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Light and Optics - 9Propagation of light

Electromagnetic waves (light) in vacuum and matterReflection and refraction of lightPolarisation of lightHuygens’ principle

Geometric opticsPlane and curved mirrorsThin lenses

InterferenceDouble slits

DiffractionSingle slitDouble slitsThin film interference

Luke Wilson (Luke.wilson@... Room E17)

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• Single slit intensity distribution

• Double slit intensity distribution

• Thin film interference

http://science.sbcc.edu/physics/flash/optics/doubleslitphasors.html

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Position of dark fringes in single-slit diffraction

amλθ =sin

If, like the 2-slit treatment we assume small angles, sinθ ≈ tan θ =ymin/R, then

aRmy λ

=min

Positions of intensity MINIMA of diffraction pattern on screen, measured from central position.

Very similar to expression derived for 2-slit experiment:

dnRymλ

= But remember, in this case ym are positions of MAXIMAin interference pattern

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Single-slit diffraction: intensity distribution

2

0 2/)2/sin(⎥⎦

⎤⎢⎣

⎡=

ββII

λθπβ sin2 a

=

2

0 /)(sin)/)(sinsin(⎥⎦

⎤⎢⎣

⎡=

λθπλθπ

aaII

Key equations:

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⎟⎠⎞

⎜⎝⎛′=

λπRdyIITOT

20 cos

“real” intensity distribution for double slits

The distribution derived earlier is for the idealised case of infinitely narrowslits:

’

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“real” intensity distribution for double slits

If we allow the slits (spacing d) to have finite width a, then the TOTAL intensity distribution is given by the 2-slit intensity distribution

⎟⎠⎞

⎜⎝⎛=

2cos4 2

0φII

‘modulated’ by the intensity distribution for a single slit of width a:

2

/)(sin)/)(sinsin( ⎥⎦

⎤⎢⎣

⎡λθπλθπα

aaI

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“real” intensity distribution for double slits

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0 2/)2/sin(

2cos4 ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

ββφII

λθπβ sin2 a

=λ

θπφ sin2 d=

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“real” intensity distribution for double slits

“ideal” pattern

Single slit pattern

“real” pattern

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Missing Orders

aRmy λ

=min dRny λ

=max

Positions of zero intensityIn diffraction pattern

Positions of maximum intensityin interference pattern

amdn

dRn

aRm

=⇒= λλ

So when:

Every nth order is missing in the 2-slit intensity distribution…….

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Example•An interference pattern is produced by 2 identical parallel slits of width aand separation (between centres) d = 3a. Which interference maxima will be missing in the observed pattern?

maman

amdn

33==

=

So, every 3rd bright fringe missing

http://lectureonline.cl.msu.edu/~mmp/kap27/Gary-TwoSlit/app.htm

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Thin film interference

Light waves reflected from front and back surfaces of thin films can constructively or destructively interfere

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Thin film interference – phase shift upon reflection

na > nb no phase shift na < nb π phase shift

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No relative phase shift:

Half cycle relative phase shift:

λ

λ

⎟⎠⎞

⎜⎝⎛ +=

=

212

2

mt

mt

(m = 0, 1, 2,…)

λ

λ

mt

mt

=

⎟⎠⎞

⎜⎝⎛ +=

2212

CONSTRUCTIVE Reflection

DESTRUCTIVE Reflection

CONSTRUCTIVE Reflection

DESTRUCTIVE Reflection

Thin film interference

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ExampleA fine metal foil separates one end of two pieces of optically flat glass. Write an expression for the phase shift for destructive interference. When light of wavelength 600 nm is normally incident, 18 dark fringes are observed. How thick is the foil?

( )

,...2,1,0 ,2

1222. of multiple odd bemust thisceinterferen edestructivFor

. surface, bottomat reflection theand ,22differencelength path fromwavesreflectedfor shift Phase

==

+=+⎟⎠⎞

⎜⎝⎛=

⎟⎠⎞

⎜⎝⎛=

mmt

mt

t

λ

πππλ

φ

π

ππλ

φ

m = 0 where glass touch, m+1 is number of fringe, so

( )( ) m1.5nm6001721 μ==t

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Thin film interference – AR coatings

Film is about ¼ wavelength thick

Light undergoes a phase shift at both reflecting surfaces

The two reflected waves emerge from the film about ½ cycle out of phase

nglass > nfilm > nair

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Summary

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0 2/)2/sin(

2cos ⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛=

ββφII

amdn

dRn

aRm

=⇒= λλ

Half cycle relative phase shift:

(m = 0, 1, 2,…)

λ

λ

mt

mt

=

⎟⎠⎞

⎜⎝⎛ +=

2212 CONSTRUCTIVE Reflection

DESTRUCTIVE Reflection

Double slits with finite slit width

Intensity distribution Missing orders

Thin film interference