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Hong Kong Physics Olympiad Lesson 15

15.1 Electromagnetic waves

15.2 Doppler effect of light

15.3 The electromagnetic spectrum

15.4 Polarization

15.5 Polarization by reflection

15.6 Polarization by scattering

15.7 Superposition revisit

15.8 Youngs double-slit experiment

15.9 Phase change due to reflection

15.10 Diffraction

15.11 Resolving power

15.1 Electromagnetic waves

Electromagnetic waves are traveling waves of oscillating electric and magnetic fields.

The electric and magnetic fields in an electromagnetic wave are perpendicular to each

other, and to the direction of propagation. They are also in phase. To find the direction

of propagation of an electromagnetic wave, point the fingers of your right hand in the

direction of E, then curl them toward B. Your thumb will be pointing in the direction

of propagation. Electromagnetic waves can be generated by an antenna, which is

connected to an AC generator.

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In fact, another way to generate electromagnetic waves is to accelerate electric charge.

As accelerated charges radiate electromagnetic waves. The intensity of radiated

electromagnetic waves depends on the orientation of the acceleration relative to the

viewer. For example, viewing the antenna perpendicular to its length, so that the

charges accelerate at right angles to the line of sight, results in maximum intensity.

Conversely, viewing the antenna straight down from above, in the same direction as

the acceleration, results in zero intensity.

According to Maxwells work in electric and magnetic fields, he proposed the light is

an electromagnetic wave. Visible light is of course an electromagnetic wave. Unlike

sound or a wave on a string requires a medium through which it can propagate, light

can propagate through a vacuum. In fact, electromagnetic waves travel through a

vacuum with the maximum speed, c, that any form of energy can have. It is also the

speed of light.

smc /1000.3 8=

A beam of light can travel around the world about seven times in a single second.

Light travels slower in a denser medium, e.g. glass and water.

Example

The distance between Earth and the Sun is 1.501011 m. How long does it take for

light to cover this distance?

Answer:

The required time is obtained by dividing the distance by the speed of light:

ssm

mt 500

/1000.3

1050.18

11

=

= .

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Example

An electromagnetic wave propagates in the positive y direction, as

shown in the figure. If the electric field at the origin is in the positive z

direction, is the magnetic field at the origin in (a) the positive x

direction, (b) the negative x direction, or (c) the negative y direction?

Answer:

According to the right hand, which has your thumb pointing the propagation direction,

the direction of magnetic field should be along positive x.

15.2 Doppler effect of light

The Doppler effect for light is different from that of sound by two ways.

(a) Sound waves require a medium through which to travel, whereas light can

propagate across a vacuum. Therefore, the Doppler effect for sound depends

on whether the medium that is, the air is moving or is still.

(b) The speed of sound can be different for different observers. For example, an

observer approaching a source of sound measures an increased speed of sound,

whereas an observer detecting sound from a moving source measures the usual

speed of sound. For this reason, the Doppler effect with sound is different for a

moving observer that it is for a moving source. In contrast, the speed of

electromagnetic waves is independent of the motion of the source and

observer. Therefore, there is just one Doppler effect for electromagnetic waves,

and it depends only on the relative speed between the observer and source.

For source speeds u that are small compared with the speed of light, the observed

frequency f from a source with frequency f is

=

c

uff 1'

Note that u is a speed and hence is always positive. The appropriate sign in front of

the term u/c is chosen for a given situation - the plus sign applies to a source that is

approaching the observer, the minus sign to a receding source, In addition, u is a

relative speed between the source and the observer, both of which may be moving.

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Example

An FM radio station broadcasts at a frequency of 88.5 MHz. If you drive your car

toward the station at 32.0 m/s, what change in frequency do you observe?

Answer:

Apply the formula

+=

c

uff 1' , that is

==

c

uffff ' , where f = 88.5 MHz, u

= 32.0 m/s and smc /1000.3 8= , as the car is moving toward the station. The

frequency changes by 9.44 Hz.

