, what you would expect to observe if a grating with a larger number of lines per centimeter is used. 3 −1 Solution : N = 56 × 10 m ; θ 2 = 3.1 ; n = 2 a. By applying the equation of diffraction grating for 2nd order 1 maximum, thus d sin θ 2 = 2λ and d =
N sin θ 2 = 2 Nλ sin 3.1 = 2 56 × 103 λ λ = 4.83 ×10 −7 m
(
)
127
PHYSICS
CHAPTER 2
Solution : b. By applying the equation of grating for maximum,
d sin θ n = nλ and sin θ n = nNλ
1 d= N
For the maximum no. of order for maximum fringe, θ n °
(
)(
sin 90 = nmax 56 ×10 4.83 × 10 nmax = 36.97 ≈ 36 3
−7
)
= 90
Therefore the number of maximum can be observed is
(36 × 2)+1 = 73 fringes
c. The fringes become farther to each another. another Reason : since 1 1 sin θ sin θ ∝ and d =
d
N
∝N
a larger number of lines per cm results in a larger diffraction angle thus the distance between two consecutive maximum fringes will increase. 128
PHYSICS
CHAPTER 2
Exercise 2.4 :
1. The first-order maximum line of 589 nm light falling on a diffraction grating is observed at an angle of 15.5°. Determine a. the slit separation on the grating. b. the angle of diffraction for third-order maximum line. (Physics for scientist & engineers ,3rd edition, Giancoli, Q32, p.914)
ANS. : 2.20 µ m; 53.4° 2. A diffraction grating has 6000 lines per cm. Calculate the angular separation between wavelengths 589.6 nm and 546.1 nm respectively after transmission through it at normal incidence, in the first-order spectrum (maximum line). ANS. : 1.60° 3. When blue light of wavelength 465 nm illuminates a diffraction grating, it produces a 1st order maximum but no 2nd order maximum. a. Explain the absence of 2nd order maximum. b. What is the maximum spacing between lines on this grating? (Physics,3rd edition, J.S.Walker, Q65, p.968)
ANS. : 930 nm
129
PHYSICS
CHAPTER 2
Next Chapter… CHAPTER 3 : Electrostatics
130
CHAPTER 2 The study of interference, diffraction and polarization of light. light Light is treated as waves rather than as rays.
CHAPTER 2: Physical optics (9 Hours)
1
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PHYSICS CHAPTER 2 Learning Outcome: 2.1 Huygen’s principle (1 hour) At the end of this chapter, students should be able to: Explain Huygen’s principle governing the propagation of wave fronts. Include spherical and plane wavefronts. Explain diffraction patterns by using Huygen’s principle.
2
PHYSICS CHAPTER 2 2.1 Huygen’s principle 2.1.1 Wavefronts
is defined as a line or surface, in the path of a wave motion, on which the disturbances at every point have the same phase. phase Figure 2.1 shows the wavefront of the sinusoidal waves. wavefront A D
Figure 2.1
B
E
C
F
v
λ
Line ing all point of adjacent wave, e.g. A, B and C or D,E and F are in phase Wavefront always perpendicular to the direction of wave 3 propagation.
PHYSICS
CHAPTER 2
Type of wavefronts Circular wavefronts as shown in Figure 2.2 are produced by a point source generates two-dimensional waves. circular wavefront
λ ray
point source
Figure 2.2 4
PHYSICS
CHAPTER 2
Spherical wavefronts as shown in Figure 2.3 are produced by a point source generates three-dimensional waves. spherical wavefronts
point source
rays Figure 2.3
5
PHYSICS
CHAPTER 2
Plane wavefronts as shown in Figures 2.4a and 2.4b are produced by a point source generates three-dimensional waves at large distance from the source.
plane wavefront rays
plane wavefront
Figure 2.4a : (3-D)
rays
Figure 2.4b : (2-D)
6
PHYSICS
CHAPTER 2
Ray is defined as a line represents the direction of travel of a wave. wave It is at right angle to the wavefronts as shown in Figure 2.5. ray wavefront
λ Figure 2.5
Beam of light is a collection of rays or a column of light. light parallel beam, e.g. a laser beam (shown in Figure 2.6a) Source of light from infinity Figure 2.6a
7
PHYSICS
CHAPTER 2
divergent beam, e.g. a lamp near you (shown in Figure 2.6b)
Figure 2.6b
convergent beam as shown in Figure 2.6c.
Figure 2.6c
8
PHYSICS
CHAPTER 2
2.1.2 Huygen’s principle
states that every point on a wavefront can be considered as a source of secondary wavelets that spread out in the forward direction at the speed of the wave. The new wavefront is the envelope of all the secondary wavelets i.e. the tangent to all of them. them secondary wavefront
wavelets
Figure 2.7
9
PHYSICS
CHAPTER 2
Application of Huygen’s principle a. Construction of new wavefront for a plane wave
P1
A
A’ Q
Q2
P3
Q3
s B
If the wave speed is v, hence in time t the distance travels by the wavelet is s = vt.
From Huygens’ Principle, points P1, P2, P3 and P4 on the wavefront AB are the sources of secondary wavelets.
From the points, draw curves of radius s.
Then draw a straight line A’B’ which is tangent to the curves at points Q1,Q2,Q3 and Q4
Hence, line A’B’ is the new wavefront after t second.
1
P2
P4
Q4 B’
Figure 2.8
10
PHYSICS
CHAPTER 2
b. Construction of new wavefront for a circular wave A’
A
Q1
P1
source
B B’
s
Q2
Explanation as in the construction of new wavefront for a plane wavefront.
But the wavefront A’B’ is a curve touching points Q1,Q2,Q3 and Q4.
The curve A’B’ is the new (circular) wavefront after t second.
P2 P3
P4
Q3 Q4
ray Figure 2.9
11
PHYSICS
CHAPTER 2
c. Diffraction of wave at a single slit
Figure 2.10
Huygens’ principle can be used to explain the diffraction of wave.
Each of the point in Figure 2.10, acts as a secondary source of wavelets (red circular arc)
The tangent to the wavelets from points 2, 3 and 4 is a plane wavefront.
But at the edges, points 1 and 5 are the last points that produce wavelets.
Huygens’ principle suggest that in conforming to the curved shape of the wavelets near the edges, the new wavefront bends or diffracts around the edges - applied to all kinds of waves. Stimulation 2.1 12
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PHYSICS CHAPTER 2 Learning Outcome: 2.2 Constructive interference and destructive interference (1 hour) At the end of this chapter, students should be able to: Define coherence. State the conditions to observe interference of light.
State the conditions of constructive and destructive interference.
13
PHYSICS
CHAPTER 2
2.2 Constructive interference and destructive interference 2.2.1 Interference of light
Light wave is an electromagnet waves (emw). It consists of varying electric field E and varying magnetic field B which are perpendicular to each other as shown in Figure 2.11.
Figure 2.11 Electric field:
E = E0 sin (ωt-kx)
Magnetic field:
B = B0 sin (ωt-kx)
14
PHYSICS
CHAPTER 2
Interference is defined as the effect of interaction between two or more waves which overlaps or superposed at a point and at a particular time from the sources. sources For light the Interference is occurred when two light waves meet at a point, a bright or a dark region will be produced in accordance to the Principle of superposition. Principle of superposition states the resultant displacement at any point is the vector sum of the displacements due to the two light waves. waves Constructive interference is defined as a reinforcement of amplitudes of light waves that will produce a bright fringe (maximum). (maximum) Destructive interference is defined as a total cancellation of amplitudes of light waves that will produce a dark fringe (minimum). (minimum)
15
PHYSICS
CHAPTER 2
2.2.2 Conditions for permanent interference
Permanent interference between two sources of light only take place if they are coherent sources. It means the sources must have the same wavelength or frequency (monochromatic). the sources must have a constant phase difference between them. The light waves that are interfering must have the same or approximately of amplitude to obtain total cancellation at minimum or to obtain a good contrast at maximum. The distance between the coherent sources should be as small as possible of the light wavelength ( ≤ λ ).
16
PHYSICS
CHAPTER 2
2.2.3 Path difference, ∆ L
is defined as the difference in distance from each source to a particular point. point
P x1 S1
x2
Figure 2.12
S2
screen
∆L
Path difference, ∆ L = |S2P − S1P| = |x2 –x1|
17
PHYSICS
CHAPTER 2
Interference of two coherent sources in phase Path difference for constructive interference S1 and S2 are two coherent sources in phase
S1
x1 P (maximum)
S2
x2
+
=
Figure 2.13
18
PHYSICS
CHAPTER 2
A bright fringe is observed at P thus Δφ = 2mπ where m = 0,±1,±2,...
