Lecture 5
tics
Phys. 281A Geometric Op rsity
ouk Unive
Physics Department Yarm
21163 Irbid Jordan
Dispersion and Prisms
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys281/Lec5
Dispersion
1
61
Dispersion
We said in the previous section that for a given
material, the index of refraction n varies with
the wavelength of the light passing through
the material.
This behavior is called dispersion.
dispersion
This implies, and according to Snell’s law of
refraction that light of different wavelengths is
bent at different angles when incident on a
refracting material.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
62
n vs. λ
The index of refraction generally decreases
with increasing wavelength.
This means that violet
light bends more than
red light does when
passing into a refracting
material.
Violet
Red
Fig. 18
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
2
63
Angle of Deviation
Fig. 19 shows a beam of light incident on a
prism.
A ray of single-wavelength light incident on
the prism from the left emerges refracted from
its original direction of travel by an angle δ,
δ
called the angle of deviation.
Original direction
Fig. 19
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
White Light
64
White light contains at least three additive
primary colors: red, green, and blue.
A surface appears black if it absorbs the light
that strikes it.
By combining red, green, and blue light (or any
two of them) at different intensities, a wide
range of other colors can be produced.
(Additive theory of light.)
Cyan (GB), magenta (RB), and yellow (RG) are
called secondary colors of light.
Each contains two primary colors and lacks a
third, complementary color.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
3
Spectrum of Electromagnetic Radiation
Region
Wavelength
Energy (eV)
Radio
Microwave
Infrared
Visible
Ultraviolet
X-Rays
Gamma Rays
> 10 cm
10 - 0.01 cm
0.01cm – 700 nm
700 nm – 400 nm
400 nm – 10 Å
10 Å – 0.01 Å
< 0.01 Å
< 10-5
10-5 - 0.01
0.01 - 2
2-3
3 - 103
103 - 105
> 105
hc
λ
12400
E (eV ) = h ν =
λ (Å )
E = hν =
65
h c = 6.626 × 10 −34 J . s × 3 × 108 m s −1
≈ 12400 eV Å
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Spectrum of Electromagnetic Radiation
Frequency
66
Wavelength
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
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67
Dispersion
Fig. 20 shows a beam of white light (a
combination
of
all
visible
wavelengths)
incident on a prism.
The rays that emerge
spread out in a series of
colors known as the
visible spectrum. These
colors,
in
order
of
decreasing wavelength,
are red, orange, yellow,
green, blue, and violet.
Fig. 20
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
δ vs. λ
68
The angle of deviation δ, clearly, depends on
wavelength. Violet light deviates the most,
red the least, and the remaining colors in
the visible spectrum fall between these
extremes.
Newton showed that each color has a
particular angle of deviation and that the
colors can be recombined to form the
original white light.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
5
‫ﻗﻮس ﻗﺰح‬
69
Rainbow
Rainbows in winter are natural examples of the
dispersion of light into a spectrum.
A rainbow is often seen by an observer
positioned between the Sun and a rain shower.
..‫ﻗﺰح ﻫﻮ اﺳﻢ أﺣﺪ اﻟﻬﺔ اﻟﻌﺮب ﻓﻲ اﻟﺠﺎﻫﻠﻴﺔ‬
.. ‫وﻛﺎﻧﻮ ﻳﻈﻨﻮﻧﻪ إﻟﻪ اﻟﺮﻋﺪ واﻟﻤﻄﺮ‬
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
70
Understanding Rainbows
To understand how
consider Figure 21.
a rainbow
is
formed,
A ray of sunlight (white
light) strikes a drop of
water in the atmosphere
and
is
refracted
and
reflected as follows:
Fig. 21
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
6
Small Wavelengths Deviate More
71
The ray is first refracted at the front surface of
the drop, with the violet light deviating the
most and the red light the least.
At the back surface of
the drop, the light is
reflected and returns to
the front surface, where
it
again
undergoes
refraction as it moves
from water into air.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
72
Small Difference of Deviation Angles
The rays leave the drop such that the angle
between the incident white light and the most
intense returning violet ray is 40°° and the angle
between the white light and the most intense
returning red ray is 42°°.
This small angular difference between the
returning rays causes us to see a colored bow.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
7
Prisms
What is a Prism?
74
Geometry: a prism is a solid having bases or
ends that are parallel, congruent polygons and
sides that are parallelograms.
In Optics: a prism is a transparent solid body,
often having triangular bases, used for
dispersing light into a spectrum or for
reflecting rays of light.
In Crystallography: a prism is a form having
faces parallel to the vertical axis and
intersecting the horizontal axes.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
8
Some Types of Prisms
75
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Minimum Angle of Deviation
76
In optics we show that the minimum angle of
deviation δmin for a prism occurs when the angle
of incidence θ1 is such that the refracted ray
inside the prism makes the same angle θ1 with
the normal to the two prism faces as shown in
the figure 22.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
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Measuring the Index of Refraction
77
From the figure we have:
θ2 = φ/2
φ
where φ is the apex angle
From Snell’s Law with n1 = 1
and n2 = n
sin θ1 = n sin θ 2