Common applications of the Doppler effect include the radar units used to measure

the speed of automobiles, and the Doppler radar that is used to monitor the weather.

In Doppler radar, electromagnetic waves are sent out into the atmosphere and are

reflected back to the receiver. The change in frequency of the reflected beam relative

to the outgoing beam provides a way of measuring the speed of the clouds and

precipitation that reflected the beam.

Example

The Doppler weather radar, Nexrad, commonly operates at a frequency of 2.7 GHz. If

a Nexrad wave reflects from an approaching weather system moving with a speed of

28 m/s, find the difference in frequency between the outgoing and returning waves.

Answer:

Two Doppler effects are involved in this system. First the outgoing wave is seen to

have a frequency

+=

c

uff 1' by the weather system, since it is moving toward the

source. The waves reflected by the weather system, then, have the frequency f. Since

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the weather system acts like a moving source of radar with frequency f, an observer

at the radar facility detects a frequency

+=

c

uff 1''

'.

Since, Hzsm

smHz

c

ufff 250

/100.3

)/28)(107.2('

8

9

=

=

= ,

and

=

c

ufff '''' ,

we have

+=+

c

ufHzfHzf )250()250('' , hence, Hzff 500'' = .

Remarks:

We focus on the difference in frequency, since f and f are both very large numbers.

15.2 The electromagnetic spectrum

15.3.1 Radio waves

The lowest-frequency electromagnetic waves of practical importance are the radio and

television waves. The frequency ranges from 106

Hz to 109 Hz.

15.3.2 Microwaves

Electromagnetic radiation with frequencies from 109 Hz to about 10

12 Hz is referred

to as microwaves. Waves in this frequency range are used to carry long distance

telephone conversations, as well as to cook our food. Microwaves, with wavelengths

of about 1 mm to 30 mm, are the highest-frequency electromagnetic waves that can be

produced by electronic circuitry.

15.3.3 Infrared waves

Electromagnetic waves with frequencies just below that of light

roughly 1012

Hz to 4.31014 Hz are known as infrared rays. Most

remote controls operate on a beam of infrared light with a

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wavelength of about 1000 nm. This infrared light is so close to the visible spectrum

and so low in intensity that it cannot be felt as heat.

15.3.4 Visible light

The full range of colors in a rainbow as seen by human eyes is visible light. The

frequency range is 4.31014 Hz to 7.51014 Hz.

15.3.5 Ultraviolet light

When electromagnetic waves have frequencies just above that of violet light from

about 7.51014 Hz to 1017 Hz they are called ultraviolet or UV rays. More prolonged

or intense exposure to UV rays can have harmful consequences, including an

increased probability of developing a skin cancer.

15.3.6 X-ray

The frequency range of X-ray is about 1017

Hz to 1020

Hz. Typically, the X-rays used

in medicine are generated by the rapid deceleration of high-speed electrons projected

against a metal target. These energetic rays, which are only weakly absorbed by the

skin and soft tissues, pass through our bodies rather freely, except when they

encounter bones, teeth, to other relatively dense material. This property makes X-rays

most valuable for medical diagnosis, research, and treatment.

15.3.7 Gamma rays

Electromagnetic waves with frequencies above about 1020

Hz are generally referred to

as gamma () rays. These rays, which are even more energetic than X-rays, are often

produced as neutrons and protons rearrange themselves

within a nucleus, or when a particle collides with its

antiparticle, and the two annihilate each other. Gamma

rays are highly penetrating and destructive to living

cells. It is for this reason that they are used to kill

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cancer cells and, more recently, microorganisms in food.

15.4 Polarization

To understand what is meant by the polarization of light, or any other electromagnetic

wave, consider the electromagnetic waves pictured in the figure. Each of these waves

has an electric field that points along a single line. For example, the electric field that

points in either the positive or negative z direction. We say, then, that this wave is

linearly polarized in the z direction.

