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Copyright © 2012 Pearson Education Inc.

PowerPoint® Lectures for

University Physics, Thirteenth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson

Chapter 35

Interference


Goals for Chapter 35

• To consider interference of waves in space

• To analyze two-source interference of

light

• To calculate the intensity of interference

patterns

• To understand interference in thin films

• To use interference to measure extremely

small distances


Introduction

• Why do soap bubbles show vibrant color patterns, even though soapy water is colorless?

• What causes the multicolored reflections from DVDs?

• We will now look at optical effects, such as interference, that depend on the wave nature of light.


Wave fronts from a disturbance

• Figure 35.1 at the right

shows a “snapshot” of

sinusoidal waves spreading

out in all directions from a

source.

• Superposition principle:

When two or more waves

overlap, the resultant

displacement at any instant

is the sum of the

displacements of each of

the individual waves.


Constructive and destructive interference

• Figure 35.2 at the right

shows two coherent wave

sources.

• Constructive interference

occurs when the path

difference is an integral

number of wavelengths.

• Destructive interference

occurs when the path

difference is a half-integral

number of wavelengths.


Two sources S1 and S2 oscillating in phase emit sinusoidal waves.

Point P is 7.3 wavelengths from source S1 and 4.3 wavelengths

from source S2. As a result, at point P there is

Q35.1

A. constructive interference.

B. destructive interference.

C. neither constructive nor destructive interference.

D. not enough information given to decide





A35.1









Q35.2









A35.2






Two-source interference of light

• Figure 35.5 below shows Young’s double-slit experiment with

geometric analysis.


Interference from two slits

• Follow the text discussion of

two-slit interference.

• Figure 35.6 at the right is a

photograph of the interference

fringes from a two-slit

experiment.


Two-slit interference

• Follow Example 35.1 using Figure 35.7 below.


In Young’s experiment, coherent light passing through

two slits (S1 and S2) produces a pattern of dark and

bright areas on a distant screen. If the wavelength of the

light is increased, how does the pattern change?

Q35.3

A. The bright areas move closer together.

B. The bright areas move farther apart.

C. The spacing between bright areas remains the

same, but the color changes.

D. any of the above, depending on circumstances

E. none of the above




bright areas on a distant screen. If the wavelength of the

light is increased, how does the pattern change?

A35.3

A. The bright areas move closer together.

B. The bright areas move farther apart.

C. The spacing between bright areas remains the

same, but the color changes.

D. any of the above, depending on circumstances

E. none of the above




bright areas on a distant screen.

What is the difference between the distance from S1 to

the m = +3 bright area and the distance from S2 to the

m = +3 bright area?

Q35.4

A. three wavelengths

B. three half-wavelengths

C. three quarter-wavelengths





bright areas on a distant screen.

What is the difference between the distance from S1 to

the m = +3 bright area and the distance from S2 to the

m = +3 bright area?

A35.4

A. three wavelengths

B. three half-wavelengths

C. three quarter-wavelengths



Broadcast pattern of a radio station

• Follow Example 35.2 using Figure 35.8 below.


Q35.5

A. 4.5l

B. 8l

C. 9l

D. both A. and B.

E. all of A., B., and C.

At what distances from point B will the receiver detect

an intensity maximum?

6l

A

B C

Two radio antennas radiating

in phase are located at points

A and B, which are 6

wavelengths apart. A radio

receiver is moved along a

line from point B to point C.


A35.5

At what distances from point B will the receiver detect

an intensity maximum?

6l

A

B C

Two radio antennas radiating

in phase are located at points

A and B, which are 6

wavelengths apart. A radio

receiver is moved along a

line from point B to point C.

A. 4.5l

B. 8l

C. 9l

D. both A. and B.

E. all of A., B., and C.


Intensity in interference patterns

• Follow the text analysis of the intensity in interference patterns using Figure 35.9 (right). Figure 35.10 (below) shows the intensity distribution for two identical slits.

• Follow Example 35.3.


Interference in thin films

• Figure 35.11 (right) shows why thin-film interference occurs, with an illustration.

• Figure 35.12 (below) shows interference of an air wedge.


Phase shifts during reflection

• Follow the text analysis of thin-film interference and phase shifts during reflection. Use Figure 35.13 below.


Wedge between two plates

• Read Problem-Solving Strategy 35.1.

• Follow Example 35.4, having air between the plates. Use

Figure 35.15 below.