= n sin φ 2
θ1 = θ 2 + α =
φ δ
+
2 2
φ
φ δ
sin θ1 = sin  +  = n sin
2
 2 2
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
78
n
φ δ
sin  + 
 2 2
n2 =
φ
sin
2
The apex angle for a prism φ/2
φ , being known,
we only measure the minimum deviation angle
and thus can calculate the index of refraction
of medium 2
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
10
Measuring n for Liquids
79
Furthermore, we can use a hollow prism to
determine the values of n for various liquids
filling the prism.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Lecture 6
tics
Phys. 281A Geometric Op rsity
ouk Unive
Physics Department Yarm
21163 Irbid Jordan
Total Internal Reflection
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys281/Lec5
11
Total Internal Reflection
Refraction from n1>n2
82
Total
internal
reflection occurs
when
light
is
directed from a
medium having a
given index of
refraction toward
one
having
a
lower index of
refraction.
Fig. 23
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
12
83
Rays are Bent Away from the Normal
Consider a light beam traveling in medium 1
and meeting the boundary between medium
1 and medium 2, where n1 is greater than n2.
In figure 23, various possible directions of
the beam are indicated by rays 1 through 5.
The refracted rays are bent away from the
normal because n1 is greater than n2. At
some particular angle of incidence θC, called
the critical angle, the refracted light ray
moves parallel to the boundary so that θ2 =
90°°.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
84
Critical Angle
According
equation:
to
Snell’s
Law
θC
verifies
the
n1 sin θC = n2 sin 90°
i.e.
n
sin θC = 2
n1
Incident light with an angle of incidence > θC
will reflect in medium 1.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
13
Total Internal Reflection
85
Fig. 24
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Optical Fibers
14
87
Fiber Optics
Total internal reflection has an
application in data transmission.
important
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
88
Optical Fiber
Glass or transparent plastic rods to “pipe”
light from one place to another. (Fig. 25)
Fig. 25
In figure 25, light is confined to
traveling within a rod, even
around curves, as the result of
successive
total
internal
reflections.
Now we use flexible thin light pipe rather than
thick rods.
A flexible light pipe is called an optical fiber.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
15
Fiber Optics
89
If a bundle of parallel fibers is used to
construct an optical transmission line, images
can be transferred from one point to another.
This technique is used in a sizable industry
known as fiber optics.
A practical optical fiber consists of a
transparent core surrounded by a cladding, a
material that has a lower index of refraction
than the core. The combination may be
surrounded by a plastic jacket to prevent
mechanical damage.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Fiber Optics
90
Figure 26 shows a cutaway view of this
construction.
Fig. 26
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
16
Fiber Optics
91
Because the index of refraction of the cladding
is less than that of the core, light traveling in
the core experiences total internal reflection if
it arrives at the interface between the core and
the cladding at an angle of incidence that
exceeds the critical angle. In this case, light
“bounces” along the core of the optical fiber,
losing very little of its intensity as it travels.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Fiber Optics
92
Any loss in intensity in an optical fiber is due
essentially to reflections from the two ends and
absorption by the fiber material. Optical fiber
devices are particularly useful for viewing an
object at an inaccessible location. For example,
physicians often use such devices to examine
internal organs of the body or to perform surgery
without making large incisions. Optical fiber cables
are replacing copper wiring and coaxial cables for
telecommunications because the fibers can carry a
much greater volume of telephone calls or other
forms of communication than electrical wires can.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
17
Lecture 7
tics
Phys. 281A Geometric Op rsity
ouk Unive
Physics Department Yarm
21163 Irbid Jordan
Fermat’s Principle
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys281/Lec7
Fermat’s Principle
94
!"
Fermat’s principle states that when a light ray
travels between any two points, its path is the
one that requires the smallest time interval.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
18
Fermat and the Ray Approximation
95
The ray approximation is an immediate
consequence of Fermat’s principle since the
straight line is the shortest distance between
two points.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Deriving Snell’s Law
96
Suppose that a light
ray is to travel from
point P in medium 1
to point Q in medium
2.
Fig. 27
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
19
97
Deriving Snell’s Law
The speed of light is c/n1 in medium 1 and c/n2
in medium 2. Using the geometry of Figure 27,
and assuming that light leaves P at t = 0, we
see that the time at which the ray arrives at Q
is
r
r
t= 1 + 2 =
v1 v 2
a2 + x2
b 2 + (d − x )
+
c n1
c n2
2
This time is minimum when we have:
dt
dx
=0
 dt 2 
sign  2  > 0
 dx  xe
and
x = xe
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
98
Minimum Time Interval
r
r
t= 1 + 2 =
v1 v 2
a2 + x2
b 2 + (d − x )
+
c n1
c n2
2
dt
1  n1 x
n2 (d − x ) 