A wave of this sort might be produced by a straight-wire antenna oriented along the z

axis. In general, the polarization of an electromagnetic wave refers to the direction of

its electric field. Polarization is an evidence for electromagnetic waves being a

transverse wave.

A beam of unpolarized light can be polarized in a number of ways,

e.g. by passing it through a polarizer. To be specific, a polarizer is

a material this is composed of long, thin, electrically conductive

molecules oriented in a specific direction. When a beam of light

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strikes a polarizer, it is readily absorbed if its electric field is parallel to the molecules;

light whose electric field is perpendicular to the molecules passes through the material

with little absorption, As a result, the light that passes through a polarizer is

preferentially polarized along a specific direction.

A simple mechanical analog of a polarizer is shown in the following figure.

Remarks:

Polaroid is made from tiny crystals of quinine iodosulphate all lined up in the same

direction in a sheet of nitrocellulose. Crystals which transmit light vibrations, i.e.

electric field variations, in one particular plane and absorb those in a mutually

perpendicular plane are said to be dichroic.

We now consider what happens when a beam of light polarized in one direction

encounters a polarizer oriented in a different direction. In the left figure in next page,

we see light with a vertical polarization and intensity I0 passes through a polarizer

with its preferred direction its transmission axis at an angle to the vertical. The

component of E along the transmission axis is cosE . Recalling that the intensity of

light is proportional to the electric field squared, we see that the intensity, I, of the

transmitted beam is reduced to

20 cosII = .

When an unpolarized beam of intensity I0 passes through a polarizer (the right figure

in next page), the transmitted beam has an intensity of 02

1I and is polarized in the

direction of the polarizer.

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An unpolarized beam of intensity I0 is polarized in the vertical direction by a polarizer

with a vertical transmission axis. Next, it passes through another polarizer, the

analyzer, whose transmission axis is at an angle relative to the transmission axis of

the polarizer. The final intensity of the beam is 20 cos2

1II = .

Example

Vertically polarized light with an intensity of 515 W/m2

passes through a polarizer

oriented at an angle to the vertical. Find the transmitted intensity of the light for (a)

o0.10= , (b) o0.45= , and (c) o0.90= .

Answer:

We apply the formula 20 cosII = .

(a) 222 /4990.10cos)/515( mWmWI o ==

(b) 222 /2580.45cos)/515( mWmWI o ==

(c) 222 /00.90cos)/515( mWmWI o ==

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Example

In the polarization experiment, the final

intensity of the beam is 0.200 I0. What is the

angle between the transmission axes of the

analyzer and polarizer?

Answer:

The intensity of the light is first reduced to 02

1I and then to 0.200 I0. That is,

0

2

0 200.0cos2

1III == . We therefore have

200.0cos2

1 2 = .

Solving for , we obtain o8.50= .

Since the analyzer absorbs part of the light as the beam passes through, it also absorbs

energy. Thus, in principle, the analyzer would experience a slight heating.

Example

Calculate the transmitted intensity for the following two cases: (a) A vertically

polarized beam of intensity I0 passes through a polarizer with its transmission axis at

60o to the vertical. (b) A vertically polarized beam of intensity I0 passes through two

polarizers, the first with its transmission axis at 30o to the vertical, and the second

with its transmission axis at 60o to the vertical. In both cases, the final beam is

polarized at 60o to the vertical.

Answer:

(a) The final intensity: 02

0

2

04

160coscos IIII o === .

(b) The final intensity: 022

02

2

1

2

016

930cos)30cos(cos)cos( IIII o === ,

where 2 is the further angle of the second polarizer from the first.

Remarks:

Thus, even though the direction of polarization is rotated by a total of 60o in each case,

the final intensity is more than twice as great when two polarizers are used instead of

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just one. In general, the more polarizers that are used - and hence the more smoothly

the direction of polarization changes the greater the final intensity.

