Wave Interference and Diffraction(Physical Optics)

( )

2Phase difference 2 2

2 wave amplitude is doubled (constructive interference)2 1 wave amplitude zero (destructive interference)

k d d

mm

πφλ

φ πφ π

= × = ×

= →

= + →

Ε − Ε = 0

φ = 0, 2π, 4π

φ = π/2

φ = π, 3π, 5π

2Ε

Ε Ε

ConstructiveInterference

Destructive interference

In between

Interference due to two coherent sources

Two sources separatedby 3 wavelengths

Path diff. = 0

Path diff. = λ

Path diff. = -λ

Young’s Double Slit ExperimentDemonstrated the wave nature of light

(peak to peak distance) Dydλ

∆ =

Effects of Slit Opening a

-20-15

-10-5

05

1015

20

0.5

1.0

-20-15

-10-5

05

1015

20

0.5

1.0

Case 10Combination of double slit interference and single slit diffraction.

d a=

4 3

Since tan , we have

where is the separation between intensity peaks.

Example: If 0.2 mm, 1 m, and 3 mm are observed, what is ? 2 10 3 10From = , we find 6 10

1

dy Dy D D m yD d

y

d D yd yD

λθ θ λ

λ

λ λ− −

−

= = = → ∆ =

∆

= = ∆ =

∆ × × ×= = × 7 m = 600 nm

Example: If one slit is covered with a fild with 1.5, what would happen?The entire interference pattern is shifted by

1

because the film causes an additional phase shift of '

n

ny Dtd

nπφλ

=

−∆ =

2= ( )

( )

1 .

2 2Th phase shift due to path difference is = sin . From '

we find 1 .

t

yd dD

Dty nd

π πφ θ φ φλ λ

−

= =

= −

Thin Film Interference

The phase difference between the waves reflected at both surfaces is2 22 2

where the additional is due to phase change at the air-film boundary,

0.

Cons

filmfilm air

film air

film air

t n t

Z ZZ Z

π πφ π πλ λ

π

= × + = × +

−Γ = <

+

1tructive interference when 2 or 22film airm n t mφ π λ = = −

Newton’s Rings

( )2

2 2 2

Newton's rings are caused by interference in thin air film snadwitched in betweena convex lens and plane glass. The air film thickness can be found from

212 is the cond2 air

rR R rR

m

δ

δ δ

δ λ

+ = + → =

= −

ition for constructive interference. Then bright rings appear at radii

12 airr m Rλ = −

Multiple slits

6 coherent sourcesVectorial sum of 6 fields

( )

0

0

0

22 sin where sin 2

2 sin 62

sin 3Eliminate

sin2

max. 6 when 2 . sin , sin in the figure.

2

50 very sharp peak

Details

Total

Total

Total

E R d

E R

R E E

E E mdx d x

N

δ πδ θλ

δ

δδ

δ πθ δ π θ π

λ λ

= =

=

→ =

=

= = =

= →

-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

0.5

1.0

x

y

-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5

0.5

1.0

x

y

-0.10 -0.08 -0.06 -0.04 -0.02 0.00 0.02 0.04 0.06 0.08 0.10

0.5

1.0

x

y

Grating SpectrometerA grating spectrometer has 50,000 grroves over a width of 9 cm. Find the angular location of intensity peaks for 400 nm and 700 nm light.

9 cmThe spacing is =1.8 m. The intensity peaks occurs at50000

d µ=

7

6

7

Re 6

sin =0, 1, 2,

4 10For =400 nm, sin 0, 0.22, 0.44, 0.66, 0.881.8 10

0, 12.7 , 26.1 , 41.4 , 61.6

7 10For =700 nm, sin 0, 0.31.8 10

Blue

Blue

d

d

n nd

n

θλ

λλ θ

θ

λ θ

−

−

° ° ° °

−

−

± ± ⋅ ⋅ ⋅

×= ± = ± → ± ± ± ±

×= ± ± ± ±

×= ± → ±

×

Re

9, 0.78.

0, 23, 51.6dθ

±

= ± ±

Diffraction by Slit

0

Increase the wave sources to a large number in

sin22' with sin

sin2

sin sinsin' , = sin

sin

Resolution limit of telescope 1.22

sDiffraction pattern

NaE EN

aaE NE Ea

D

δπδ θ

δ λ

π θα πλ α θπ α λθ

λλθ

= = = =

∆

2in is shown.αα

-10 -8 -6 -4 -2 0 2 4 6 8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

x

y

2

2

22

0

sinInterference cos

sin sinDiffraction ,

sin sinCombine these cos

d

a

dI I

π θλ

α π θαα λ

π θ αλ α

=

=

d = 3a

d = 4a

d = 10a

7

Resolving Power of Telescope

If the diameter is 2 m (Hubble), the resolution limit for green light is550 nm1.22 1.22 3.4 10 rad

2 mThis is the minimum angular resolution.The Hubble telescope is

Dλθ −∆ = = = ×

free from atmospheric disturbance which is themain advantage.

Example: If a telescope carried by a sattelite 200 km above is to resolve an object 1 macross, what is the diameter? Assume =550 nm.

Augul

λ

-65

9

-6

1ar spread is =5 10 rad. Then 2 10

550 101.22 =0.13 m = 13 cm.5 10

This is for ideal case without air disturbance.

D

θ

−

∆ = ××

×=

×

8

Example: Ruby laser beam ( =690 nm) shot to the moon has a diameter of 1 cm on the earth. Find the spot size on the moon. The eartth-moon distance is 3.8 10 m.

690The angular spread is 1.22 1.22D

λ

λθ

×

×∆ = =

95

8 5 4

10 8.4 10 rad. Then0.01

the diameter on the moon is 3.8 10 8.4 10 3.2 10 m. D

−−

−

= ×

= × × × = ×

Knife edgediffraction

Scattering of E&M waves

Electromagnetic waves are scattered by electrons in matter.Electrons are accelerated by the electric field in the incident wave and re-radiate.

Rayleigh scatterring: Scattering by molecules much small4 4

er than the wavelength. Radiation power proportional to or 1/ . This explains why the sky is blue at daytime. At sunset, the sun appears red because blue light is scattered away fromthe light path

ω λ

.

Mie scattering: Scattering by larger molecules (e.g., water particles in fog). Independent of or . Clouds appear white. ω λ