⇒
= 
−
2
2
2
2
dx c  a + x
b + (d − x ) 

This time is minimum when we have:
dt
=0 ⇔
dx
n1 x
a2 + x2
=
n2 (d − x )
b 2 + (d − x )
2
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
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99
Snell and Fermat
It can be easily seen
that:
sin θ1 =
sin θ 2 =
x
a2 + x2
d−x
b 2 + (d − x )
2
Fig. 27
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Value for xe
100
dt
= 0 ⇔ n1 sin θ1 = n2 sin θ 2
dx
Exercise: Check that:
xe = a tan θ1 = d − b tan θ 2
 d 2t 
sign  2  > 0
 dx  xe
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
21
Lecture 7
tics
Phys. 281A Geometric Op rsity
ouk Unive
Physics Department Yarm
21163 Irbid Jordan
Problems
© Dr. Nidal Ershaidat
http://ctaps.yu.edu.jo/physics/Courses/Phys281/Lec7
Problem 1
102
(35.35)
The index of refraction for violet light in silica
flint glass is 1.66, and that for red light is 1.62.
What is the angular dispersion of visible light
passing through a prism of apex angle 60.0°° if
the angle of incidence is 50.0°°? (See Fig.
P35.35).
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
22
Solution 1
The
103
(35.35)
0.766
= 27.48°
1.66
0.766
sin 50° = 1.62 sin θ R ⇒ θ R = sin −1
= 28.22°
1.62
sin 50° = 1.66 sin θV ⇒ θV = sin −1
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Problem 2
104
(35.71)
A light ray enters a rectangular block of plastic at
an angle θ1 = 45.0°° and emerges at an angle θ2 = 76.0°,
as shown in Figure P35.71.
n
(a) Determine the index of refraction of the
plastic.
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
23
Solution 2
105
(35.71)
The triangle OAC is rectangular!
sin θ1 = n sin θ 3
n
n cos θ 3 = sin θ 2
=
sin 45.0°
⇒ θ 3 = 36.1°
sin 76.0°
n=
C
O
sin θ1
⇒ tan θ 3 =
sin θ 2
A
sin θ1
sin 76.0°
=
sin θ 3
cos 36.1°
⇒ n ≈ 1.2
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Problem 2
106
(35.71)
(b) If the light ray enters the plastic at a point
L = 50.0 cm from the bottom edge, how long does it
take the light ray to travel through the plastic?
L cos θ 3
l
=
c
c
0.5 × 0.8
≈
≈ 8.6 ns
8
3 × 10
t=
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
24
Problem 3
107
(35.72)
Students allow a narrow beam of laser light to
strike a water surface. They arrange to measure
the angle of refraction for selected angles of
incidence and record the data shown in the
accompanying table.
θ1
10
20
30
40
50
60
70
80
θ2
7.5
15.1
22.3
28.7
35.2
40.3
45.3
47.7
sin θ1
0.17365
0.34202
0.5
0.64279
0.76604
0.86602
0.93969
0.98481
sin θ2
0.13053
0.2605
0.37946
0.48022
0.57643
0.64679
0.7108
0.73963
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
Solution 3
108
(35.72)
Use the data to verify Snell’s law of refraction by
plotting the sine of the angle of incidence versus
the sine of the angle of refraction.
Snell’s Law sin θ1 = n sinθ
θ2
sin θ1 is linear in sinθ
θ2.
The slope is n
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
25
Solution 3
109
(35.72)
Use the resulting plot to deduce the index of
refraction of water.
The best (fit) line is obtained using the Least
Squares Method (A linear regression method)
which gives the slope, i.e. n
n = 1.32965 ± 0.00281
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
110
Problem 4
The speed of sound in air is vA = 340 m s-1 and
vW = 1320 m s-1 in water
Compute the critical angle for a sound wave
incident on a lake’s surface.
Solution:
The index of refraction of a medium is
inversely proportional to the speed of sound in
this medium, i.e.
n (λ ) ∝
1
> 1
v (λ )
© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
26
Problem 4 – Solution
111
Thus we see that:
n sound in air
v
1320
= W =
= 3 . 88 > 1
n sound in water
vA
340
For sound waves, the index of refraction in water
is smaller that the index of refraction in air.
nA sin θC = nW
n
θ C = sin − 1  W
 nA
 v

 = sin − 1  A

 vW



 340 
θ C = sin − 1 
 = sin − 1 0 . 25 ≈ 14 0 3 0 ′
1320


© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics
End of Chapter 11
27