Example

Consider a set of three polarizers, P1, P2 and P3. P1 has

a vertical transmission axis, and P3 has a horizontal

transmission axis. Taken together, P1 and P3 are a pair

of crossed polarizers. P2, with a transmission angle 45o

to the vertical, is placed between P1 and P3. A beam of

unpolarized light shines on P1 from the left. Is the

transmission of light through the three polarizers (a)

zero, (b) nonzero?

Answer:

Since P1 and P3 are still crossed, it might seem that no light can be transmitted.

However, a polarizer causes a beam to have a polarization in the direction of its

transmission axis, it becomes clear that transmission indeed possible. Light is

polarized at 45o to the vertical through P2 and then passes through P3. Specially, the

intensity of the incident light is reduced by a factor of 2 when it passed through P1, by

a factor of 2 when it passes through P2 (cos2 45

o = ), and again by a factor of 2

when it passes through P3. The final intensity, then, is one-eight the incident intensity.

The applications of crossed polarizers include photoelastic stress analysis and LCD.

(b) On

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The operation of a liquid-crystal display (LCD) is simply ON and OFF. The

structure of LCD is shown in the figure. There are basically three essential elements

to an LCD two crossed polarizers and a thin cell that holds a fluid formed of long,

thin molecules known as a liquid crystal. The liquid crystal is selected for its ability to

rotate the direction of polarization, and the thickness of the liquid-crystal cell is

adjusted to give a rotation of 90o. Thus, in its OFF state, the liquid crystal rotates

the direction of polarization, and light passes through the crossed polarizers. In this

state, the LCD is transparent. When a voltage is applied to the liquid crystal it no

longer rotates the direction of polarization, and light is no longer transmitted. Thus, it

is in the ON state. The advantage of LCD is its energy efficient. A very little energy

is required to give the voltage necessary to turn a liquid crystal cell ON. The light

that the LCD uses is already present in the environment.

15.5 Polarization by reflection

Unpolarized light can be partially polarized by reflection. When unpolarized light

strikes a reflecting surface between two optical materials, preferential reflection

occurs for those waves in which the electric-field vector is parallel to the reflecting

surface. At one particular angle of incidence, called the polarization angle p, only the

light for which the E vector is perpendicular to the plane of incidence is reflected. The

reflected light is therefore linearly polarized perpendicular to the plane of incidence.

Sir David Brewster noticed that when the angle of incidence is equal to the polarizing

angle p, the reflected ray and the refracted ray are perpendicular to each other, so

pr = 90 or pr cossin = . From the law of refraction,

rnn p sinsin 21 = ,

p p

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which gives pp nn cossin 21 = ,

and thus the Brewsters law: 1

2tann

np = .

Remark:

When light is incident at the polarizing angle, none of the E field component parallel

to the plane of incidence is reflected; this component is transmitted 100% in the

refracted beam. So the reflected light is completely polarized. The refracted light is a

mixture of the component parallel to the plane of incidence, all of which is refracted,

and the remainder of the perpendicular component; it is therefore partially polarized.

Example

Sunlight reflects off the smooth surface of an unoccupied swimming pool. (a) At what

angle of reflection is the light completely polarized? (b) What is the corresponding

angle of refraction for the light that is transmitted (refracted) into the water? (c) At

night an underwater floodlight is turned on in the pool. Repeat parts (a) and (b) for

rays from the floodlight that strike the smooth surface from below.

Answer:

(a),(b) The light is first in the air, then in the water. Take na = 1.0 (air) and nb = 1.33

(water). The Brewsters law a

b

pn

n=tan gives

o

a

b

pn

n1.53

0.1

33.1tantan

11 =

=

= .

The reflected and refracted rays are perpendicular, so o

p r 90=+ and thus

ooo

p

or 9.361.539090 === .

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(c) Now the light is first in the water, then in the air. Take na = 1.33 (water) and nb

= 1.33 (air). The Brewsters law a

b

pn

n=tan gives

o

a

b

pn

n9.36

33.1

0.1tantan

11 =

=

= .

The reflected and refracted rays are perpendicular, so o

p r 90=+ and thus

ooo

p

or 1.539.369090 === .