At P, then
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 )
Δφ = (ωt − kx2 ) − (ωt − kx1 ) Δφ = k ( x1 − x2 ) since k = 2π and ( x1 − x2 ) = ∆L λ 2π order Δφ = ∆L λ 2π ∆L = mλ therefore 2mπ = ∆L Note : λ When
m=0
m=±1 m=±2
where
m = 0,±1,±2,.....
λ : wavelength Central bright fringe (zeroth order bright) 1st bright fringe (1st order bright) 2nd bright fringe (2nd order bright)
19
PHYSICS
CHAPTER 2
Path difference for destructive interference S1 and S2 are two coherent sources in phase
x1
Q (minimum)
S1 x2 S2
+
=
Figure 2.14
20
PHYSICS
CHAPTER 2
A dark fringe is observed at Q thus Δφ = 2m + 1 π where m = 0,±1,±2,...
At P,
(
then
)
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 )
Δφ = (ωt − kx2 ) − (ωt − kx1 ) Δφ = k ( x1 − x2 ) 2π Δφ = ∆L λ 2π ∆L therefore ( 2m + 1)π = λ Note : When
m=0 m=±1 m=±2
1 ∆L = m + λ 2 where m = 0,±1,±2,.....
1st dark fringe (zeroth order dark) 2nd dark fringe (1st order dark) 3rd dark fringe (2nd order dark)
21
PHYSICS
CHAPTER 2
Interference pattern for two coherent sources in phase Fringe 2nd bright 2nd dark 1st bright 1st dark Central bright 1st dark 1st bright 2nd dark 2nd bright
S1 S2
∆φ
m
∆L
4π 3π 2π π 0 π 2π 3π 4π
2 1 1 0 0 0 −1 −1 −2
2λ 3 2
λ
1 2
λ
λ 0 1 2
λ
3 2
λ
λ
2λ
screen Figure 2.15 22
PHYSICS
CHAPTER 2
Interference of two coherent sources in antiphase Path difference for constructive interference S1 and S2 are two coherent sources in antiphase
P (maximum)
x1 S1
x2
S2
+
=
Figure 2.16
23
PHYSICS
CHAPTER 2
A bright fringe is observed at P thus Δφ = 2mπ where m = ±1,±2,...
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 − π ) then Δφ = (ωt − kx − π ) − (ωt − kx ) 2 1 Δφ = k ( x1 − x2 ) − π 2π Δφ = ∆L − π λ 1 2π ∆L = m + λ ∆L − π therefore 2mπ = 2 λ Note : where m = 0,±1,±2,..... When
At P,
m=0 m=±1 m=±2
1st bright fringe (zeroth order bright) 2nd bright fringe (1st order bright) 3rd bright fringe (2nd order bright)
24
PHYSICS
CHAPTER 2
Path difference for destructive interference S1 and S2 are two coherent sources in antiphase
S1
x1 Q (minimum)
S2
x2
+
=
Figure 2.17
25
PHYSICS
CHAPTER 2
A dark fringe is observed at Q thus Δφ = 2m + 1 π where m = 0,±1,±2,...
(
)
E1P = E0 sin(ωt − kx1 ) E2 P = E0 sin(ωt − kx2 + π ) then Δφ = (ωt − kx2 + π ) − (ωt − kx1 ) Δφ = k ( x1 − x2 ) + π 2π Δφ = ∆L + π λ 2π ∆L = mλ ∆L + π therefore ( 2m + 1)π = λ Note : where m = 0,±1,±2,..... When
At P,
m=0 m=±1 m=±2
Central dark fringe (zeroth order dark) 1st dark fringe (1st order dark) 2nd dark fringe (2nd order dark)
26
PHYSICS
CHAPTER 2
Interference pattern for two coherent sources in antiphase Fringe 2nd dark 2nd bright 1st dark 1st bright Central dark 1st bright 1st dark 2nd bright 2nd dark
S1 S2
∆φ
m
∆L
5π 4π 3π 2π π 2π 3π 4π 5π
2 1 1 0 0 0 −1 −1 −2
2λ 3 2
λ
1 2
λ
λ 0 1 2
λ
3 2
λ
λ
2λ
screen Figure 2.18 27
PHYSICS
CHAPTER 2
Table 2.1 shows the summary of chapter 2.2.3. Two Coherent Bright fringe Dark fringe sources
In phase
∆L = mλ
m = 0,±1,±2,... Δφ = 2mπ m = 0,1,2,... Antiphase
1 ∆L = m + λ 2 m = 0,±1,±2,... Δφ = 2mπ m = 1,2,... Table 2.1
1 ∆L = m + λ 2 m = 0,±1,±2,... Δφ = (2m + 1)π m = 0,1,2,... ∆L = mλ
m = 0,±1,±2,... Δφ = (2m + 1)π m = 0,1,2,... 28
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PHYSICS CHAPTER 2 Learning Outcome: 2.3 Interference of transmitted light through doubleslits (2 hours) At the end of this chapter, students should be able to: Derive with the aid of a diagram and use
mλD ym = d
xm
for bright fringes (maxima)
( m + 12 ) λD = d
for dark fringes (minima),
where m = 0, ±1, ±2, ±3, … .
Use expression
λD ∆y = d
and
explain the effect of changing any of the variables.
29
PHYSICS
CHAPTER 2
2.3 Interference of transmitted light through double-slits 2.3.1 Methods of obtaining two coherent sources Division of wavefront
S
S1 S2
monochromatic light source single slit
double slits
A slit S is placed at equal distance from slits S1 and S2 as shown in figure.
Light waves from S that arrived at S1 and S2 are in phase.
Therefore, both slits S1 and S2 are two new coherent sources, e.g. in Young’s double slit experiment
Figure 2.19 30
PHYSICS
CHAPTER 2
Division of amplitude incident ray
1 2 partial reflection
air film
t
Both reflected waves 1 and 2 are coherent and will result in interference when they superpose.
e.g. Newton’s ring, air wedge fringes and thin film interference.
partial transmission
air
The incident wavefront is divided into two waves by partial reflection and partial transmission.
Figure 2.20
31
PHYSICS
CHAPTER 2
2.3.2 Young’s double-slit experiment Figure 2.21 shows the schematic diagram of Young’s double-slit Intensity experiment.
Max
m=2
Min Max S monochromatic light beam single slit Picture 2.1
S1
Min
S2
Min
Max Max
double slits Figure 2.21
m=1 m=0 m = −1
Min Max screen
m = −2 interference pattern 32
PHYSICS
CHAPTER 2
Explanation of Young’s double-slit experiment by using Huygens’ principle Wavefront from light source falls on a narrow slit S and diffraction occurs. Every point on the wavefront that falls on S acts as sources of secondary wavelets that will produce a new wavefront that propagate to slits S1 and S2 .
S1 and S2 are produced two new sources of coherent waves in phase because they originate from the same wavefront and their distance from S are equal. An interference pattern consisting of bright and dark fringes is formed on the screen as shown in Figure 2.21. The bright fringes are occurred when the light from slits S1 and S2 superposes constructively. The dark fringes are occurred when the light from slits S1 and S2 superposes destructively. 33
PHYSICS
CHAPTER 2
Derivation of Young’s double-slit equations Equation for separation between central bright fringe and mth bright fringe (m+1)th bright M
∆y
P
θ
ym
S1
d
θ
Q
O
N
S2
mth bright
y m +1 Central bright
mλ
D double-slit
screen Figure 2.22
34
PHYSICS
CHAPTER 2
Suppose P in Figure 2.22 is the mth order bright fringe, thus
S2 P − S1P = mλ
Let OP = ym = distance from P to O .
