Upload
others
View
26
Download
0
Embed Size (px)
Citation preview
Chapter 8
1
Chapter 8: Physical Optics
8.1 Huygens’ Principle
L.O 8.1.1 State Huygens’ principle
L.O 8.1.2 Sketch and explain the wave front of light after passing through a single slit
and obstacle using Huygens’ principle
Huygens’ Principle:
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.
Plane wave Spherical wave Single slit Obstacle
When applying Huygens’s principle, show
the centres of the wavelets
the wavelets from these centres
the line touching these wavelets
draw an arrow to show the direction of the ray (normal to wavefront)
Example
Question Solution
The figure shows a point light source P on the
ground.
Draw the wavefront from point P at time t = 1
s and t = 2 s.
Wavefront at
time t = 0
New wavefront
at time t
P
Chapter 8
2
8.2 Constructive Interference and Destructive Interference
L.O 8.2.1 Define coherence
A stable interference pattern can be produced if the sources of wave are coherent.
The two sources of wave are coherent if they have:
i. the same phase difference (constant)
ii. the same wavelength (monochromatic)
L.O 8.2.2 State the conditions for interference of light
Interference
When two or more 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
The resultant displacement at any point is the vector sum of the displacements due to the
two light waves.
Conditions for fixed interference
two coherent sources
same or approximately same amplitude
distance between the coherent sources, d ≤ λ
L.O 8.2.3 State the conditions of constructive and destructive interference
Constructive interference is
defined as a reinforcement of
amplitudes of light waves that
will produce a bright fringe
(maximum).
Destructive interference is
defined as a total cancellation of
amplitudes of light waves that will
produce a dark fringe (minimum).
Chapter 8
3
Path difference is the difference between two paths
of waves from two different sources at a point (a
difference in path length).
Conditions of constructive and destructive interference (based on path difference):
S1 and S2 are two coherent sources in phase
Bright fringe (Constructive) Dark fringe (Destructive)
Path Difference, L
= |S2P - S1P|
= |x2 –x1|
,.....3,2,1,0
,.....3,2,,0
mwhere
m
L
,.....3,2,1,0
2
1
,.....2
5,
2
3,
2
mwhere
m
L
Chapter 8
4
S1 and S2 are two coherent sources in phase
Bright fringe (Constructive) Dark fringe (Destructive)
Example
Question Solution
Two point sources X and Y emit waves of
wavelength 2.0 cm in phase. The point P is 6.0
cm from X and 10.0 cm from Y. Another point
Q is 7.0 cm from X and 4.0 cm from Y. What
is the path difference of waves from X and Y
at
a. The point P and
b. The point Q?
Hence deduce whether constructive
interference or destructive interference occurs
at P and Q.
,.....3,2,1,0
,.....3,2,,0
mwhere
m
L
,.....3,2,1,0
2
1
,.....2
5,
2
3,
2
mwhere
m
L
Chapter 8
5
8.3 Interference of Transmitted Light through Double-Slits
L.O 8.3.1 Derive and use 𝒚𝒎 =𝒎𝝀𝑫
𝒅 for bright fringes and 𝒚𝒎 =
(𝒎+𝟏 𝟐⁄ )𝝀𝑫
𝒅 for dark
fringes where m = 0, ±1, ±2,±3, …
L.O 8.3.2 Use ∆𝒚 =𝝀𝑫
𝒅 and explain the effect of changing any of the variables
Double-slits interference: Young’s Double Slits Experiment
The double slits, S1 and S2, act as coherent sources of light waves.
Under these conditions, the waves emerging from S1 and S2 have the same frequency
and amplitude and are in phase because they originate from the same light source
and their distance from S is equal.
Constructive interference and destructive interference occur at different points on
the screen to produce a pattern of alternating bright and dark fringes.
Chapter 8
6
Derivation of Young’s double-slit equations
From the figure:
Big triangle: Small triangle:
D
ymtan d
Lsin
since θ is very small, tan θ ≈ sin θ , hence
d
L
D
ym
For bright fringes, mL For dark fringes,
2
1mL
**m = 0, ±1, ±2, ...
Separation between two consecutive (successive) dark or bright fringes:
y depends on :
i) the wavelength of light, λ y
ii) the distance apart, d of the double slits
dy
1
iii) distance between slits and the screen, D dy
d
Dmym
d
Dm
ym
2
1
mm yyy 1
d
Dm
d
Dmy
1
d
Dy
Chapter 8
7
Example
Question Solution
An interference pattern is formed on a screen
when light of wavelength 550 nm is incident
on two parallel slits 50 μm apart. The second-
order bright fringe is 4.5 cm from the center
of the central maximum. How far from the
slits is the screen?
In a Young’s double experiment, the slits
separation is 1.0 mm. The distance between
the slits and the screen is 1.0 m. The
wavelength of the sodium light used is
5.9×10-5 cm.
a. Calculate the separation between two
consecutive dark fringes.
b. If the sodium light is replaced with a blue
light, what are the changes to the
interference pattern on the screen?
