April 18 Physics 54 LectureProfessor Henry Greenside

Key Points from Previous Lecture

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Today’s TopicsChapter 35: “The Wave Nature of Light: Interference”• Huygens's principle• Young’s double-slit experiment which demonstrates that light is

a wave and that the wave is transverse.• Interference in thin films.• How f and for light change in a new medium.• Applications of thin films: non-reflective coatings, measuring

thickness of small wires.

Chapter 35: Interference of Waves

Please review Sections 15-6 through 15-8 of Volume I Giancoli.

Huygens’ Principle Explains Diffraction

Waves in isotropic mediumr = v t

Huygens' Principle

Huygens’ Principle can be used to:1. Explain the law of reflection.2. Explain refraction of waves at an interface.3. Explain diffraction of waves around a corner.

As you might expect, the heuristic idea of Huygens can be fully justified through various derivations associated with the Maxwell equations.

Applying Huygens’ Principle to Two Slits

To Understand Interference, Understand Superposition of Waves

Nota bene: it is the relative phase that counts here.

Interference of Coherent Waves In the Double-Slit Experiment

Waves at slits have to be coherent for interference to occur! Two different light bulbs in front of each slit will not give interference pattern.

Light Particles or Waves? Young’s Double-Slit Experiment!

particles waves

Interference Preconditions

1. Light must be monochromatic, i.e., involve just a single frequency and single wavelength.

2. Light sources must be coherent, the relative phase is always the same.

3. Light sources must have the same amplitudes.

If these conditions do not hold, one still gets constructive and destructive interference but the interference pattern can change with time or not be complete (destructive interference leads to a decrease in amplitude but not to zero amplitude).

Conditions for Constructive and Destructive Interference of Two Coherent Waves

Constructive and destructive interference (of two coherent monochromatic equal-amplitude waves emitted from sources S1 and S2) occur at point P in space when difference l in distances l1 and l2 of P to sources S1 and S2 respectively satisfies:

l = l1 – l2 = mconstructive l = l1 – l2 = (m-1/2)destructive

where is wavelength of waves and m is any integer.

For points for which these conditions don’t hold, you get intermediate results, partial constructive or partial destructive interference.

Two-slit Experiment: Light, Electrons Are Neither Waves Nor Particles

Akira Tonomurahttp://www.pnas.org/content/vol0/issue2005/images/data/0504720102/DC1/04720Movie1.mpg

sin( ) , 0, 1, 2, constructive

sin( ) ( 1/ 2) , 0, 1, 2, destructive

d m m

d m m

m is the order of an interference fringe

Whiteboard: Derivation of Criteria for Dark and Light Bands in Young Expt

Proving that Light is a Transverse Wave: Two-slit Experiment Plus Two Linear Polarizers

Consider Young’s double-slit interference experiment but now put a linear polarizing filter in front of each slit. Then as transmission axis of one polarizer is slowly rotated about its axis while transmission axis of other polarizer is kept fixed, one finds experimentally that the bright regions become dimmer, the dark regions become brighter, until the fringes disappear into a smoothly varying region that is brightest behind the slits and gets dimmer off to the sides.

This proves that light is transverse, i.e., can be polarized. What is happening is that the light that is polarized vertically can not interfere with light that is polarized horizontally since the E vectors can’t line up to cancel to zero.

PRS Question:

If distance d between slits is decreased, then the angles corresponding to the bright fringes will

1. all become smaller.2. all become larger3. some will become larger, some smaller.4. remain unchanged but the fringes will all

become dimmer.5. remain unchanged but the fringes will all

become brighter.

sin( ) , 0, 1, 2,

sin( ) ( 1/ 2) , 0, 1, 2,

d m m

d m m

Worked Example at WhiteboardA screen contains two slits distance d = 0.100 mm apart and is length L = 1.20 m from a viewing screen. Monochromatic light of wavelength = 500 nm falls on the slits from a distant source. About how far apart x will the bright interference fringes be on the screen?

Answer: About 6 mm. Use small angle approximations to simplify algebra, avoid using sin and tan functions.

PRS Question:

If wavelength of monochromatic light impinging on two-

slits experiment increases, then bright fringes

1. all become closer.2. all spread further apart.3. some become closer, some further apart.

sin( ) , 0, 1, 2,

sin( ) ( 1/ 2) , 0, 1, 2,

d m m

d m m

Wavelengths of LightFrom Double-Slit Interference

White light (not monochromatic light) impinges on slits

two slits are distance d = 0.50 mm apart screen is length L = 2.5 m away.

Violet and red m=1 fringes are 2.0 and 3.5 mm from central fringe. What are the wavelengths of the violet

and red light?

Answers: 400 nm and 700 nm respectively.

Application of Interference: Thin Films

Thin Film Interference Versus Two-Slit Interference

Two-slit interference involves light from two separate sources S1 and S2.

Thin-film interference involves light from single source that interferes with itself by reflecting off different interfaces between two different media, which then produces different sources of light that interfere with each other.

Thin-film interference of great importance in optics, precision machinery, and biology.

Whiteboard Discussion of Thin-Film Interference

1. Review how wave on rope reflects at interface between less dense and more dense sections of rope. Can get inversion of wave which corresponds to a shift of wave by ½ wavelength.

2. Same occurs for light: when light passes from less dense medium (vacuum or air) to more dense medium, reflected light will be shifted by ½ wavelength.

3. Need to understand index of refraction n that characterizes how light behaves in a medium. Speed slows down to c/n < c, wavelength decreases to n=/n, and frequency is unchanged as light passes into new medium.

4. Criterion for bright reflection (constructive interference) is then the usual one: difference in distances traveled must be integer multiple of wavelength n in the medium. Get no reflection (destructive interference) if difference in distances traveled is half-integer multiple of wavelength n.

Section 15-7 on Waves: Reflection

When wave hits discontinuity, part is reflected and part is transmitted. If new medium is “stiffer”, wave inverts (/2 shift in wavelength, /2 shift in phase)

Light Reflection From Denser Media:/2 Shift In Position of Wave

Inversion with /2 shift in position No inversion

Measure Small Thicknesses with Air Wedge

2 1, 1, 2,3, bright fringe

22 1 1

, 0,1,2,3, dark fringe2 2

dm m

dm m

Interference From Thin FilmCount total number of half-wavelength shifts:1. One at air-oil interface.2. One at oil-water interface.

Two inversions cancel each other so get constructive interference when difference in length of paths is integer multiple of wavelength n in the oil!

Next get total extra length of path 2t that light traverses.

edestructiv 2/12

econtructiv 2

n

n

mt

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What is minimum thickness t of soap bubble if it appears green (=540 nm) at the point of front surface closest to viewer? Assume nsoap = 1.35.

Solution: Total number of inversions (phase shifts of /2) is 1 so criterion for constructive interference:

Get smallest t for m = 0, gives answer t=100 nm.

Worked Example 35-7:Thickness of Soap Bubble Skin

nmt )2/1(2

PRS: How Many Phase Shifts In Example 35-8?

nair = 1, nMgF2 = 1.38nglass = 1.50

? with )2/1( 2 6.

? with )2/1(2 .5

? with )2/1(2 4.

? with 2 3.

? with 2 .2

? with 2 1.

:condition is ce,interferen edestructivFor

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