In practice d is very small (<1mm) and D>>d, then S1N meets PQ at right angle. Hence NP = S P then S N = S P− NP = mλ . 1 2 2 angle PQO = angle S2S1N = θ From the figure, S2 N mλ sin θ = = ΔS2S1N
S2S1 d PO ym tanθ = = ΔPQO QO D Since θ is small, thus sinθ = tanθ
mλ ym = d D
35
PHYSICS
CHAPTER 2
Therefore, the separation between central bright and mth bright fringes, ym is given by
mλD ym = d
where
(2.1)
m : order = 0,±1,±2,... λ : wavelength D : distance between double - slits and the screen d : separation between double - slits
Note: For bright fringes Central bright fringe (Zeroth order maximum) m=0
m = ±1 m = ±2 m = ±3
1st bright fringe (1st order maximum) 2nd bright fringe (2nd order maximum) 3rd bright fringe (3rd order maximum) 36
PHYSICS
CHAPTER 2
Equation for separation between central bright fringe and mth dark fringe R
mth order dark
∆y
(m−1)th order dark
θ
xm
S1
d
θ
Q S2
N
1 m + λ 2
double-slit
O
Central bright
D screen Figure 2.23 37
PHYSICS
CHAPTER 2
Suppose R in Figure 2.23 is the mth order dark fringe, thus
1 S2 R − S1R = m + λ 2 Let OR = xm = distance from R to O .
In practice d is very small (<1mm) and meets RQ at right angle. Hence,
D>>d, then S1N
1 λ NR = S R then S2 N = S2 R − NR = m + 1 2 angle RQO = angle S S N = θ 2 1 S2 N 1λ From the figure, sin θ = = m + ΔS2S1N S2S1 2 d RO xm tanθ = = ΔRQO QO D 1 λ xm Since θ is small, thus sinθ = tanθ m + = 2 d D
38
PHYSICS
CHAPTER 2
Therefore, the separation between central bright and mth order dark fringes, xm is given by
1 λD xm = m + 2 d
where
(2.2)
m : order = 0,±1,±2,...
Note: For dark fringes 1st dark fringe (Zeroth order minimum) m=0
m = ±1 m = ±2 m = ±3
2nd dark fringe (1st order minimum) 3rd dark fringe (2nd order minimum) 4th dark fringe (3rd order minimum)
39
PHYSICS
CHAPTER 2
Equation for separation between successive (consecutive) bright or dark fringes, ∆ y (Figure 2.22) is given by
λD mλD and y m +1 = ( m + 1) ∆y = ym +1 − ym where ym = d d
λD mλD ∆y = ( m + 1) − d d λD ∆y = d where
(2.3)
∆y : separation between consecutive bright or dark fringes λ : wavelength D : distance between double - slits and the screen d : separation between double - slits 40
PHYSICS
CHAPTER 2
Appearance of Young’s double-slit experiment From the equation (2.3), (2.3)
λD ∆y = d
∆ y depends on :
the wavelength of light, λ
the distance apart, d of the double slits,
distance between slits and the screen, D Explanation for the above factors:
if λ is short and thus Δy decreases for fixed D and d. The interference fringes are closer to each other and vice-versa. if the distance apart d of the slits diminished, Δy increased for fixed D and λ and vice-versa. if D increases Δy also increases for fixed λ and vice-versa. 41
PHYSICS
CHAPTER 2
if a source slit S (Figure 2.21) is widened the fringes gradually disappear. The slit S then equivalent to large number of narrow slits, each producing its own fringe system at different places. The bright and dark fringes of different systems therefore overlap, giving rise to a different illumination. if one of the slit, S1 or S2 is covered up, the fringes disappear. if the source slit S is moved nearer the double slits, Δy is unaffected but their intensity increases. if the experiment is carried out in a different medium, for example water, the fringe separation Δy decreased or increased depending on the wavelength, λ of the medium. if white light is used the central bright fringe is white, and the fringes on either side are coloured. Violet is the colour nearer to the central fringe and red is farther away as shown in Figure 2.24. 42
PHYSICS
CHAPTER 2
Figure 2.24 Table 2.2 shows the range of wavelength for colours of visible light. Colour Range of λ/ nm
Stimulation 2.2 Stimulation 2.3
Violet
400 – 450
Blue
450 – 520
Green
520 – 560
Yellow
560 – 600
Orange
600 – 625
Red
625 - 700 Table 2.2
43
PHYSICS
CHAPTER 2
Example 1 : A double-slits pattern is view on a screen 1.00 m from the slits. If the third order minima are 25.0 cm apart, determine a. the ratio of wavelength and separation between the slits, b. the distance between the first order minimum and fourth order maximum on the screen. Solution : a.
d
D = 1.00 m; ∆x3 = 0.25 m; m = 3
3rd order minimum
S1
∆x3
S2
D
x3 zeroth order maximum
x3 3rd order minimum
From the figure,
∆x3 0.25 x3 = = 2 2 x3 = 0.125 m
44
PHYSICS
CHAPTER 2
Solution : D = 1.00 m; ∆x3 = 0.25 m; m = 3 a. By using the equation of separation between central bright and
1 λD x3 = 3 + 2 d 1 λ (1.00 ) 0.125 = 3 + 2 d λ = 3.57 ×10 − 2 d
mth order dark 1 λDthus fringes,
xm = m + 2 d
b. The separation between central max and the 1st order min. is
1 λD x1 = 1 + 2 d λD x1 = 1.5 d
45
PHYSICS
CHAPTER 2
Solution : b. and the separation between central max and the 4th order max. (m = 4) is given by
mλD ym = d
λD y4 = 4 d
Therefore the distance between the first order minimum and fourth order maximum on the screen is
∆d = y4 − x1 λD λD ∆d = 4 − 1. 5 d d λD ∆d = 2.5 d ∆d = 2.5( 3.57 ×10 −2 )(1.00 ) ∆d = 8.93 ×10 −2 m
46
PHYSICS
CHAPTER 2
Example 2 : a. How would you expect the interference pattern of a double-slit experiment to change if white light is used instead of monochromatic light? b. Describe the changes that would be observed in a double-slit interference pattern if the entire experiment were submerged in water. (Physics, 3rd edition, J. S. Walker, Q4&Q6, p.963)
Solution : a. The locations of bright and dark fringes depends on the wavelength of light. light Therefore, if white light is used in a double-slit experiment, each bright fringe will show some separation into colours, colours giving a “rainbow” effect. effect
47
PHYSICS
CHAPTER 2
Solution : b. Submerging the double-slit experiment in water would reduce the wavelength of the light from λ to λ /n, where n = 1.33 is the refraction index of water. Therefore, the bright or dark fringe separation would be reduced, reduced according to the λD equation below:
∆y =
d
It follows that the interference pattern fringes get closer to each other. other
48
PHYSICS
CHAPTER 2
Example 3 : In a Young’s double-slit experiment, when a monochromatic light of wavelength 600 nm shines on the double slits, the fringe separation of the interference pattern produced is 7.0 mm. When another monochromatic light source is used, the fringe separation is 5.0 mm. Calculate the wavelength of the second light. Solution : λ1
= 600 ×10 −9 m; ∆y1 = 7.0 ×10 −3 m; ∆y2 = 5.0 ×10 −3 m
1st case:
d
∆y1
S1 S2
∆y1 Central of interference pattern
D By applying the fringe separation equation, thus
λ1 D ∆y1 = d
7.0 × 10 −3
( 600 ×10 ) D = −9
d
(1) 49
PHYSICS
CHAPTER 2
Solution : λ1 2nd case:
= 500 ×10 −9 m; ∆y1 = 7.0 ×10 −3 m; ∆y2 = 5.0 ×10 −3 m
d
S2
λ2 D ∆y2 = d (2)÷ (1):
∆y2
S1
∆y2 Central of interference pattern
D
λ2 D 5.0 ×10 = d −3 5.0 ×10 λ2 = −3 7.0 ×10 600 ×10 −9 −3
(2)
λ2 = 429 ×10 −9 m OR 429 nm 50
PHYSICS
CHAPTER 2
Example 4 :
A
S1 1.2 mm S2
2.5 m Figure 2.25
B
Figure 2.25 shows two coherent sources (S1 and S2) of light in phase. The separation of S1 and S2 is 1.2 mm and the screen is 2.5 m from the sources. a. The frequency of the light is 5.77 × 1014 Hz. Calculate i. the wavelength of the light used ii. the separation between two consecutive bright fringes if the experiment is carried out in air. b. If the experiment is carried out in water of refractive index 1.33, calculate the separation of two consecutive dark fringes. (The speed of light in vacuum, c = 3.00 × 108 m s−1)
51
PHYSICS
CHAPTER 2 −3
Solution : d = 1.2 × 10 m; D = 2.5 m 14 a. i. Given f = 5.77 × 10 Hz By applying the wave speed equation, thus
c = λf 8 14 3.00 ×10 = λ 5.77 ×10 λ = 520 ×10 −9 m OR 520 nm
(
)
ii. By using the equation of fringe separation, thus
λD ∆y = d −9 520 ×10 ( 2.5) ∆y = −3 1.2 × 10 ∆y = 1.08 × 10 −3 m
(
)
52
PHYSICS
CHAPTER 2 −3
Solution : d = 1.2 × 10 m; D = 2.5 m b. Given n = 1.33 The wavelength of light in water is given by
λ n= λw
520 ×10 −9 1.33 = λw −7 λw = 3.91×10 m
Therefore the dark fringes separation is
λD ∆x = d 3.91×10 −7 ( 2.5) ∆x = 1.2 ×10 −3
(
)
∆x = 8.15 × 10 −4 m 53
PHYSICS
CHAPTER 2
Exercise 2.1 : 1.
