Dispersion and Prisms - CTAPS · Dispersion and Prisms Lecture 5 ... (Additive theory of light.) Cyan (GB), magenta (RB), ... Furthermore, we can use a hollow prism to

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Dispersion and Prisms

Lecture 5

Phys. 281A Geometric Optics

Physics Department Yarmouk University

21163 Irbid Jordan

http://ctaps.yu.edu.jo/physics/Courses/Phys281/Lec5

© Dr. Nidal Ershaidat

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DispersionDispersion

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© Dr. N. Ershaidat Phys. 281A: Geometric Optics Chapter 1: Nature of Light - Laws of Geometric Optics

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DispersionWe 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 dispersiondispersion.

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.


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n vs. λλλλThe index of refraction generally decreases with increasing wavelength.

Fig. 18

This means that violet light bends more than red light does when passing into a refracting material.

Violet

Red

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Angle of DeviationFig. 19 shows a beam of light incident on a prism.

Fig. 19

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 δ δ, called the angle of deviation.

Original direction


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White LightWhite 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.

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Gamma RaysX-Rays

UltravioletVisibleInfraredMicrowave

Radio

Region

Spectrum of Electromagnetic Radiation

> 105< 0.01 Å103 - 10510 Å – 0.01 Å3 - 103400 nm – 10 Å2 - 3700 nm – 400 nm

0.01 - 20.01cm – 700 nm10-5 - 0.0110 - 0.01 cm< 10-5> 10 cm

Energy (eV)Wavelength

λλλλ====νννν====

chhE

(((( ))))(((( ))))λλλλ

====νννν====12400

heVEÅ

12400

1031062661834

≈≈≈≈

××××××××××××==== −−−−−−−− sms.J.ch

eV Å


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Spectrum of Electromagnetic Radiation

Frequency Wavelength

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DispersionFig. 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


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δδδδ vs. λλλλThe angle of deviation δδδδ, 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.

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Rainbows in winter are natural examples of the dispersion of light into a spectrum.

Rainbow

A rainbow is often seen by an observer positioned between the Sun and a rain shower.

قوس قزحقوس قزحقوس قزحقوس قزح

........في الجاهليةفي الجاهليةفي الجاهليةفي الجاهلية العرب العرب العرب العرب الهةالهةالهةالهةأحد أحد أحد أحد اسم اسم اسم اسم قزح هو قزح هو قزح هو قزح هو .. .. .. ..له الرعد والمطرله الرعد والمطرله الرعد والمطرله الرعد والمطرإإإإ يظنونهيظنونهيظنونهيظنونه وكانووكانووكانووكانو


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Understanding RainbowsTo understand how a rainbow is formed, consider Figure 21.

Fig. 21

A ray of sunlight (white light) strikes a drop of water in the atmosphere and is refracted and reflected as follows:

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Small Wavelengths Deviate MoreThe 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.


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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.

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PrismsPrisms


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What is a Prism?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.

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Some Types of Prisms


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Minimum Angle of DeviationIn optics we show that the minimum angle of deviation δδδδmin

for a prism occurs when the angle of incidence θθθθ1

is such that the refracted ray inside the prism makes the same angle θθθθ1

with the normal to the two prism faces as shown in the figure 22.

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From Snell’s Law with n1

= 1

and n2

= n

Measuring the Index of RefractionFrom the figure we have:

θθθθ2= φ φ φ φ/2

where φφφφ is the apex angle

2

21

φφφφ====

θθθθ====θθθθ

sinn

sinnsin

2221

δδδδ++++

φφφφ====αααα++++θθθθ====θθθθ

2221

φφφφ====

δδδδ++++

φφφφ====θθθθ sinnsinsin


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n

2sin

22sin

2 φφφφ

δδδδ++++

φφφφ

====n

The apex angle for a prism φ φ φ φ/2, being known, we only measure the minimum deviation angle and thus can calculate the index of refraction of medium 2

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Measuring n for LiquidsFurthermore, we can use a hollow prism to determine the values of n for various liquids filling the prism.

Total Internal Reflection

Lecture 6



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Refraction from n1>n

2

Total internal reflection occurs when light is directed from a medium having a given index of refraction toward one having a lowerlower index of refraction.

Fig. 23

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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 n

1is greater than n

2. At

some particular angle of incidence θθθθC, called

the critical angle, the refracted light ray moves parallel to the boundary so that θθθθ

2=

90°°°°.

Rays are Bent Away from the Normal

Consider a light beam traveling in medium 1and meeting the boundary between medium 1 and medium 2, where n

1is greater than n

2.


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Critical AngleAccording to Snell’s Law θθθθC verifies the equation:

°°°°====θθθθ 9021 sinnsinn C

i.e.

1

2

n

nsin C ====θθθθ

Incident light with an angle of incidence > θθθθC

will reflect in medium 1.

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Fig. 24

�� Optical Fibers

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Fiber OpticsTotal internal reflection has an important application in data transmission.


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Glass or transparent plastic rods to “pipe”light from one place to another. (Fig. 25)

In figure 25, light is confined to traveling within a rod, even around curves, as the result of successive total internal reflections.

Optical Fiber

Fig. 25

Now we use flexible thin light pipe rather than thick rods.