15.6 Polarization by scattering

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15.7 Superposition Revisit

The simple addition of two or more waves to give a resultant wave is referred to as

superposition. When waves are superposed, the result may be a wave of greater

amplitude (constructive interference) or of reduced amplitude (destructive

interference).

Constructive interference occurs: mll = 12 , where ....,3,2,1,0 =m

Destructive interference occurs: )2

1(12 = mll , where ....,3,2,1,0 =m

The source waves that we often used:

Monochromatic light: it consists of waves with a single frequency and hence a

single color.

Coherent/incoherent light: Light waves that maintain a constant phase

relationship with one another are referred to as coherent. Light waves in which

the relative phases vary randomly with time are said to be incoherent.

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Example

Two friends tune their radios to the same frequency and

pick up a signal transmitted simultaneously by a pair of

antennas. The friend who is equidistant from the

antennas, at P0, receives a strong signal. The friend at

point Q1 receives a very weak signal. Find the

wavelength of the radio waves if d = 7.50 km, L = 14.0

km, and y = 1.88 km. Assume that Q1 is the first point

of minimum signal as one moves away from P0 in the y

direction.

Answer:

The first minimum in the y direction from the maximum at point P0, we know that the

path difference, l2 l1, is half a wavelength.

The path length l1:

kmkmkm

kmyd

Ll 1.1488.12

)50.7()0.14(

2

2

2

2

2

1 =

+=

+=

The path length l2:

kmkmkm

kmyd

Ll 1.1588.12

)50.7()0.14(

2

2

2

2

2

2 =

++=

++=

By the relation 2

12

= ll , it gives km0.2= . This wavelength corresponds to a

frequency of about 150 kHz.

15.8 Youngs double-slit experiment

Monochromatic light from a narrow vertical slit S falls on two other narrow slits S1

and S2 which are very close together and parallel to S. S1 and S2 act as two coherent

sources (both being derived from S) and if they (as well as S) are narrow enough,

diffraction causes the emerging beams to spread into the region beyond the slits.

Superposition occurs in the shaded area where the diffracted beams overlap. Alternate

bright and dark equally-spaced vertical bands (interference fringes) can be observed

on a screen. Note that if either S1 or S2 is covered the bands disappear.

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An expression for the separation of two bright (or dark) fringes can be obtained as

follows.

For the central bright fringe: since OSOS 21 = , the two waves are in phase and

constructive interference occurs.

At P, distance x1 from O, there will be a bright fringe if the path difference is a whole

number wavelength. The path difference between waves

nPSPS = 12 ,

where n is an integer. We say the nth bright fringe is formed at P. If d is the distance

from the screen to the double-slit and a is the slit separation. Hence,

4/)2/()( 212

1

22

1

22

2 aaxxdaxdPS +++=++=

4/)2/()( 212

1

22

1

22

1 aaxxdaxdPS ++=+=

Therefore, 12

1

2

2 2)()( axPSPS =

But, )()()()( 12122

1

2

2 PSPSPSPSPSPS += , so we have

112 22)( axdPSPS = or 112 )( axdPSPS =

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For the nth bright fringe at P we have

1axdn = or d

axn 1= . (1)

For the (n+1)th bright fringe at Q, say, we have

2)1( axdn =+ or d

axn 2)1( =+ . (2)

(2) (1) gives d

ay= , where 12 xxy = .

Remarks:

1. a

y1

, if and d are constant.

2. dy , if and a are constant.

3. y , if a and d are constant.

In fact, as the double-slit is far from the screen, the paths for bright fringes or dark

fringes are parallel. Refer to the following figure and note that the parameters d and y

are defined differently from the above discussion. Hence,

for bright fringe: md =sin , ....,3,2,1,0 =m

for dark fringe: )2

1(sin = md , ....,3,2,1,0 =m

The linear distance from central fringe is tanLy = .

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Example

Red light ( = 750 nm) passes through a pair of slits with a separation of 6.2 105 m.