Young’s double-slit experiment is performed with 589-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits. (Physics for scientists and engineers,6th edition,Serway&Jewett, Q37.5, p.1198)
ANS. : 1.54 mm 2. A Young’s interference experiment is performed with monochromatic light. The separation between the slits is 0.500 mm, and the interference pattern on a screen 3.30 m away shows the first side maximum 3.40 mm from the centre of the pattern. What is the wavelength? (Physics for scientists and engineers,6th edition,Serway&Jewett, Q37.2, p.1197)
ANS. : 515 nm 54
PHYSICS
CHAPTER 2
Exercise 2.1 : 3.
A coherent light that contains two wavelength, 660 nm (red) and 470 nm (blue) es through two narrow slits separated by 0.3 mm and the interference pattern is observed on a screen 5.00 m from the slits. Determine the distance between the first order bright fringes for each wavelength. (University physics,11th edition, Young&Freedman, Q35.14, p.1362)
ANS. : 3.17 mm 4. A monochromatic light of wavelength 560 nm es through a Young’s double-slit system of unknown slit separation. After that, the slits is illuminated by a monochromatic light of unknown wavelength. It was observed that the 4th order minimum of the known wavelength light overlapped with the 5th order maximum of the unknown wavelength light on a screen. Calculate the wavelength of the unknown wavelength light. ANS. : 504 nm 55
www.kmph.matrik.edu.my/physic s
PHYSICS CHAPTER 2 Learning Outcome: 2.4 Interference of reflected light in thin films (1 hour) At the end of this chapter, students should be able to: Explain with the aid of a diagram the interference of light in thin films for normal incidence. For non-reflective coating: Constructive interference : 2nt = mλ
Destructive interference : 2nt = (m + ½ )λ For reflective coating: Constructive interference : 2nt = (m + ½ )λ
Destructive interference : 2nt = mλ where m = 0, ±1, ±2, ±3, …
56
PHYSICS
CHAPTER 2
2.4 Interference of reflected light in thin films
Interference due to reflected waves is observed in many everyday circumstances such as bright colours reflected from oil film on water and soap bubble. The reflected waves can change their phase in two ways: The phase changes in proportion to the distance of the waves travel. travel The phase changes as a result of the reflection process itself. Optical path is defined as the product between a distance travelled by light and the refractive index of the medium OR
L = nl
where
L : optical path n : refractive index of a medium l : distance travelled by light in the medium 57
PHYSICS
CHAPTER 2
2.4.1 Phase changes due to reflection
A light wave travelling in a medium of lower refractive index (n1) when reflected from a medium’s surface of higher refractive index (n2) undergoes a π radian phase change as shown in Figure 2.26a.
π radian phase change
incident wave
n1 reflected wave
incident pulse
n2 transmitted wave
transmitted pulse
Note:
∆φ = π rad λ ∆L = 2
n1
n2
Figure 2.26a
reflected pulse Figure 2.26b: string analogous 58
PHYSICS
CHAPTER 2
A light wave travelling in a medium of higher refractive index (n2) when reflected from a medium’s surface of lower refractive index (n1) undergoes no phase change as shown in Figure 2.27a. no phase change incident wave
n2 reflected wave
incident pulse
n1 transmitted wave
transmitted pulse
Note:
∆φ = 0 ∆L = 0
n2
n1
Figure 2.27a Stimulation 2.4
reflected pulse Figure 2.27b: string analogous 59
PHYSICS
CHAPTER 2
2.4.2 Interference from thin films on a denser medium
Figure 2.28 shows the light waves reflected from the upper and lower surfaces of a thin film (refractive index, n) on a denser medium. π rad phase π rad phase change change E F A 1
n1 = 1.0
D
B
n = 1.5
2
t : thickness of thin film
C
n2 = 3.5 Figure 2.28
60
PHYSICS
CHAPTER 2
When an incident ray falls on a thin film surface almost normal to the surface (point B) division of amplitude occurs, part of ray are reflected (ray 1→ ray ABE), part of ray are refracted and reflected (ray 2→ ray ABCDF), point D very close to B (BC and CD become straight line). At B,
the reflected ray (ray 1) undergoes π radian phase change. change because the ray 1 reflected from a surface of higher refractive index (denser medium). medium) At C,
the reflected ray (ray 2) undergoes π
radian phase change. change Therefore both rays 1 and 2 are two coherent sources in phase because the phase difference, ∆ φ is
∆φ = interference π − π = 0 pattern. and meet at a point produces 61
PHYSICS
CHAPTER 2
The optical path difference between rays 1 and 2 is given by
∆L = ABCDF− ABE ∆L = BC + CD ∆L = nt + nt ∆L = 2nt
Constructive interference: where
2nt = mλ
λ : wavelength of light in vacuum m = 0,±1,±2,...
Destructive interference:
1 where 2nt = m + λ 2 m = 0,±1,±2,...
Example of thin film on a denser medium: Non-reflective (anti-reflective) coating Oil film on water and etc… Picture 2.2
Picture 2.3
Picture 2.4 62
PHYSICS
CHAPTER 2
2.4.3 Interference from thin films on a less dense medium
Figure 2.29 shows the light waves reflected from the upper and lower surfaces of a thin film (refractive index, n) in a less dense medium. π rad phase no phase change change E F A 1
nair = 1.0 n = 1.33
2
D
B
t : thickness of thin film
C
nair = 1.0 Figure 2.29
63
PHYSICS
CHAPTER 2
When an incident ray falls on a thin film surface almost normal to the surface (point B) division of amplitude occurs, part of ray are reflected (ray 1→ ray ABE), part of ray are refracted and reflected (ray 2→ ray ABCDF), point D very close to B (BC and CD become straight line). At B, the reflected ray (ray 1) undergoes π radian phase change. change because the ray 1 reflected from a surface of higher refractive index (denser medium). medium) At C, the reflected ray (ray 2) undergoes no phase change. change Therefore both rays 1 and 2 are two coherent sources antiphase because the phase∆difference, is φ = π − 0∆= φπ rad
and meet at a point produces interference pattern.
64
PHYSICS
CHAPTER 2
The optical path difference between rays 1 and 2 is given by
∆L = ABCDF− ABE λ ∆L = BC + CD + λ 2 λ ∆L = nt + nt + ∆L = 2nt + 2 2
Constructive interference:
λ 2nt + = mλ 2
1 where 2nt = m + λ m = 0,±1,±2,... 2
Destructive interference:
λ 1 2nt + = m + λ 2 2
where
2nt = mλ m = 0,±1,±2,...
