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Physical (Wave) Optics:

Diffraction &

Interference of Light

AP Physics 2

Serway – Chapter 24

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The Wave Behavior of Light

� Geometric (Ray) Model

� Uses rays and ray diagrams to understand how light reflects and refracts.

� Fails to describe interference and diffraction of light.

� Physical (Wave) Model

� Picks up where the ray model fails.

� Helps explain how CDs and DVDs operate and how images are formed on TV screens.

RECALL: Superposition ..AKA….Interference

One of the characteristics of a WAVE is the ability to

undergo INTERFERENCE. There are TWO types.

We call these waves IN PHASE which

results in CONSTRUCTIVE INTERFERENCE.

We call these waves OUT OF PHASE which results in

DESTRUCTIVE INTERFERENCE.

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Interference – Two SourcesWhen light OR sound is produced by TWO sources a

pattern results as a result of interference.

Interference – Two Sources

Example: WATER WAVES

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Interference – One Sources

Example: Light passing thru a narrow opening.

Diffraction

Diffraction is normally taken to refer to various phenomena which occur

when a wave encounters an obstacle. It is described as the apparent

bending of waves around small obstacles and the spreading out of waves

past small openings

• Diffraction is wavelength dependent.

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Monochromatic light

� Waves start off in phase at two slits

� Travel different distances to screen

� Relative phase at screen

� 1st experiment that measured λ of light

r1

r2

Light and dark alternating fringes

Screen

Young’s Double-Slit Interference Experiment

Young’s Experiment – Light Sources

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For sustained interference between two sources of light to be observed, the following conditions must be met.

1. The sources must be coherent. Which means they must maintain a constant phase relationship with one another.

2. The sources should be monochromatic. Which means they must have a single color and a single frequency.

Conditions for Interference of Light

The Central Maximum

Where,

d - the distance between the slits

L - the distance from the slit spacing

to the screen

• As the light waves pass through the slit

they cover the SAME DISTANCE.

• A very bright intense constructive

interference pattern will form on the

screen, called the CENTRAL MAXIMUM.

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The Central Maximum

In reality there are many waves of light passing through the slit.

FRINGES are the bright spots on the screens with the brightest being the

central maximum.

Here is the pattern you will see. Notice in figure 2 that there are several

bright spots and dark areas in between. The spot in the middle is the

BRIGHTEST and thus the CENTRAL MAXIMUM. We call these spots

FRINGES.

Figure 1 Figure 2

Orders, mWe use the letter, m, to

represent the ORDER of

the fringe from the bright

central.

Central

Maximum

First Order

Bright Fringe

m = 1

First Order

Bright Fringe

m = 1

Second Order

Bright Fringe

m = 2

Second Order

Bright Fringe

m = 2

It is important to understand that we see these bright fringes as a result of

CONSTRUCTIVE INTERFERENCE.

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Bright FringesThe reason you see additional

bright fringes is because the

waves CONSTRUCTIVELY build.

There is a difference however in

the intensity as you saw in the

previous slide.

As you can see in the picture, the

BLUE WAVE has to travel farther than the RED WAVE to reach the

screen at the position shown. For

the BLUE WAVE and the RED

WAVE to build constructively they

MUST be IN PHASE.

Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO TRAVEL SO THAT THEY BOTH HIT THE SCREEN IN PHASE?

Path difference (∆L)

Notice that these 2 waves are IN

PHASE. When they hit they screen

they both hit at the same relative

position, at the bottom of a crest.

How much farther did the red wave

have to travel?

Exactly – ONE WAVELENGTHλλλλ

They call this extra distance the PATH DIFFERENCE. The path difference

and the ORDER of a fringe help to form a pattern.

∆� � �λ

L – distance (m)

m – order number (integer)

λ – wavelength (m)

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Path Difference – Bright Fringes

1λλλλ1λλλλ

2λλλλ 2λλλλ

The bright fringes you see

on either side of the central

maximum are multiple

wavelengths from the bright

central. And it just so

happens that the multiple is

the ORDER.

Therefore, the PATH DIFFERENCE is equal

to the ORDER times the WAVELENGTH

P.D. = mλλλλ

((((Constructive)

Dark Fringes

We see a definite DECREASE in intensity between the bright fringes. In

the pattern we visible notice the DARK REGION. These are areas where

DESTRUCTIVE INTERFERENCE has occurred. We call these areas

DARK FRINGES or MINIMUMS.

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Dark Fringes

Central

Maximum

ZERO Order

Dark Fringe

m = 0

ZERO Order

Dark Fringe

m = 0

First Order

Dark Fringe

m = 1

First Order

Dark Fringe

m = 1

It is important to understand that we see these dark fringes as a result of

DESTRUCTIVE INTERFERENCE.

Dark Fringes

On either side of the bright

central maximum we see areas

that are dark or minimum

intensity.