• Follow Example 35.5, having water between the plates.

• Follow Example 35.6, another variation on the plates.


A. l/2.

B. 3l/4.

C. l.

D. either A. or C.

E. any of A., B., or C.

Q35.6

An air wedge separates two

glass plates as shown. Light of

wavelength l strikes the upper

plate at normal incidence. At a

point where the air wedge has

thickness t, you will see a

bright fringe if t equals


A35.6

An air wedge separates two

glass plates as shown. Light of

wavelength l strikes the upper

plate at normal incidence. At a

point where the air wedge has

thickness t, you will see a

bright fringe if t equals

A. l/2.

B. 3l/4.

C. l.

D. either A. or C.

E. any of A., B., or C.


Newton’s rings

• Figure 35.16 below illustrates the interference rings (called Newton’s rings) resulting from an air film under a lens.


Using interference fringes to test a lens

• The lens to be tested is placed on top of the master lens. If the two surfaces do not match, Newton’s rings will appear, as in Figure 35.17 at the right.


Nonreflective coatings

• The purpose of the nonreflecting film is to cancel the reflected light. (See Figure 35.18 at the right.)

• Follow Example 35.7.


Michelson interferometer

• The Michelson interferometer is used to make precise measurements of wavelengths and very small distances.

• Follow the text analysis, using Figure 35.19 below.


Conditions for Light Interference

Sources must be coherent (constant phase)

Wavelengths must be identical


Double-Slit Interference

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=RU1KoX8w_MeNGM&tbnid=d2C_-PcSuzuFsM:&ved=0CAUQjRw&url=http://www.wikipremed.com/01physicscards.php?card=876&ei=els1UbSRH6qGyQHu-4CgAQ&bvm=bv.43148975,d.aWc&psig=AFQjCNF5UUfp6BG7lELFt9bc075QNogdWA&ust=1362537672262491

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=g9ul5QD3ChUt3M&tbnid=T2J3HtcSD0bhKM:&ved=0CAUQjRw&url=http://www.visualphotos.com/image/1x8465433/double-slit_diffraction&ei=5Vs1UZLEOq6GyQHQk4CgAQ&bvm=bv.43148975,d.aWc&psig=AFQjCNF5UUfp6BG7lELFt9bc075QNogdWA&ust=1362537672262491


Thin Films

Assume air is on either

side of the thin film,

and the film is plastic

Is there any difference in

phase between

reflected rays 1 & 2?

Which reflected ray

travels farther?

1 2

MUST consider both questions to determine if

rays constructively or destructively interfere!!!


Thin Films

If rays 1 & 2 are out of phase, then the path difference must be λn/2

If the thickness of the film is “t” then ray 2 also travels an extra distance of 2t before the waves recombine

What if 2t = λn/2? Meaning, the extra distance traveled is exactly equal to the original path difference due to reflection of ray 1??

Then rays 1 & 2 would recombine IN PHASE!!!!!


Thin Films

Condition for constructive interference of two

light rays interacting with a thin film:

2t = (m + ½)λn

Condition for destructive interference of two

light rays interacting with a thin film:

2t = mλn


Wavelength of Light in a Medium

Index of refraction -> n = c/v

Using v = fλ, can write λ1n1 = λ2n2

Assuming that n1 = 1 (a vacuum) can write…

n = λo/λn

(Where λo is the wavelength of the light in a

vacuum and λn is the wavelength of the light

in the thin film)


Thin Film Equations

SO, you can rewrite the constructive & destructive equations in terms of n and

the wavelength of the light in a vacuum

Const. -> 2nt = (m + ½)λo

Dest. -> 2nt = mλo

WARNING!! These equations are ONLY valid for ONE PHASE REVERSAL (when

the n of the film is typically greater than the n of the material on either side of it)

If n of film is in between the n’s of the sandwiching materials, then simply

REVERSE THE EQUATIONS (const. vs. dest.)


Pretty Colors

If you were blowing bubbles outside, what kind of light would be shining on the bubbles?

Is the wavelength and/or speed of different colors of light the same in a given thin film?

RED has longer wavelength, faster speed, so will emerge with less phase change compared to VIOLET

Film thickness determines which colors are cancelled are which are reinforced in a certain region


Example Question

A thin soap film is formed by dipping a plastic

rectangular wand into bubble solution (n =

1.4). When viewed in daylight, one part of the

film reflects blue light (475 nm).

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