Example of a thin film on a less dense medium: Soap bubbles Reflective coating and etc… Picture 2.5
Picture 2.6
65
PHYSICS
CHAPTER 2
Example 5 : A non-reflective coating of magnesium fluoride of refractive index 1.38 covers the camera lens of refractive index 1.52. The coating prevents reflection of yellow-green light of wavelength in vacuum 565 nm. Determine the minimum non zero thickness of the magnesium fluoride. (Physics,7th edition, Cutnell&Johnson, Q48, p.886)
Solution :
λ = 565 ×10 −9 m π rad phase change
n1 = 1.0 n = 1.38
n2 = 1.52
1
2
π rad phase change
2 coherent sources in phase
t min 66
PHYSICS
CHAPTER 2
Solution : λ = 565 × 10 −9 m By using the condition of destructive interference for non-reflective coating, thus
1 2nt = m + λ 2
For minimum thickness of MgF2,
2nt min 2(1.38) t min
m=0
1 = 0 + λ 2 1 = ( 565 ×10 −9 ) 2
t min = 1.02 ×10 −7 m OR
102 nm 67
PHYSICS
CHAPTER 2
Example 6 : White light is incident on a soap film of refractive index 1.30 in air. The reflected light looks bluish because the red light of wavelength 670 nm is absent in the reflection. a. State the condition for destructive interference. b. What is the minimum thickness of the soap film? (Physics,3rd edition, J.S.Walker, Q26, p.966)
Solution :
λ = 670 ×10 −9 m π rad phase change
n1 = 1.0 n = 1.30
n1 = 1.0
1
2
no phase change 2 coherent sources antiphase
t min 68
PHYSICS
CHAPTER 2
Solution : λ = 670 × 10 −9 m a. The condition of destructive interference is given by
2nt = mλ b. For minimum thickness of soap film,
2nt min = (1) λ
(
2(1.30 ) t min = 670 ×10 −9
m =1
)
−7
t min = 2.58 × 10 m OR
258 nm
69
www.kmph.matrik.edu.my/physic s
PHYSICS CHAPTER 2 Learning Outcome: 2.5 Interference of reflected light in air wedge and Newton’s rings (1 hour) At the end of this chapter, students should be able to: Explain with the aid of a diagram the interference in air wedge. Explain with the aid of a diagram the formation of Newton’s rings. Use 2t = (m + ½)λ for bright fringes (maxima) 2t = mλ for dark fringes (minima), where m = 0, 1, 2, 3, …
70
PHYSICS CHAPTER 2.5 Interference of reflected light2in air wedge and Newton’s ring 2.5.1 Air wedge
S
travelling microscope
α X glass plate
l
monochromatic light source
L Q
O P B
t
T
Y
L Figure 2.31
m= 0 0 1 1 2 2 3 3 4 4 5 glass slide thin foil
1st dark fringe Figure 2.30: Apparatus setup
∆x
Figure 2.32
air
71
PHYSICS
CHAPTER 2 Ray S falls almost normal to the surface of a glass slide. At point O, Ray S is partially reflected (ray OL) partially refracted (OB) and then reflected at B (ray PQ) The two refracted rays (OL and PQ) are coherent since both have originated from the same source O. OL and PQ produces interference pattern if it is brought together as shown in Figure 2.32. Since the incidence is nearly normal (point P very close to O), the path difference between the rays at O (ray OL and ray OBPQ) is given by, path difference, ∆
L = OB + BP = nt + nt = 2nt
where n is refractive index of air = 1.0
At X, t = 0 and thus the path difference = 0 and a bright fringe is expected, but a dark fringe is observed at X. This is due to the phase change of π radian for ray PQ (reflected on a denser medium at B). 72
PHYSICS
CHAPTER 2 Hence, ray PQ is in antiphase with ray OL and when brought together (by the retina or lens) to interfere, a dark fringe is obtained. Constructive interference (bright fringe):
1 2t = mλ + λ 2
1 2t = m + λ 2 where
(2.4)
m = 0,1,2,...
Destructive interference (dark fringe):
2t = mλ Note:
where
A phase change of difference of ½ λ
(2.5)
m = 0,1,2,...
π radian is equivalent to a path 73
PHYSICS
CHAPTER 2 From equation (2.4), When
m = 0; t = 14 λ m = 1; t = 34 λ m = 2; t = 54 λ
1st bright fringe (Zeroth order maximum) 2nd bright fringe (1st order maximum) 3rd bright fringe (2nd order maximum)
i.e. bright fringes are formed when the thickness of air film, t = 14 λ , 34 λ , 54 λ ,.......
From equation (2.5), When
m = 0; t = 0 m = 1; t = 12 λ m = 2; t = λ
1st dark fringe (Zeroth order minimum) 2nd dark fringe (1st order minimum) 3rd dark fringe (2nd order minimum)
i.e. dark fringes are formed when the thickness of air film, t = 0, 12 λ , λ , 32 λ ,.......
74
PHYSICS
CHAPTER 2 Equation for separation between the 1st dark fringe and the mth order dark fringe, l
From Figure 2.31,
T t tan α = = L l
t (2.6) l= tan α mλ Rearrange eq. (2.5): t = substitute into eq. (2.6) 2 mλ l= (2.7) 2 tan α where
m : order = 0,1,2,... λ : wavelength of light in vacuum
α : angle of inclination of glass slide 75
PHYSICS
CHAPTER 2 Equation for separation between the 1st dark fringe and the mth order bright fringe, l
Rearrange eq. (2.4):
( m + 12 ) λ substitute into eq. (2.6) t=
( m + 12 ) λ l= 2 tan α
where
2
(2.8)
m : order = 0,1,2,...
Equation for separation between adjacent dark fringes or bright fringes, ∆ x
Put m = 1 into eq. (2.7),
λ ∆x = 2 tan α
(2.9) 76
PHYSICS
CHAPTER 2 C
2.5.2 Newton’s ring Figure 2.34
travelling microscope
monochromatic light source
A
t
glass plate
L
R R −t
Q
S
Y O X
d
P B
plano-convex lens glass block
Figure 2.33: Apparatus setup
Figure 2.35: Newton’s ring
77
PHYSICS
CHAPTER 2 Ray S falls almost normal to the surface of a plano-convex. At point O, Ray S is partially reflected (ray OL) partially refracted (OB) and then reflected at B (ray PQ) The two refracted rays (OL and PQ) are coherent since both have originated from the same source O. OL and PQ produces interference pattern if it is brought together as shown in Figure 2.35. The pattern is a series of circular interference fringes called Newton’s ring. This because of a curved piece of glass with a spherical cross section. Since the incidence is nearly normal (point P very close to O), the path difference between the rays at O (ray OL and ray OBPQ) is given by, path difference, ∆
L = OB + BP = nt + nt = 2nt
where n is refractive index of air = 1.0 78
PHYSICS
CHAPTER 2 At X, t = 0 and thus the path difference = 0 and a bright spot is expected, but a dark spot is observed at X. This is due to the phase change of π radian for ray PQ (reflected on a denser medium at B). Hence, ray PQ is in antiphase with ray OL and when brought together (by the retina or lens) to interfere, a dark spot is obtained. Constructive interference (bright ring):
1 2t = mλ + λ 2
1 2t = m + λ 2
(2.8)
where m = 0,1,2,... Destructive interference (dark ring):
2t = mλ where
m = 0,1,2,...
(2.9) 79
PHYSICS
CHAPTER 2 Relationship between diameter of ring, d and thickness of air gap, t
From Figure 2.34, C
By using the Phytogorean theorem, thus the distance AY is
AY 22 = AC2 − YC 2
R R −t A
Y
d 2
d 2 2 = R − ( R − t) 2 d2 = 2 Rt + t 2 4
Since t is very thin thus t2 ≈ 0
d2 = 2 Rt 4
(2.10)
80
PHYSICS
CHAPTER 2 Equation for diameter of dark ring
mλ substitute into eq. (2.10) Rearrange eq. (2.9): t = 2 d2 mλ = 2 R 4 2 When
m = 0; t = 0 m = 1; t = 12 λ m = 2; t = λ m = 3; t = 32 λ
d 2 = 4 Rmλ where
(2.11)
m : order = 0,1,2,...
Central dark spot (zeroth order minimum), d = 0 1st dark ring (1st order minimum) 2nd dark ring (2nd order minimum) 3rd dark ring (3rd order minimum)
81
PHYSICS
CHAPTER 2 Equation for diameter of bright ring
( m + 12 ) λ substitute into eq. Rearrange eq. (2.9): t = d ( m + )λ = 2R 4 2 2
1 2
2
d = 4 R( m + 12 ) λ 2
where When
(2.10)
(2.12)
m : order = 0,1,2,...
m = 0; t = 14 λ m = 1; t = 34 λ m = 2; t = 54 λ m = 3; t = 74 λ
1st bright ring (zeroth order maximum) 2nd bright ring (1st order maximum) 3rd bright ring (2nd order maximum) 4th bright ring (3rd order maximum) 82
PHYSICS
CHAPTER 2 From Figure 2.35,
The rings become more closely spaced as one moves farther from the centre of the Newton’s ring.