Find the angles corresponding to (a) the first bright fringe and (b) the second dark

fringe above the central bright fringe.

Answer:

(a) The relation md =sin with m = +1, gives

o

m

m

d

m69.0

102.6

)10750)(1(sinsin

5

911 =

=

=

.

(b) The relation )2

1(sin = md , with m = +2, gives

o

m

m

dm 04.1

102.6

10750)

2

12(sin)

2

1(sin

5

911 =

=

=

.

Example

Two slits with a separation of 8.5 105 m create an interference pattern on a screen

2.3 m away. If the tenth bright fringe above the central fringe is a linear distance of 12

cm from it, what is the wavelength of light used in the experiment?

Answer:

The relation tanLy = gives

o

m

m

L

y0.3)

3.2

12.0(tan)(tan 11 ===

For bright fringe: md =sin which gives

nmm

m

d o 440104.4)0.3sin(10

105.8sin 7

5

==

==

.

The light used in the experiment is blue in color.

Example

A two-slit experiment is performed in the air. Later, the same apparatus is immersed

in water and the experiment is repeated. When the apparatus is in water, are the

interference fringes (a) more closely spaced, (b) more widely spaced, or (c) spaced the

same as when the apparatus is in air?

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Answer:

The wavelength is smaller in water than in air, e.g. aw

a

wn

= , where naw > 1.

According to the equation md =sin for the bright fringe, a decreased wavelength

implies a decreased , and hence the fringes are closely packed.

15.9 Phase change due to reflection

There is no phase change when light reflects from a region with a lower index

of refraction.

There is half-wavelength (/2 or ) phase change when light reflects from a

region with a higher index of refraction, or a solid surface.

15.9.1 Air wedge

Interference effect is between light (ray 1) reflected from the bottom surface of the top

glass plate and light (ray 2) reflected from the upper surface of the lower plate. Ray 1

experience no phase change. Ray 2, that is the reflection from air to glass results in a

half-wavelength (/2 or ) phase change, thus the effective path difference of ray 2

and ray 1 is 2

22

+=++= dddleff . For constructive interference (bright fringe):

mleff = , i.e.

md =+2

2 , where m = 1, 2, 3,

For destructive interference (dark fringe):

21

2

mleff = , i.e.

222

md =+ , where m = 1, 3, 5,

Or md =2 , where m = 0, 1, 2, 3,

When m = 0, d = 0, it corresponds to the first dark fringe located at the joint of the

glass plates.

Remarks:

The bright (or dark) fringes are equally spaced. Take a look at adjacent bright fringes:

)1(2

2 1 +=++ mdm ,

mdm =+2

2

The difference of this two equations gives =+ )(2 1 mm dd , hence

=+ sin)(2 1 mm ll where lm+1 and lm are the lengths measured from the joint of

glass plates to the (m+1)th and mth bright fringes respectively. The angle between

glass plates is . Thus, we have = sin2 l , which shows that the fringes are

equally spaced.

Example

An air wedge is formed by placing a human hair between two glass plates on one end,

and allowing them to touch on the other end. When this wedge is illuminated with red

light ( = 771 nm) it is observed to have 179 bright fringes. How thick is the hair?

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Answer:

Setting that the thickness of hair as t which corresponds to the 179th

bright fringe, we

have from the relation

mt =+2

2 ,

)10771)(179(2

107712 9

9

mm

t

=

+

Thus, we obtain the thickness of hair = 68.8 106 m = 68.8 m.

Remarks:

If a thicker hair is used, then by the relation

md =+2

2 again, )2

1(2 = mdm , the

number of fringes increased under the same frequency of light.

Example

When light is reflected from a medium of several layers as shown in the right figure,

the first destructive interference is given by the path difference af

a

nt

4

= . Given that

the film is less dense than the glass.

