S-110 A.What does the term Interference mean when applied to waves? B.Describe what you think would...

S-110

A. What does the term Interference mean when applied to waves?

B. Describe what you think would happened when light interferes constructively.

C. Describe what would happen if light went through total destructive interference.

Objectives

I can explain Young’s double slit experiment.

I can calculate wavelength using interference patterns.

Interference and Diffraction

9.1 Superposition and Interference

9.1 Superposition and Interference

Review – Superposition

net displacement caused by a combination

of waves – the algebraic sum

Phase difference

180o – out of step by half a wavelength

Coherent – maintains a constant phase difference

Applet

9.2 Young’s Two-Slit Experiment

9.2 Young’s Two-Slit Experiment

1801 – Thomas Young

Shot monochromatic, coherent light

through two slits

The result was bands of color

called bright fringes

Applet

9.2 Young’s Two-Slit Experiment

What happened

Hugyen’s principle – every point on a wave front can be treated as the source of a new wave

When a wave is sent

through a barrier

two identical wave

fronts are created

9.2 Young’s Two-Slit Experiment

So when a wave is sent through double slits

The fringes are caused by

constructive interference – bright fringes

destructive interference – dark fringes

When we look at the pathway

of two rays

Ripple Tank

9.2 Young’s Two-Slit Experiment

For constructive interference, the path difference must be a multiple of So

m – any whole

number

dsin

md sin

9.2 Young’s Two-Slit Experiment

For dark fringes – the waves must be 180o out of phase, so

To measure , we need to go back to the experiment

An look at the triangle

One side is L, we will

call the other side

x

)(sin 21 md

x

L

9.2 Young’s Two-Slit Experiment

We can calculate the angle using the law of tangents

x

L

L

xtan

S-111

A light is shined through two slits that are 1.2 mm apart. If the third order minima is produced on a screen 1.5, how far from the central maxima will it be. Assume that the light used has a frequency of 7.2 x1014 hz.

S-112

A light is shined through two slits that are 0.50 mm apart. If the fourth order maxima is produced on a screen 0.80 m from the slits, how far from the central maxima will it be. Assume that the light used has a wavelength of 1.8 x10-7 m.

9.3 Interference in Reflected Waves

9.3 Interference in Reflected Waves

Waves that reflect of objects at different locations can interfere

Waves can go through a phase change due to reflection

9.3 Interference in Reflected Waves

When waves reflect at a boundary as they go

From higher n to lower n – no phase change

From lower n to higher n – 180o phase change

9.3 Interference in Reflected Waves

Lets look at what happens in an air wedge

There is no phase change

at the first boundary

There is a phase change

at the second boundary

The paths are essentially the same length in a real interference pattern (or they wouldn’t hit the same part of eye)

9.3 Interference in Reflected Waves

So the length of the path is

For constructive

interference

For destructive interference

tttl 221

21

mt

2

21

21

21 2

mt

9.3 Interference in Reflected Waves

Thin Films – soap bubbles or oil slicks

First ray (phase change)

Second ray (no

phase change)

Solving in terms of

211 l

tl 22

nvac

n

9.3 Interference in Reflected Waves

Combining we get

So for constructive interference

211 l tl 22 nvac

n

nn

tl

22

vacn

ntt

22

mnt

vac

212

9.3 Interference in Reflected Waves

And for destructive interference

mnt

vac

2

S-113

Light of wavelength 615 nm strikes the surface of an oil film (n=1.55) that is floating on water (n=1.33). The light strikes the surface of the oil at an angle of 22o. What is the minimum thickness of the oil that would produce a destructive interference pattern?

9.4 Diffraction

9.4 Diffraction

Waves diffract (bend) when they pass through barriers

Single Slit Diffraction – monochromatic light sent through a single slit will cause an interference pattern

The pattern occurs because of the diffraction of light around the edge of the slit

Diffraction

Applet

9.4 Diffraction

Similar to double slit geometry

If a is the width of

the slit, the

first minimum

would occur

In general

sina

ma sin

9.4 Diffraction

If the distance D is much greater than the slit distance y, then

we can use the

Approximation

Combining with

D

ytanD

ysin sin

D

y

ma sin

mD

ya

a

Dmy

9.5 Resolution

9.5 Resolution

Resolution – the ability to visually separate closed spaced objects

Depends on the aperture (size of lens)

In a slit the first dark fringe would

be

For a circular aperture of

diameter D produces a

central bright and a

dark fringe at an angle

a

sin

D

22.1sin

9.5 Resolution

Due to this interference, a point source of light will be viewed as a circular image

Rayleigh’s Criterion: If the first dark fringe of one circular diffraction patter passes through the center of a second pattern, the two sources responsible for the patterns will appear to be a single source.

Applet

9.5 Resolution

Examples: pixels on TV’s and Computer

9.6 Diffraction Gratings

9.6 Diffraction Grating

Diffraction Grating – a system with a large number of slits

3D glasses

Produced most often by

taking pictures of slits

and putting them on a slide

Produces sharp, widely spaced fringes

9.6 Diffraction Grating

Patterns are caused by multiple waves interfering

This example shows a

maxima produced by

10 slits

There would be other areas of constructive interference

9.6 Diffraction Grating

Because different colors of light have different wavelengths, and diffract differently

We can separate colors

Equations are the same

as double slits

Constructive

Destructive md sin

)(sin 21 md