The reason is that the convex surface of the lens moves away from the lower glass block at a progressively faster rate therefore the thickness of air film increases rapidly. rapidly
Newton’s ring can be used to test the accuracy with which a lens has been ground. ground
The rings are not circular if the surface is not spherical (or the glass block is not flat). flat)
83
PHYSICS
CHAPTER 2
Example 7 : An air wedge is formed by placing a human hair between two glass slides of length 44 mm on one end, and allowing them to touch on the other end. When this wedge is illuminated by a red light of wavelength 771 nm, it is observed to have 265 bright fringes. Determine a. the diameter of hair, b. the angle of air wedge, c. the thickness of the air film for 99th dark fringe to be observed, d. the separation between two consecutive bright fringes. −9 −3 Solution : λ = 771 × 10 m; L = 44 × 10 m
d
α
L
84 265th bright fringe
PHYSICS
CHAPTER 2
Solution : λ = 771 × 10 −9 m; L = 44 × 10 −3 m a. Assuming the diameter of the hair, d = the thickness of air film, t at 265th bright fringe Therefore the diameter of the hair is given by
1 2t = m + λ and m = 264 2 1 2d = 264 + 771×10 −9 2 d = 1.02 × 10 −4 m
(
)
b. The angle of air wedge is
d tan α = L
1.02 ×10 −4 tan α = −3 44 ×10 α = 0.13 85
PHYSICS
CHAPTER 2
Solution : λ = 771 × 10 −9 m; L = 44 × 10 −3 m c. By applying the equation for dark fringe (air wedge), thus 2t = mλ and m = 98
(
2t = 98 771×10 −9
)
t = 3.78 ×10 −5 m d. The separation between two consecutive bright fringes is
λ ∆x = 2 tan α −9 771×10 ∆x = 2 tan 0.13 ∆x = 1.70 ×10 −4 m
86
PHYSICS
CHAPTER 2
Example 8 : a. Explain why the central spot in Newton’s ring is dark. b. In a Newton’s ring experiment, the radius of the qth bright ring is 0.32 cm and the radius of the (q+19)th dark ring is 0.67 cm. Determine the radius of curvature of the plano-convex used in the experiment if the wavelength of light used is 589 nm. Solution : a. A ray of light reflected from the lower surface of the convex surface has no phase change. Meanwhile, a ray of light reflected from the top surface of glass block undergoes a π radian phase change. Thus the two reflected rays are two coherent sources in antiphase. At the centre of the interference pattern, the thickness of the air film is zero, hence the path difference for these two rays goes to zero. These resulting a destructive interference at the central of the Newton’s ring. 87
PHYSICS
CHAPTER 2
= 0.32 × 10 −2 m; r( q +19 ) = 0.67 × 10 −2 m λ = 589 ×10 −9 m b. For qth bright ring, m = q − 1 Solution : rq
d 2 = 4 R( m + 12 ) λ
( 2r ) ( 2r )
2
q
and
= 4 R[ ( q − 1) + 12 ]λ
d = 2rq
= 4 R( q − 0.5) λ For (q+19) dark ring, m = q + 19 d 2 = 4 Rmλ and d = 2r( q +19 ) 2
q
(1)
th
( 2r( ( 2r(
) ))
q +19 ) q +19
2
= 4 R( q + 19 ) λ
2
= 4 R( q + 19 ) λ
(2)
88
PHYSICS
CHAPTER 2
= 0.32 × 10 −2 m; r( q +19 ) = 0.67 × 10 −2 m λ = 589 ×10 −9 m 2 r( q +19 ) b. (2) ÷ (1) : = q + 19 r q − 0.5 q −2 2 0.67 ×10 q + 19 = q = 6.27 −2 0.32 ×10 q − 0.5 Solution : rq
By substituting q = 6.27 into eq. (1) thus
( 2( 0.32 ×10 ) ) −2
(
= 4 R( 6.27 − 0.5) 589 ×10 −9 R = 3.01 m 2
89
)
PHYSICS
CHAPTER 2
Exercise 2.2 : 1. A thin film of gasoline floats on a puddle of water. Sunlight falls almost perpendicularly on the film and reflects into your eyes. Although the sunlight is white, since it contains all colours, the film has a yellow hue, because destructive interference has occurred eliminating the colour of blue (λ =469 nm) from the reflected light. If the refractive indices for gasoline and water are 1.40 and 1.33 respectively, Calculate the minimum thickness of the film. ANS. : 168 nm 2. White light is incident normally on a thin soap film (n =1.33) suspended in air. a. What are the two minimum thickness that will constructively reflect yellow light of wavelength 590 nm? b. What are the two minimum thickness that will destructively reflect yellow light of wavelength 590 nm? (Physics,3rd edition, J.S.Walker, Q34, p.966)
ANS. : 110 nm, 330 nm ; 220 nm, 440 nm ;
90
PHYSICS
CHAPTER 2
3. Two plane glass plates which are in at one edge are separated by a piece of metal foil 12.5 cm from that edge. Interference fringes parallel to the line of are observed in reflected light of wavelength 546 nm and are found to be 1.50 mm apart. Determine the thickness of the foil. ANS. : 2.27× 10−5 m 4. Newton’s rings are formed by reflection between an biconvex lens of focal length 100 cm made of glass of refractive index 1.50 and in with a glass block of refractive index 1.60. Calculate the diameter and thickness of air film for fifth bright ring using light of wavelength 6000 Å . Given 1 angstrom (Å) = 10−10 m ANS. : 3.28 mm; 1.35 µ m 5. Newton’s rings are formed with light of wavelength 589 nm between the plano-convex lens of radius of curvature 100 cm and a glass block, in perfect . a. Determine the radius of the 20th dark ring from the centre. b. How will this ring move and what will its radius become if the lens and the block are slowly separated to a distance apart 5.00 × 10−4 cm? 91 ANS. : 3.43 mm; inwards, 1.26 cm
www.kmph.matrik.edu.my/physic s
PHYSICS CHAPTER 2 Learning Outcome: 2.6 Diffraction by a single slit (1 hour) At the end of this chapter, students should be able to: Explain with the aid of a diagram the diffraction of a single slit. Derive and use formula
nλD yn = a
yn
for dark fringes (minima)
( n + 12 ) λD = a
for bright fringes (maxima),
where n = ±1, ±2, ±3, ... Explain with the aid of a diagram the effect of changing wavelength on the resolution of single slit from two coherent sources. 92
PHYSICS
CHAPTER 2
2.6 Diffraction by a single slit 2.6.1 Diffraction of light
is defined as the bending of light waves as they travel around obstacles or through an aperture or slit comparable to the wavelength of the light waves. waves Figures 2.36a, 2.36b and 2.36c show the bending of plane wavefront.
λ
Figure 2.36a: obstacle
λ
Figure 2.36b: slit, a > λ
λ
Figure 2.36c: slit, a ≈ λ
93
PHYSICS
CHAPTER 2
2.6.2 Diffraction by a single silt
Figure 2.37 shows an apparatus setup of diffraction by a single slit. 3rd minimum 2nd maximum 2nd minimum 1st maximum 1st minimum Central maximum 1st minimum 1st maximum 2nd minimum 2nd maximum 3rd minimum
θ1 θ2 S
θ1
θ2
Animation 2.1 Picture 2.7 Picture 2.8
single slit
intensity Figure 2.37
where
screen
θ : angle of diffraction
94
PHYSICS
CHAPTER 2
Explanation of single slit diffraction experiment Wavefront from light source falls on a narrow slit S and diffraction occurs. Every point on the wavefront that falls on S acts as sources of secondary wavelets and superposed each another to form an interference pattern on the screen as shown in Figure 2.37. The central fringe is bright (maximum) and widen compare to other bright fringes. The central fringe has the highest intensity compare to the other bright fringes. The intensity of bright fringes reduce as the distance increase from the central bright fringe.
Other rays with angle θ2 and θ1 will produce minimum and maximum on both sides of the central maximum.
95
PHYSICS
CHAPTER 2
Derivation of single slit diffraction equations Equation for separation between central maximum (bright) and nth minimum (dark) fringes P nth minimum
yn
p st r i 1 θ1 nd strip θ1 2 aE st
a
where
a 2
A
a 2
C B
2
Central Q maximum
sin θ1
a sin θ1 2
D
a : slit width D : distance between single slit and screen Figure 2.38
screen 96
PHYSICS
CHAPTER 2
A single slit is split into two equal parts, AC and CB. A,C and B are new sources of secondary wavelets. (Huygen’s principle) When the wavelets from A, C and B superpose, interference will occur at P. As AB is very small, small thus AE is perpendicular to and AP = EP, EP the outgoing rays are considered parallel, parallel and therefore the path difference at P between ray AP a and is :
∆L = CE =
2
sin θ1
Consider two narrow strips as shown in Figure 2.38, for the two strips superposed destructively thus both strip of light must in antiphase to each another which is equivalence to a path difference of ½λ . If the 1st minimum (1st order minimum) is at P, hence :
a λ ∆L = sin θ1 = 2 2
a sin θ1 = λ
97
PHYSICS
CHAPTER 2
a
For the 2nd minimum and 3rd minimum, minimum AB is split into 4 equal parts, 6 equal parts and so on as shown in Figures 2.39 and 2.40. ip p r t p st s tri i r t st s 1 nd s rip t a a 1 p 2 i r rd s rip t nd s 3 th st rip 6 4 2 p ri 4 th st trip rd st 3 5 th s rip a th st 6 4
2λ 2nd minimum (2nd order minimum)
a λ sin θ 2 = 4 2 a sin θ 2 = 2λ
Figure 2.39
θ3
θ2
3λ 3rd minimum (3rd order minimum)
a λ sin θ 3 = 6 2 a sin θ 3 = 3λ
Figure 2.40 98
PHYSICS
CHAPTER 2
In general, for minima (dark fringes)
a sin θ n = nλ
where n : order = ±1,±2,±3,.. If the distance of single slit to the screen is D, and D>>a yn then:
sin θ n = tan θ n =
D
Therefore the distance of nth minimum from central maximum is:
When
yn a = nλ D
n = ±1 n = ±2 n = ±3
nλ D yn = a
1st minimum fringe (1st order minimum) 2nd minimum fringe (2nd order minimum) 3rd minimum fringe (3rd order minimum) 99
PHYSICS
CHAPTER 2
Equation for separation between central maximum (bright) and nth maximum (bright) fringes R nth maximum
yn
p stri ip 1 d str n 2 rd strip 3 θ1 st
a
a 3 a 3 a 3
θ1 A C aE sin θ1 D 3 B a sin θ
Central Q maximum
1
D screen Figure 2.41
100
PHYSICS
CHAPTER 2
A single slit is split into three equal parts, AC,CD and DB. A,C,D and B are new sources of secondary wavelets. (Huygen’s principle) When the wavelets from A,C,D and B superpose, interference will occur at R. As AB is very small, thus AE is perpendicular to and AP = EP, the outgoing rays are considered parallel, and therefore the path difference at P between ray AP a and is :
∆L = CE =
3
sin θ1
Consider three narrow strips as shown in Figure 2.41, the first two strips (pair) superposed destructively at which the path difference is ½λ and leave the third strip. strip The 3rd strip produces the maximum (bright) fringe at R.