Answer:

As the reflected rays 1 and 2 has both a phase change of . The effect in the path

difference is cancelled. But there is a path difference due to the thickness of film. So,

22 aaftn

= or

af

a

nt

4

= . Thus, we get the result. A glass (n = 1.52) is often coated

with a thin film of magnesium fluoride (n = 1.38) to minimize reflected light. For

yellow-green light nm565= , hence, we have the thickness of thin film equals to

102 nm. The figure shows no phase change in the reflected ray from the third medium.

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15.9.2 Newtons ring

The thickness of the air film gradually increases outwards from zero at the point of

contact, as shown in figure. Interference occurs between light reflected from the lower

surface ABC of the lens and the upper surface DBE of the plate. A series of bright and

dark rings is seen through G when a traveling microscope is focused on the air film.

The rings are fringes which have equal thickness for the same ring. As the radius

increases, the separation decreases. At the center B of the fringe system there is a dark

spot.

As shown in figure, the triangles SLO and BPO are similar, the corresponding sides

have relations PO

LO

BO

SO= or now, we rewrite it as BOLOPOSO = , that is

llRrr nn )2( =

22 2 lRlrn =

Since R >> l, we have Rlrn 22

= . For the dark rings,

we have nl =2 , thus nRrn =2

, where n = 0,1,

2, . The central dark sport corresponds to n = 0.

For bright rings,

nl =+2

2 , thus

=

2

12nRrn , where n = 1, 2, . The first bright

ring corresponds to n = 1.

Remarks:

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The wavelength of light can be obtained by measuring the diameter of dark rings. The

equation nRrn =2

becomes nRd n =

2

2 or nRd n 4

2= . A plot of dn

2 against n

gives a straight line, for which the slope is 4R.

Example

Calculate the radius of curvature of a plano-convex lens used to produce Newtons

rings with a flat glass plate if the diameter of the tenth dark ring is 4.48 mm, viewed

by normally reflected light of wavelength 5.00107m. What is the diameter of the

20th

bright ring?

Answer:

Apply the formula for the nth dark ring nRrn =2

, where we plug in n = 10, =

5.00107m and r10 = 4.48103

m / 2. We obtain R = 1.00m.

Apply the formula for the nth bright ring

=

2

12nRrn , where we plug in n = 20,

= 5.00107m and R = 1.00m. We obtain r20 = 3.12103

m = 3.12 mm. That is

d20 = 3.12 mm 2 = 6.24 mm.

15.10 Diffraction

The waves are initially traveling directly to the right. After passing through the gap in

the barrier, however, they spread out and travel in all possible forward directions, in

accordance with Huygens principle. Thats why an observer located at P (not along

the line with the incoming wave and the gap) detects the wave. In general, waves

always bend, or diffract when they pass by a barrier or through an opening (or slit).

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A familiar example of diffraction is the observation that you can hear a person talking

even when that person is out of sight around a corner. The sound waves from the

person bend around a corner. It might seem, then, that light cannot be wave, since it

does not band around a corner along with the sound. In fact, the observation is due to

the great difference between the sound and light waves of many orders. For sound:

the wavelength is in the order of meter, but for light, the wavelength is in the order of

107

meter.

Remarks:

According to Huygens, every point on a wavefront may be regarded as a source of

secondary spherical wavelets which spread out with the wave velocity. The new

wavefront is the envelope of these secondary wavelets.

26

15.10.1 Single-slit diffraction

As shown in figure, since the screen is distant, the waves from 1 and 1 travel on

approximately parallel paths to reach the screen. So, the path difference for these

waves is sin)2/(W . When = 0, constructive interference occurs and bright fringe

is observed on the screen. As is increased, the path difference is also increased,

when 2

sin)2/(

=W , we observe the first dark fringe (the first minimum). Or we

can write

=sinW .

The second dark fringe is obtained by considering the following figure.

2/sin)4/( =W , hence we have 2sin =W . In general, we have mW =sin ,

where ...,3,2,1 =m .

27

For the first bright fringe, we have 2/sin)3/( =W , giving W2

3sin

= .

Similarly, the second bright can be obtained by 2/sin)5/( =W , giving

W2

5sin

= .