Once again we notice that the

BLUE WAVE had to travel

farther than the RED WAVE to

reach the screen. At this point ,

however, they are said to

destructively build or that they

are OUT OF PHASE.

Here is the question: HOW MUCH FARTHER DID THE BLUE WAVE HAVE TO TRAVEL SO THAT THEY BOTH HIT THE SCREEN OUT OF PHASE?

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Path difference

Notice that these 2 waves are OUT OF

PHASE. When they hit they screen they

both hit at the different relative positions,

one at the bottom of a crest and the other

coming out of a trough. Thus their

amplitudes SUBTRACT.

How much farther did the red wave have to

travel?

Exactly – ONE HALF OF A WAVELENGTH

0.5 λ0.5 λ0.5 λ0.5 λ

The call this extra distance the PATH DIFFERENCE. The path difference and

the ORDER of a fringe help to form another pattern.

Path Difference – Dark Fringes

0.5λλλλ0.5λλλλ

1.5λλλλ 1.5λλλλ

The dark fringes you see on

either side of the central

maximum are multiple

wavelengths from the bright

central. And it just so

happens that the multiple is

the ORDER.

Therefore, the PATH DIFFERENCE is equal

to the ORDER plus a HALF, times A

WAVELENGTH

P.D. = (m+1/2)λλλλ((((Destructive)

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Path Difference - Summary

For CONSTRUCTIVE INTERFERENCE or

MAXIMUMS use:

For DESTRUCTIVE INTERFERENCE or

MINIMUMS use:

P.D. = mλλλλ

P.D. = (m+1/2)λλλλ

Where “m”, is the ORDER and “λ”, is the WAVELENGTH (m).

Young’s ExperimentIn 1801, Thomas Young successfully showed that light does produce an

interference pattern. This famous experiment PROVES that light has WAVE PROPERTIES.

Suppose we have 2 slits separated by a distance, d and a

distance L from a screen. Let point P be a bright fringe.

d

L

P

θθθθ

y

We see in the figure that we can

make a right triangle using, L, and

y, which is the distance a fringe is

from the bright central. We will

use an angle, θθθθ, from the point in

the middle of the two slits.

We can find this angle using

tangent!

L

y

θθθθ

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Another way to look at “path difference”

d

P

θ d

P.D.

θ

P.D.

P.D. = dsinθθθθ

Notice the blue wave travels farther. The

difference in distance is the path difference.

This right triangle is

SIMILAR to the one

made by y & L.

L

y

d

θθθθ

P.D.

Similar Triangles lead to a path difference

θ

L

y

d

θ

dsinθ

These angles

Are EQUAL.

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Putting it all together

Path difference is equal to the following:

� mλ

� (m+1/2)λ

� dsinθ

Therefore, we can say:

Will be used to find the

angle!

AP

ExampleA viewing screen is separated from a double slit source by 1.2 m. The distance between the two slits is 0.030 mm. The 2nd order bright fringe (m=2) is 4.5 cm from the central maximum. Determine the wavelength of light.

veConstructi bright""

045.02

100.32.1 5

=

==

== −

mym

mxdmL

)(tan 1

L

y−=θ

λθ md =sinGiven:

?=λ

Find:

Solution:

== − )2.1

045.0(tan 1θ

Find θ:

o15.2

Find λ :

λ2)15.2sin()103( 5 =− ox

2

)15.2sin()103(5 o−

=x

λ

m1062.5 7−×=λ

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Diffraction Grating

� A diffraction grating is the tool of choice for separating the colors in incident light.

� A typical grating contains several thousand lines per centimeter.

� To determine d, for a diffraction grating find the inverse of the number of lines per unit length.

� The tracks of a compact disc act as a diffraction grating, producing a separation of the colors of white light.

Example

m 0.0021

m 1.80m10450 9

=

=×= −

y

Lλ

λθ )2

1(sin += md

A light with wavelength, 450 nm,

falls on a diffraction grating

(multiple slits). On a screen 1.80 m

away the distance between dark

fringes on either side of the bright

central is 4.20 mm. a) What is the

separation between a set of slits?

b) How many lines per meter are on

the grating?

line

metersd =

Given:

Find:

Solution:

?rLines/mete

?

=

=d

Find θ:

)(tan 1

L

y−=θ

== − )8.1

0021.0(tan 1θ o067.0

Find d:

910450)2

10()067.0sin( −×+=d

m 0.0001924=d

== −

dord

meter

lines 11lines/m 5197.2

Find lines/meter:

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Small Angle Approximation

d

Lmy

λ=

L

y=≈ θθ tansin

For fringes between O and P

Position of bright fringes from O (ym)

Position of dark fringes from O (ym)

+=

2

1m

d

Ly

λ

m

m Lx

d

λ≈ AP

Where,distance to fringe (m)

order #

wavelength (m)

distance from filter to screen (m)

distance between slits (m)

m mx y

m

L

d

λ

= −

−

−

−

−