If the 1st maximum (1st order maximum) is at R, hence :
a λ ∆L = sin θ1 = 3 2
3λ a sin θ1 = 2
101
PHYSICS
CHAPTER 2
a
For the 2nd maximum and 3rd maximum, maximum AB is split into 5 equal parts, 7 equal parts and so on as shown in Figures 2.42 and 2.43. ip r t st s trip p ri 1 nd s trip st st p 2 rd s trip 1 nd stri a a 3 th s trip p i 2 r 7 4 th s tripp 5 rd st p 3 th stri 5 th s tri 6 th s 4 th strip a 7 5
5λ 2
2nd maximum (2nd order maximum)
θ3
θ2
a λ sin θ 2 = 5 2 5λ a sin θ 2 = 2
Figure 2.42
3rd maximum (3rd order minimum)
7λ 2 a λ sin θ 3 = 7 2 7λ a sin θ 3 = 2
Figure 2.43 102
PHYSICS
CHAPTER 2
In general, for maxima (bright fringes)
1 where a sin θ n = n + λ n = ±1,±2,±3,.. 2 If the distance of single slit to the screen is D, and D>>a then:
yn sin θ n = tan θ n = D
Therefore the distance of nth maximum from central maximum is:
1 yn a = n + λ 2 D
When
n = ±1 n = ±2 n = ±3
yn
( n + 12 ) λD = a
1st maximum fringe (1st order maximum) 2nd maximum fringe (2nd order maximum) 3rd maximum fringe (3rd order maximum) 103
PHYSICS
CHAPTER 2
Equation for central maximum (bright) fringe
D 1st minimum
θ1 θ1
a single slit
y1 y1
Q Central maximum 1st minimum
Figure 2.44 A C D E B
Figure 2.45
screen
104
PHYSICS
CHAPTER 2
Figure 2.45 shows five sources of Huygen’s wavelets and the screen is to be so far from the slit (D>>a) thus the rays from each source are nearly parallel. parallel
All the wavelets from each source travel the same distance to the point Q (Figure 2.44) and arriving there in phase. phase
Therefore, the constructive interference is occurred at the central of the single slit diffraction pattern. pattern
The angular width of central maximum, θ
w
is given by
θ w = 2θ1 and θ1 : 1st minimum diffraction angle −1 λ θ1 = sin a
−1 λ θ w = 2 sin a 105
PHYSICS
CHAPTER 2
The width of central maximum, w is given by
w = 2 y1
and
y1 : separation of 1st minimum and central maximum λD y1 = a
λD w = 2 a
Note:
To calculate the maximum number of orders observed, observed take the diffraction angle,θ = 90°.
From both equations for minima and maxima, we obtain
sin θ n ∝ λ
and
yn ∝ λ
By using this two relations, the changes of single slit diffraction pattern can be explained. 106
PHYSICS
CHAPTER 2
Example 9 : A sodium light of wavelength 580 nm shines through a slit and produces a diffraction pattern on a screen 0.60 m away. The width of the central maximum fringe on the screen is 5.0 cm. Determine a. the width of the slit, b. the angular width of the central maximum fringe, c. the number of minimum that can be observed on the screen.
−9 −2 λ = 580 × 10 m; D = 0 . 60 m, w = 5 . 0 × 10 m Solution :
w
a a. Since
w = 2 y1
λD w = 2 a
and
1st minimum Central maximum 1st minimum
λD y1 = a
5.0 × 10
−2
( 580 × 10 )( 0.60) =2 −9
a
a = 1.39 × 10−5 m
107
PHYSICS
CHAPTER 2
Solution :λ = 580 × 10 m; D = 0.60 m, w = 5.0 × 10 b. The angular width of the central maximum fringe is given by −9
−2
λ θ w = 2θ1 and θ1 = sin a −9 580 × 10 −1 λ −1 θ w = 2 sin θ w = 2 sin −5 1.39 × 10 a θ w = 4.78 −1
c. By applying the equation for minimum fringe,
a sin θ = nλ
For the maximum no. of order for minimum fringe, θ
(1.39 ×10 ) sin 90 −5
nmax
(
= nmax 580 ×10 −9 = 23.97 ≈ 23
)
= 90
Therefore the number of minimum that can be observed is
23 × 2 = 46 fringes
108
m
PHYSICS
CHAPTER 2
Example 10 : a. State the similarities and differences of double-slit interference and single slit diffraction patterns. b. How many bright fringes will be produced on the screen if a green light of wavelength 553 nm is incident on a slit of width 8.00 µ m? Solution : a. The similarities are Double-slit interference pattern
Single slit diffraction pattern
Both patterns consist of alternating dark and bright fringes. The central for both patterns is bright fringe.
109
PHYSICS
CHAPTER 2
a. The differences are Double-slit interference pattern
The width of each fringe is the same. The intensity of each bright fringe is constant.
Single slit diffraction pattern The central fringe is wider than the other fringes. The intensity of bright fringes reduce as a distance increase from the central bright.
b. Given λ = 553 × 10 −9 m; a = 8.00 × 10 −6 m By applying the equation for bright (maximum) fringe,
1 a sin θ = n + λ 2
For the maximum no. of order for bright fringe, θ = 90 Central bright −6 −9 8 . 00 × 10 sin 90 = n + 0 . 5 553 × 10 fringe max
(
)
(
nmax = 13.97 ≈ 13
)(
)
Therefore the number of bright that can be observed is
(13 × 2)+1 = 27 fringes
110
PHYSICS
CHAPTER 2
Exercise 2.3 : 1. Monochromatic light of wavelength 689 nm falls on a slit. If the angle between first bright fringes on either side of the central maximum is 38°, calculate the slit width. (Physics for scientist & engineers ,3rd edition, Giancoli, Q4, p.913)
ANS. : 3.2 µ m 2. Light of wavelength 633 nm from a distant source is incident on a single slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. Determine the distance between the two dark fringes on either side of the central bright fringe. (University physics,11th edition, Young&Freedman, Q36.4, p.1396)
ANS. : 5.91 mm 3. A screen is placed 1.00 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.60 cm wide. What is the distance between the two second order minima? (Physics,3rd edition, J.S.Walker, Q45, p.967)
ANS. : 3.20 cm
111
www.kmph.matrik.edu.my/physic s
PHYSICS CHAPTER 2 Learning Outcome: 2.7 Diffraction grating (2 hours) At the end of this chapter, students should be able to: Explain with the aid of a diagram the formation of diffraction. Apply formula,
d sinθ n = nλ
where
1 d= N
Describe with the aid of diagram the formation of spectrum by using white light.
112
PHYSICS
CHAPTER 2
2.7 Diffraction grating
is defined as a large number of equally spaced parallel slits. slits Diffraction grating can be made by ruling very fine parallel lines on glass or metal by a very precise machine. machine The untouched spaces between the lines serve as the slits as shown in Figure 2.46.
d
lines
Light es through the slit because it is transparent.
