CHAPTER 26 INTERFERENCE - Texas A&M Universitypeople.physics.tamu.edu/adair/phys202/CHAPTER 26...

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CHAPTER 26 INTERFERENCE AND DIFFRACTION

INTERFERENCE

CONSTRUCTIVE

DESTRUCTIVE

YOUNG’S EXPERIMENT

THIN FILMS

NEWTON’S RINGS

DIFFRACTION

SINGLE SLIT

MULTIPLE SLITS

RESOLVING POWER

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IN PHASE 1800 OUT OF PHASE

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OR THE BOOKS DISCRIPTION

CONSTRUCTIVE INTERFERENCE

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DESTRUCTIVE INTERFERENCE

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When two waves arrive at a point where path

differences from the sources differ by one

wavelength, two wavelengths, three wavelengths,

etc. there will be constructive interference.

Or mathematically:

�� −�� = ��ℎ���� = 0,±1,±2,±3,….

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When two waves arrive at a point where path

differences from the sources differ by ����������ℎ, ����������ℎ, ����������ℎ, ���. (Or half integer number of wavelengths) there will

be destructive interference.

Or mathematically:

�� −�� = �� +��! �

Where ℎ���� = 0,±1,±2,±3,….

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Many ways to produce two rays of coherent light.

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Consider path differences

"��ℎ#$%%������ = #&$�'

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Constructive Interference

#&$�' = �� ℎ���� = 0,±1,±2,±3,….

Destructive Interference

#&$�' = �� + ��! �

ℎ���� = 0,±1,±2,±3,….

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We can obtain the distance ( a bright line is below

(or above) the center of the screen.

Use

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If ) ≫ #

tan ' ≈ ()

Therefore

( = ) tan '

And for the �/0 bright line

(1 ≈ ) tan '1

For small angles

tan ' ≈ sin '

Therefore

(1 ≈ ) sin '1

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Use this with

#&$�' = ��

Or

&$�'1 = ��#

To get

(1 ≈ )��#

ℎ���� = 0,±1,±2,±3,…

This is the equation that Young used to measure

the wave length of light.

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As was said many ways of producing two rays of

coherent light. We have been discussing

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Can also get two coherent rays from thin films.

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We will discuss the interference between rays

coming from the two surfaces on opposite sides of

a space filled with liquid (thin film) or air.

First: we need to know that when light reflects

from a surface sometimes it undergoes a phase

shift of 1800.

For example if a ray is in air 4� = 1.005and reflects

at the surface with water 4� = 1.335back into the

air it does not undergo a phase shift.

But if a ray is in glass 4� = 1.335and reflects at the

surface with air 4� = 1.005 back into the glass it

undergoes a 1800 phase shift.

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Rule

If ray goes from low n material and reflects at

surface of high n material no phase shift.

If ray goes from high n material and reflects at

surface with low n material 180 degree phase

shift.

6%�7 89:;9</9=:8>1?@@@@@@@@@@@A�B��# �B > �7

No phase shift.

But

6%�B 89:;9</9=:8>1?@@@@@@@@@@@A�7��# �B > �7

1800

(half wavelength) phase shift

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Now consider we have a film as shown

Assume perpendicular.

Film thickness where shown �.

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No phase shift at either surface or phase shift at

both surfaces

Constructive interference if

2� = ��� = 0, 1, 2, 3, …

But if one surface has phase shift and other does

not

Destructive interference.

Or

No phase shift at either surface or phase shift at

both surfaces

Destructive interference if

2� = �� +��! �� = 0, 1, 2, 3, …

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Example 26.3

Find separation of interference fringes.

One reflection (top) – no phase shift other

(bottom) – phase shift.

Glass

Air

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Equation for destructive interference

2� = ��� = 0, 1, 2, 3, …

Where glass touches – destructive interference.

From figure

�D =

ℎ�

So

� = Dℎ�

Therefore

2 EDℎ� F = ��

D = � E �2ℎF � = ���2ℎ

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D = � G40.1�54500D10IJ�5240.02D10IK�5 L

D = �41.25��5

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EXAMPLES OF INTERFERENCE PHENOMONO

Newton’s Rings

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Nonreflective Coatings

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DIFFRACTION

Light goes in straight line.