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 the
distance between the first order minimum and
fourth order maximum on the screen.
A monochromatic light of wavelength 600
nm falls on a system of double-slits of
unknown slit separation. At the same time,
the double-slits is illuminated by a
monochromatic light of unknown
wavelength. It was observed that the 4th order
maximum of the known wavelength light
overlapped with the 5th order maximum of the
unknown wavelength light. Find the
wavelength of the unknown wavelength light.
Chapter 8
8
Question Solution
Suppose you pass the light from a He-Ne
laser through two slits separated by 0.0100
mm and find that the third bright line is
formed at an angle of 10.95o relative to the
incident beam. What is the wavelength of the
light?
Exercise
Question
In a lab experiment, monochromatic light passes through two narrow slits that are 0.050 mm
apart. The interference pattern is observed on a white wall 1.0 m from the slits, and the
second-order bright is 2.4 cm from the center of the central maximum.
a. What is the wavelength of the light?
b. What is the distance between the second-order and third-order bright fringes?
Answer: 6.7×10-7 m, 1.2×10-2 m
Monochromatic light illuminates a double-slit system with a slit separation d = 0.30 mm. The
second-order maximum occurs at y = 4.0 mm on a screen 1.0 m from the slits. Find
a. the wavelength.
b. the distance y on the screen between the central maximum and the third order-minimum.
c. the angle, θ of the first-order maximum.
Answer: 600 nm, 7mm, 0.11°
A screen containing two slits 0.100mm apart is 1.20 m from the viewing screen. Light of
wavelength λ = 500 nm falls on the slits from a distant source.
a. Approximately how far apart will the bright interference fringes be on the screen?
b. What happens to the interference pattern if the incident light (500 nm) if replaced by light
of wavelength 700 nm.
c. What happens instead if the slits are moved farther apart?
Answer: 6 mm, u think
What is the highest-order maximum for 400 nm light falling on double slits separated by 25.0
μm?
Answer: 62
Chapter 8
9
8.4 Interference of Reflected Light in Thin Films
L.O 8.4.1 Identify the occurrence of phase change upon reflection
L.O 8.4.2 Explain with the aid of a diagram the interference of light in thin films at
normal incidence
L.O 8.4.3 Use 𝟐𝒏𝒕 = 𝒎𝝀 and 𝟐𝒏𝒕 = (𝒎 + 𝟏𝟐⁄ )𝝀
Less dense →Denser Denser →Less dense
Phase change = π rad Phase change = 0 rad
Non-reflective coating Reflective coating
Phase difference between ray 1 and ray 2,
∆𝜙 = 𝜋 − 𝜋 = 0 2 sources in phase
Path difference between ray 1 and ray 2,
∆𝐿 = 2𝑛𝑡
Constructive interference:
mnt 2
Destructive interference:
2
12 mnt
Phase difference between ray 1 and ray 2,
∆𝜙 = 𝜋 − 0 = 𝜋 2 sources anti phase
Path difference between ray 1 and ray 2,
∆𝐿 = 2𝑛𝑡
Constructive interference:
2
12 mnt
Destructive interference:
mnt 2
**m = 0, ±1, ±2, ...
Reflected Reflected Transmitted Transmitted
Chapter 8
10
Example
Question Solution
Light is at normal incidence on a thin soap
film of refractive index 1.30 and thickness
0.15 µm. Determine the maximum
wavelength of the reflected light that
undergoes
a. constructive interference
b. destructive interference
White light is incident normally on a lens
(n = 1.52) that is coated with a film of MgF2
(n = 1.38). For what minimum thickness of
the film will yellow light (λvacuum = 550 nm)
be missing in the reflected light?
A lens appears greenish yellow (λvacuum = 570
nm is strongest) when white light reflects
from it. What minimum thickness of coating
(n = 1.25) do you think is used on such a
(glass) lens?
Exercise
Question
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. What is the
minimum thickness of the soap film?
Answer: 2.58×10-7 m
A non-reflective coating of MgF2 (n = 1.38) covers the glass (n = 1.52) of a camera lens.
Assuming that the coating prevents reflection of yellow-green light (λvacuum = 565 nm) ,
determine the minimum nonzero thickness that the coating can have.
Answer: 102 nm
A plastic film with index of refraction 1.80 is put on the surface of a car window to increase
the reflectivity and thereby to keep the interior of the car cooler. The window has index of
refraction 1.60. What minimum thickness required if light of wavelength 600 nm in air
reflected from the two sides of the film is to interfere constructively.