A flexible light pipe is called an optical fiber.

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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.

Fiber Optics


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Fiber OpticsFigure 26 shows a cutaway view of this construction.

Fig. 26

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Fiber OpticsBecause 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.


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Fiber OpticsAny 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.

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Fermat’s Principle

Lecture 7



21163 Irbid Jordan




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Fermat’s PrincipleFermat’s principle states that when a light ray travels between any two points, its path is the

one that requires the smallest time interval.

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Fermat and the Ray Approximation

The ray approximation is an immediate consequence of Fermat’s principle since the

straight line is the shortest distance between two points.


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Deriving Snell’s LawSuppose that a light ray is to travel from point P in medium 1 to point Q in medium 2.

Fig. 27

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The speed of light is c/n1in medium 1 and c/n

2

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

This time is minimum when we have:

Deriving Snell’s Law

0======== exx

dx

dt0

2

2

>>>>

exdx

dtsignand

(((( ))))

2

22

1

22

2

2

1

1

nc

xdb

nc

xa

v

r

v

rt

−−−−++++++++

++++====++++====


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(((( ))))

(((( ))))22

2

22

10

xdb

xdn

xa

xn

dx

dt

−−−−++++

−−−−====

++++⇔⇔⇔⇔====

This time is minimum when we have:

Minimum Time Interval

(((( ))))

(((( ))))

−−−−++++

−−−−−−−−

++++====⇒⇒⇒⇒

22

2

22

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xdb

xdn

xa

xn

cdx

dt

(((( ))))

2

22

1

22

2

2

1

1

nc

xdb

nc

xa

v

r

v

rt

−−−−++++++++

++++====++++====

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Snell and FermatIt can be easily seen that:

Fig. 27

221

sin

xa

x

++++====θθθθ

(((( ))))222

sin

xdb

xd

−−−−++++

−−−−====θθθθ


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2211sinsin0 θθθθ====θθθθ⇔⇔⇔⇔==== nn

dx

dt

Exercise: Check that:

Value for xe

21 θθθθ−−−−====θθθθ==== tanbdtanaxe

02

2

>>>>

exdx

tdsign

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Problems

Lecture 7



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Problem 1 (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).

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Solution 1 (35.35)The

°°°°========θθθθ⇒⇒⇒⇒θθθθ====°°°° −−−−4827

661

766066150

1 ..

.sinsin.sin VV

°°°°========θθθθ⇒⇒⇒⇒θθθθ====°°°° −−−−2228

621

766062150

1 ..

.sinsin.sin RR


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Problem 2 (35.71)A light ray enters a rectangular block of plastic at an angle θθθθ

1= 45.0°°°° and emerges at an angle θθθθ

2= 76.0°,

as shown in Figure P35.71.

(a) Determine the index of refraction of the plastic.

n

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Solution 2 (35.71)The triangle OAC is rectangular!

O

A

Cn

31 θθθθ====θθθθ sinnsin

23 θθθθ====θθθθ sincosn

°°°°====θθθθ⇒⇒⇒⇒°°°°

°°°°====

θθθθ

θθθθ====θθθθ⇒⇒⇒⇒

136076

045

3

2

13

..sin

.sin

sin

sintan

n

21

136

076

3

1

.n

.cos

.sin

sin

sinn

≈≈≈≈⇒⇒⇒⇒

°°°°

°°°°====

θθθθ

θθθθ====


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Problem 2 (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?

ns...

c

cosL

c

lt

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103

8050

8

3

≈≈≈≈××××

××××≈≈≈≈

θθθθ========

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Problem 3 (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.

10 7.5

20 15.1

30 22.3

40 28.7

50 35.2

60 40.3

70 45.3

80 47.7

θθθθ1

θθθθ2

sin θθθθ1

sin θθθθ2

0.17365 0.13053

0.34202 0.2605

0.5 0.37946

0.64279 0.48022

0.76604 0.57643

0.86602 0.64679

0.93969 0.7108

0.98481 0.73963


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Solution 3 (35.72)

Snell’s Law sin θθθθ1

= n sinθθθθ2

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.

sin θθθθ1is linear in sinθθθθ

2.

The slope is n

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Solution 3 (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


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The index of refraction of a medium is inversely proportional to the speed of sound in this medium, i.e.

Problem 4The speed of sound in air is v

A= 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:

(((( ))))(((( ))))

11

>>>>λλλλ

∝∝∝∝λλλλv

n

27


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Problem 4 – Solution

For sound waves, the index of refraction in water is smaller that the index of refraction in air.

nA

sin θθθθC= n

W

====

====θθθθ −−−−−−−−

W

A

A

WC

v

vsin

n

nsin

11

03142501320

340 011 ′′′′≈≈≈≈====

====θθθθ −−−−−−−− .sinsinC

Thus we see that:

1883340

1320>>>>============ .

v

v

n

n

A

W

waterinsound

airinsound

End of Chapter End of Chapter End of Chapter End of Chapter 1111End of Chapter End of Chapter End of Chapter End of Chapter 1111

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Dispersion and Prisms - CTAPS · Dispersion and Prisms Lecture 5 ... (Additive theory of light.) Cyan (GB), magenta (RB), ... Furthermore, we can use a hollow prism to