The spaces between the lines are the slits, for example : if there are four lines then we have 3 slits.
slit
Figure 2.46
113
PHYSICS
CHAPTER 2
If there N lines per unit length, then slit separation, d is given by: 1
d=
N
e.g. if a diffraction grating has 5000 lines per cm, then
1 1 d= = N 5000 cm
d = 2 × 10−4 cm
The light that es through the slits are coherent . The Interference pattern is narrower and sharper than doubleslits. There are two type of diffraction grating which are transmission grating (usual diffraction grating) reflection grating e.g. CD and DVD Diffraction grating is used in spectrometer to determine the wavelength of light and to study spectra. spectra 114
PHYSICS
CHAPTER 2
2.7.1 Explanation of diffraction by using Huygen’s principle for diffraction grating Figure 2.47 shows an incident lights fall on the transmission diffraction grating. C first order wavefront A source of secondary wavelets E second order wavefront
incident lights
third order wavefront D
grating
F zeroth order wavefront
Figure 2.47
B
115
PHYSICS
CHAPTER 2
Using Huygens’ principle, each maximum is located by taking the tangent of the wavelets from the slits. If the wavelets from each of the slits are drawn and a tangent AB is drawn, a plane wavefront parallel to the diffraction grating is obtained. This represents the zeroth-order maximum (n = 0). If the wavelets are grouped such that the first wavelet from one slit is combined with the second wavelet from the next slit, the third wavelet from the third slit and so on, the tangent CD will represent the first-order maximum (n =1). For the second-order maximum, the wavelets are grouped are such that the second wavelet of one slit is combined with the fourth wavelet of the next slit, the sixth wavelet from the third slit and so on. (tangent EF) EF Similarly, the third-, fourth-,…. order maximum may be obtained.
116
PHYSICS
CHAPTER 2
2.7.2 Equation of diffraction grating
Figure 2.48 illustrates light travels to a distant viewing screen from five slits of the grating. first order maximum (n = 1)
incoming plane wavefront of light
central or zeroth order maximum (n = 0) first order maximum (n = −1)
diffraction grating
d sin θ
Figure 2.48
d Figure 2.49
θ θ 117
PHYSICS
CHAPTER 2
The maximum (bright) fringes are sometimes called the principal maxima or principal fringes since they are placed where the light intensity is a maximum. Since the screen is far so that the rays nearly parallel while the light travels toward the screen as shown in Figure 2.49. In reaching the place on the screen while the 1st order maximum is located, light from one slit travels a distance of one wavelength farther than light from adjacent slit. Therefore the path difference for maximum fringe (constructive interference) is given by
d sinθ n = nλ n : order = 0,±1,±2,±3,.. θ n : n th order of diffraction angle
where
When
n=0 n = ±1 n = ±2
Central maximum fringe (0th order maximum) 1st maximum fringe (1st order maximum) 118 2nd maximum fringe (2nd order maximum)
PHYSICS
CHAPTER 2
The maximum fringes produce by a grating are much narrower and sharper than those from a double-slit as the intensity graph in Figures 2.50a and 2.50b.
Figure 2.50a
n = −2
−1
0
1
2
n = −2
−1
0
1
2
Figure 2.50b
119
PHYSICS
CHAPTER 2
Figures 2.51 shows the diffraction grating pattern.
n=2
Parallel beam of monochromatic light
θ2 θ1
second-order maximum
first-order n = 1 maximum
n=0
zero-order maximum
n = −1first-order
maximum
grating
n = −2 second-order maximum
Figure 2.51 120
PHYSICS
CHAPTER 2
If the white light is falls on the grating, a rainbow colours would be observed to either side of the central fringe on the screen which is white as shown in Figure 2.51. This because the white light contains wavelengths between violet and red. red
n = −2
−1
1
0
Rainbow Rainbow
white Rainbow
2 Rainbow
White light Figure 2.51
121
PHYSICS
CHAPTER 2
Note:
To calculate the maximum number of orders for bright fringes observed, observed take the diffraction angle,θ = 90°. Therefore
d sin 90 = nmaxλ d nmax = λ
where nmax : maximum number of orders that can be observed. From the equation for maxima, we obtain
sin θ n ∝ λ
and
1 sin θ n ∝ d
By using this two relations, the changes of diffraction grating pattern can be explained. 122
PHYSICS
CHAPTER 2
Example 11 : A monochromatic light of unknown wavelength falls normally on a diffraction grating. The diffraction grating has 3000 lines per cm. If the angular separation between the first order maxima is 35°. Calculate a. the wavelength of the light, b. the angular separation between the second-order and thirdorder maxima. Solution : N
= 3000 cm −1 ; 2θ1 = 35 ; n = 1
35
θ1
1st order max. max
1st order max. max
123
PHYSICS
CHAPTER 2
Solution : N = 3000 cm ; 2θ1 = 35 ; n = 1 a. The diffraction angle for 1st order maximum is −1
2θ1 = 35 θ1 = 17.5
And the slit separation, d is given by
1 d= −1 3000 cm d = 3.33 × 10 −4 cm OR 3.33 × 10−6 m
1 d= N
Therefore the wavelength of the light is
d sin θ n = nλ
(3.33 ×10 ) sin θ (3.33 ×10 ) sin 17.5 −6
−6
=λ =λ λ = 1.00 × 10 −6 m 1
124
PHYSICS
CHAPTER 2
Solution : b.
n=3 θ3 θ2
∆θ 23
3rd order maximum
2nd order n = 2 maximum
n=0
0th order maximum
2nd order n = −2 maximum 3rd order n = −3 maximum By using the equation of diffraction grating for maxima,
d sin θ n = nλ 125
PHYSICS
CHAPTER 2
Solution : b. For 2nd order maximum,
n=2 d sin θ 2 = 2λ (3.33 ×10−6 ) sin θ 2 = 2(1.00 ×10−6 ) θ 2 = 36.9 For 3 order maximum, n = 3 d sin θ 3 = 3λ (3.33 ×10−6 ) sin θ3 = 3(1.00 ×10−6 ) θ 3 = 64.3 rd
Therefore the angular separation,
∆θ 23 = θ 3 − θ 2 ∆θ 23 = 64.3 − 36.9 ∆θ 23 = 27.4 126
PHYSICS
CHAPTER 2
Example 12 : The second-order maximum produced by a diffraction grating with 560 lines per centimeter is at an angle of 3.1°. a. What is the wavelength of the light that illuminates the grating? b. Determine the number of maximum can be observed on a screen. c. State and giving reason, what you would expect to observe if a grating with a larger number of lines per centimeter is used. 3 −1 Solution : N = 56 × 10 m ; θ 2 = 3.1 ; n = 2 a. By applying the equation of diffraction grating for 2nd order 1 maximum, thus d sin θ 2 = 2λ and d =
N sin θ 2 = 2 Nλ sin 3.1 = 2 56 × 103 λ λ = 4.83 ×10 −7 m
(
)
127
PHYSICS
CHAPTER 2
Solution : b. By applying the equation of grating for maximum,
d sin θ n = nλ and sin θ n = nNλ
1 d= N
For the maximum no. of order for maximum fringe, θ n °
(
)(
sin 90 = nmax 56 ×10 4.83 × 10 nmax = 36.97 ≈ 36 3
−7
)
= 90
Therefore the number of maximum can be observed is
(36 × 2)+1 = 73 fringes
c. The fringes become farther to each another. another Reason : since 1 1 sin θ sin θ ∝ and d =
d
N
∝N
a larger number of lines per cm results in a larger diffraction angle thus the distance between two consecutive maximum fringes will increase. 128
PHYSICS
CHAPTER 2
Exercise 2.4 :
1. The first-order maximum line of 589 nm light falling on a diffraction grating is observed at an angle of 15.5°. Determine a. the slit separation on the grating. b. the angle of diffraction for third-order maximum line. (Physics for scientist & engineers ,3rd edition, Giancoli, Q32, p.914)
ANS. : 2.20 µ m; 53.4° 2. A diffraction grating has 6000 lines per cm. Calculate the angular separation between wavelengths 589.6 nm and 546.1 nm respectively after transmission through it at normal incidence, in the first-order spectrum (maximum line). ANS. : 1.60° 3. When blue light of wavelength 465 nm illuminates a diffraction grating, it produces a 1st order maximum but no 2nd order maximum. a. Explain the absence of 2nd order maximum. b. What is the maximum spacing between lines on this grating? (Physics,3rd edition, J.S.Walker, Q65, p.968)
ANS. : 930 nm
129
PHYSICS
CHAPTER 2
Next Chapter… CHAPTER 3 : Electrostatics
130
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