Therefore would expect sharp shadow of object on

a screen.

But

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Consider light going through single slit.

This shows a ray from the top of the slit and one

from half way down the slit (middle).

Or looking only at the slit

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The difference in path distance to the screen is

"��ℎM$%%. = �2 sin '

If the path difference is one half wave length –

destructive interference.

Therefore the requirement for a dark fringe is

�2 sin ' = ±

�2

sin ' = ± ��

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If you divide the slit into 4 regions

If the path difference is one half wave length –

destructive interference.

Therefore the requirement for a dark fringe is

�4 sin ' = ±

�2

sin ' = ±2 ��

Becomes a/4

Becomes O/Q sinθ

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Divide the slit into more sections and can show

The requirements for dark fringes are

sin ' = � ��4� = ±1,±2,±3,… 5

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If the wavelength of light is much smaller than the

slit width (and normally it is) then

��� ≪ 14���(&����5

Therefore

' ≪ 14���(&����5

And

sin ' ≈ '

Therefore for dark fringes

' ≈ ��� 4� = ±1,±2,±3,… 5

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Also for

Use for distance to screen ) = D

Then tan ' = ST

And for the ��ℎ dark band

tan '1 =(1)

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If the screen is a distance from the slit such that

) ≫ (1

And it normally is

Then ' ≪ 14���(&����5

And tan '1 ≈'1

Therefore

'1 ≈(1)

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Then using

' ≈ ��� 4� = ±1,±2,±3,… 5

'1 ≈(1) ≈ ��� 4� = ±1,±2,±3,… 5

Solve for (1

(1 ≈ )��� 4� = ±1, ± 3, ± 5,… 5

This is equation 26.10 in the book.

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Multiple Slits

Path difference will be # sin '

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Constructive interference occurs for

# sin ' = ��4� = 0, ±1,±2,±3,… . 5

More complicated than two slit diffraction

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Eight Slit

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More – 16 slits

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DIFFRACTION GRATINGS

Prisms refract light different amounts depending

on the wavelength of the light.

Note short wavelengths (blue) are refracted more

than he long wavelengths (red).

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Diffraction Gratings will do much the same.

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The equation for the maxima is the same as for

multiple slits.

# sin ' = ��4� = 0, ±1,±2,±3,… . 5

Note

sin ' = ��#

Thus long wavelengths diffracted more than short

wavelengths. (Opposite to prisim.)

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X-RAY DIFFRACTION

First experiments in 1912.

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Crystal is uniform arrangement of atoms forming

planes for diffraction.

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Gives pattern

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Consider the atoms as all lined up on the sites and

represented by the dots shown.

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Look at top row for example

Path lengths are equal when 'B ='8

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Therefore when the two angles are equal get

constructive interference from diffraction from

rows.

Then considering two rows one beneath the other.

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Additional constructive interference when path

lengths are equal to integer times wavelength of

x-rays.

2#&$�' = ��4� = 1, 2, 3, … . . 5

When both conditions are satisfied

1. 'B ='8

And

2. 2#&$�' = ��

When these conditions are not satisfied radiation

interferes destructively.

Therefore obtain diffraction pattern characteristic

of crystal.

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Scattering and thus diffraction can occur from

various planes within the crystal.

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CIRCULAR APERTURES

Light travelling through a small circular aperture is

diffracted also.

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The angular locations of the dark rings are given

by:

sin '� = 1.22 �M

sin '� = 2.23 �M

sin 'K = 3.24 �M

OBSERVING OBJECTS THROUGH VARIOUS SIZE

APERTURES.

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If the aperture in (a) were smaller 3 and 4 would

appear as one object.

The larger the aperture the better the resolving

power.

Lord Rayleigh proposed a criterion for telescopes

(and other optical instruments) to give a measure

for their capability.

Two objects can only be resolved if the center of

the second just falls at a distance of the center of

the first dark ring of the first.

The limit of resolution, '89U for a telescope (or

other optical instrument):

'89U = 1.22 �M