Remark:

The width of the central bright fringe can be estimated by first considering the first

dark fringes which has angular separation W

2 (approximately). It is because the first

dark fringes locate at W

=sin , and for small , sin .

Example

If the width of the slit through which light passes is reduced, does the central bright

fringe (a) become wider, (b) become narrower, or (c) remain the same size?

Answer:

The answer is wider due to the above discussion. In addition, we know that a

narrower slit will give more diffraction.

Example

Light with a wavelength of 510 nm forms a diffraction pattern after passing through a

single slit of width 2.2106 m. Find the angle associated with (a) the first and (b) the

second dark fringe above the central bright fringe.

28

Answer:

We apply the formula for the dark fringe,

=

W

m 1sin

When m = 1, o13

102.2

)10510)(1(sin

6

91 =

=

.

When m = 2, o28

102.2

)10510)(2(sin

6

91 =

=

.

15.10.2 Diffraction grating

A system with a large number of slits is referred to as

diffraction grating. In some cases, it is possible to produce

gratings with as many as 40,000 slits per centimeter. The

interference pattern formed by a diffraction grating

consists of a series of sharp, widely spaced bright

fringes called principal maxima separated by relatively

dark regions with a number of weak secondary maxima. In

the limit of a large number of slits, the principal maxima

become more sharply peaked, and the secondary maxima

become insignificant.

For constructive interference in a diffraction grating

md =sin , where ...,3,2,1 =m

Notice that a smaller d gives a larger spread angle . Thats why we observe a widely

spread of light if a grating of great number of lines per centimeter is passed with light

beam.

29

Example

Find the slit spacing necessary for 450-nm light to have a first-order (m = 1) principal

maximum at 15o.

Answer:

For constructive interference in a diffraction grating, we have

md =sin ,

Now, plug in the given data, we have mmm

do

69

107.1)15sin(

)10450)(1(

sin

=

==

.

That means the diffraction grating has 588000107.1/1 6 lines per meter, or 5,880

lines per cm.

Example

When 546-nm light passes through a particular diffraction grating a second-order

principal maximum is observed at an angle of 16.0o. How many lines per centimeter

does this grating have?

Answer:

For constructive interference in a diffraction grating, we have

md =sin ,

Now, plug in the given data, we have mmm

do

69

1096.3)0.16sin(

)10546)(2(

sin

=

==

.

That means the diffraction grating has 000,2531096.3/1 6 = lines per meter, or

2,530 lines per cm.

30

15.11 Resolving power

The first dark fringe for the diffraction pattern of a circular opening is given by

D

22.1sin =

It is different from the case of single slit W

=sin

Diffraction-induced smearing also makes it difficult to visually separate objects that

are close to one another. In particular, if two closely spaced sources of light are

smeared by diffraction, the circles they produce may overlap, making it difficult to

tell if there are two sources or only one.

Rayleighs criterion states that if the first dark fringe of one circular diffraction

pattern passes through the center of a second diffraction pattern, the two sources

responsible for the patterns will appear to be a single source. Thus, two objects can be

seen as separate only if their angular separation is greater than the following

minimum:

D

22.1min =

as is always small. Note also that the wavelength in the above equation corresponds

to the wavelength of light in the eyes, when the light is observed directly through

pupil. In such case we should consider the refractive index of our eyes (n = 1.36).

31

Example

The linear distance separating the headlights of a car is 1.1 m. Assuming light of 460

nm, a pupil diameter of 5.0 mm, and an average index of refraction for the eye of 1.36,

find the maximum distance at which the headlights can be distinguished as two

separate sources of light.

Answer:

According to the Rayleighs criterion, we have

D

22.1min = , where m

m

n

a 99

1033836.1

10460

=

==

,

hence, radm

m 53

9

min 103.8100.5

1033822.1

=

= .

The maximum distance L is obtained by

mrad

myL 000,13

)103.8tan(

1.1

tan 5min=

==

.