Answer: 83 nm
Chapter 8
11
8.5 Interference of Reflected Light in Air Wedge and Newton’s Rings
L.O 8.5.1 Explain with the aid of a diagram the interference of light in air wedge
L.O 8.5.2 Use 𝟐𝒕 = 𝒎𝝀 for dark fringes and 𝟐𝒕 = (𝒎+ 𝟏𝟐⁄ )𝝀 for bright fringes
L.O 8.5.3 Use diagram to explain qualitatively the formation of Newton’s rings and
the centre dark spot
Air wedge Newton’s rings
˃ Ray 1 (OL): From denser to less dense, no phase change
˃ Ray 2 (BQ): From less dense to denser, π rad phase change
˃ The two rays (ray 1 & 2) are coherent since both have originated from the
same source and produces a produces interference pattern. When thickness
is equal to zero, destructive interference occurred. Hence a dark fringe (Air
wedge)/ central dark spot (Newton’s ring) is obtained.
Equations for air wedge:
For bright fringes, For dark fringes,
2
12 mnt mnt 2
**m = 0, 1, 2, ...
Chapter 8
12
Example
Question Solution
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
A plate of glass 10.0 cm long is placed in
contact with a second plate and held at small
angle with it by a metal strip 0.100 mm thick
placed under one end. The space between the
plates is filled with air. The glass is
illuminated from above with light having a
wavelength of 635 nm. How many
interference fringes are observed per
centimeter in the reflected light?
Exercise
Question
A fine metal foil separates one end of two pieces of optically flat glass. When light of
wavelength 670 nm is incident normally, 28 dark lines are observed (with one at each).
a. How thick is the foil?
b. How far apart are the dark fringes if the glass plates are each 26.5 m long?
Answer: 9.05×10-6 m, 9.8 mm
White light is incident normally on a thin soap film (n =1.33) suspended in air.
a. What are the two minimum thicknesses that will constructively reflect yellow light of
wavelength 590 nm?
b. What are the two minimum thicknesses that will destructively reflect yellow light of
wavelength 590 nm?
Answer: 110 nm, 330 nm, 220 nm, 440 nm
Chapter 8
13
8.6 Diffraction by a Single Slit
L.O 8.6.1 Define diffraction
L.O 8.6.2 Explain with the aid of a diagram the diffraction of a single slit
L.O 8.6.3 Use 𝒚𝒏 =𝒏𝝀𝑫
𝒅 for dark fringes and 𝒚𝒏 =
(𝒏+𝟏 𝟐⁄ )𝝀𝑫
𝒅 for bright
fringes where n = ±1, ±2,±3, …
Diffraction of light is defined as the bending of waves as they travel around obstacles or pass
through an aperture comparable to the wavelength of the waves.
Diffraction by a single slit
Formation of diffraction by a single slit:
According to Huygen’s principle, wavefront from light
source falls on a narrow slit and diffraction occurs. Every
point on the wavefront that falls on the slit acts as sources
of secondary wavelets and superposed each another to
form an interference pattern on the screen.
Chapter 8
14
For bright fringes, For dark fringes,
**n = ±1, ±2, ...
Example
Question Solution
A monochromatic light of wavelength 6 x10-7
m passes through a single slit of width 2 x 10-
6 m.
a. Calculate the width of central maximum:
i. in degrees;
ii. in centimeters,
on a screen 5 cm away from the slit
b. Find the number of minimum that can be
observed.
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 µm?
Exercise
Question
Visible light of wavelength of 550 nm falls on a single slit and produces its second diffraction
minimum at an angle of 45.0o relative to the incident direction of the light.
a) What is the width of the slit?
b) At what angle is the first minimum produced?
Answer: 1.56×10-6 m, 20.7°
Important!!
» The width of central maximum is
2y1 (equation for first dark)
» To calculate the maximum number
of orders observed, θ = 90o
nn sina
2
1sina nn
a
Dnyn
)(21
a
Dnyn
Chapter 8
15
8.7 Diffraction Grating
L.O 8.7.1 Explain with the aid of a diagram the formation of diffraction
L.O 8.7.2 Apply 𝒅𝐬𝐢𝐧 𝜽 = 𝒏𝝀 where 𝒅 =𝟏
𝑵
Diffraction grating is defined as a large number of equally spaced parallel slits.
Pattern of diffraction grating:
Formation of diffraction:
A diffraction grating is a plate containing many parallel lines/slits at uniform distance between
one another. According to Huygen’s principle, when light is incident on a diffraction grating,
each slit will become a secondary source of light so that superposition of light waves from each
source will produce diffraction images of regular orders on a screen
Path difference
3,...2,1,0,n
sin
nd n
Nd
1
Slit separation Number of lines
per unit length
Chapter 8
16
Example
Question Solution
A diffraction grating with 600 lines per mm is
illuminated normally with a monochromatic
light of wavelength 589 nm. Calculate
a. the angles of the first-order and second-
order maximum lines from the zero-order
maximum line.
b. the number of orders that can be
observed.
When a 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?
How many bright fringes are produced when
a grating with a spacing of 2.00 x 10-6 m is
illuminated normally with light of
wavelength 6.44 x 10-7 m?
Exercise
Question
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,
a) the angular separation between the second-order and third- order maxima.
Answer: 1.00×10-6 m, 27.4°
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.
Answer: 4.83×10-7 m, 37, u think