Physical Optics

The wave nature of light

Interference

Diffraction

Polarization

Huygens’ Principle• Every point on a propagating

wavefront serves as the source of spherical wavelets, such that the wavelets at sometime later is the envelope of these wavelets.

• If a propagating wave has a particular frequency and speed, the secondary wavelets have that same frequency and speed.

“Isotropic”

Diffraction

• Diffraction – Bending of light into the shadow region• Grimaldi - 17th Century observation of diffraction• Diffraction vs. Refraction?

a

a

Explanation of Snell’s Law

11

BD v tsin

AD AD

22

AC v tsin

AD AD

1 1 1

2 2 2

sin v t c / n

sin v t c / n

1 1 2 2n sin n sin

Superposition of waves

Constructive Interference

Destructive Interference

Conditions for Interference

• To observe interference in light waves, the following two conditions must be met:1) The sources must be coherent

• They must maintain a constant phase with respect to each other

2) The sources should be monochromatic• Monochromatic means they have a single

wavelength

Young’s Experiment

Young’s Experiment

Young’s Experiment

0m

1m

1m

2m

2m

maxima1S

2S

1r

2r

1S

2S

d

1 2 r r

1 2r r m

Maxima occur when:

y

s

s a

d

sind

dsin m

Minima occur when:

1 2

1r r m

2

1dsin m

2

What the pattern looks like

Intensity Distribution, Electric Fields

• The magnitude of each wave at point P can be found– E1 = Eo sin ωt

– E2 = Eo sin (ωt + φ)

– Both waves have the same amplitude, Eo

2 2

sin π π

φ δ d θλ λ

Intensity Distribution, Resultant Field

• The magnitude of the resultant electric field comes from the superposition principle– EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)]

• This can also be expressed as

– EP has the same frequency as the light at the slits

– The magnitude of the field is multiplied by the factor 2 cos (φ / 2)

2 cos sin2 2P o

φ φE E ωt

A B A Bsin A sin B 2sin cos

2 2

Intensity Distribution, Equation• The expression for the intensity comes from the

fact that the intensity of a wave is proportional to the square of the resultant electric field magnitude at that point

• The intensity therefore is

2 2max max

sin cos cosI I I

πd θ πdy

λ λL

Resulting Interference Pattern

• The light from the two slits forms a visible pattern on a screen

• The pattern consists of a series of bright and dark parallel bands called fringes

• Constructive interference occurs where a bright fringe occurs

• Destructive interference results in a dark fringe

Example

A He-Ne Laser has a wavelength of 633 nm. Two slits are placed immediately in front of the laser and an interference pattern is observed on a screen 10 m away. If the first bright band is observed 1 cm from the central bright fringe, how far apart are the two slits?

At what angle is the fourth dark band found?

Thin Films

ABC nd m

Constructive Interference (maxima)

on n

Destructive Interference (minima)

ABC n

1d m

2

Chapter 35 - Problem 45

Stealth aircraft are designed to not reflect radar whose wavelength is 2 cm, by using an anti-reflecting coating. Ignoring any change in wavelength in the coating, estimate its thickness.

Phase shift on reflection

External Reflection Internal Reflection

Now, If one reflection is internal and one reflection is external half wavelength path differences will result in constructive interference

Phase Changes Due To Reflection

• An electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was traveling– Analogous to a pulse on

a string reflected from a rigid support

Lloyd’s Mirror

• An arrangement for producing an interference pattern with a single light source

• Waves reach point P either by a direct path or by reflection

• The reflected ray can be treated as a ray from the source S’ behind the mirror

Lloyd’s Mirror

Phase shift on reflection

2sin 2ma m

1sin 2 1

2 2m

m m

ya m m a a

s

S

1S

a

y

s as

1 2r r

Interference in Thin Films Again

• Assume the light rays are traveling in air nearly normal to the two surfaces of the film

• Ray 1 undergoes a phase change of 180° with respect to the incident ray

• Ray 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave

• For constructive interference

=2t = (m + ½)λn (m = 0, 1, 2 …)

• This takes into account both the difference in optical path length for the two rays and the 180° phase change

• For destructive interference

=2t = mλn (m = 0, 1, 2 …)

Newton’s Rings

m

12d m

2

Maxima - bright Minima - dark

m2d m

o

n

Chapter 37 – Problem 62

Show that the radius of the mth dark Newton’s ring as viewed from directly above is given by:

om

Rmx

n

Where R is the radius of curvature of the curved glass surface and is the wavelength of the light used. Assume that the thickness of the air gap is much less than R at all points and that x<<R

Newton’s Rings

x

d

RR d

22 2R x R d

22 2 2 2 2 2x R R d R R d Rd

2 2 2 2x d Rd Rd

n2d m

Maxima

2omx

2 m2R n

om

R 1x m

n 2

Minima

om

Rmx

n

Newtons rings are a special case of Fizeau fringes. They are useful for testing surface accuracy of a lens.

o

n

Wedges – Fringes of Equal Thickness

x

d

t x 12t m

2

Minima (destructive)

m

1x m

2 2

mx2

Fringes of this type are alsoknown as Fizeau fringes

o2t m mn

Maxima (constructive)

Example: thin film of air

Michelson Interferometer• A lens can be used to form

fringes of equal inclination (rings)

• Tilting the mirrors can cause fringes of equal thickness.

• Accurate length measurements are accomplished by fringe counting as one of the mirrors is moved.

Det

ecto

r

CompensatorPlate

d

2 d m

2

md

42. Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced 0.382 mm, causing the interferometer pattern to reproduce itself 1 700 times. Determine the wavelength of the light. What color is it?

74.49 10 m 449 nm

5. Young’s double-slit experiment is performed with 589-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits.

17. In Figure 37.5, let L = 120 cm and d = 0.250 cm. The slits are illuminated with coherent 600-nm light. Calculate the distance y above the central maximum for which the average intensity on the screen is 75.0% of the maximum.

7. Two narrow, parallel slits separated by 0.250 mm are illuminated by green light (λ = 546.1 nm). The interference pattern is observed on a screen 1.20 m away from the plane of the slits. Calculate the distance (a) from the central maximum to the first bright region on either side of the central maximum and (b) between the first and second dark bands.

32. A thin film of oil (n = 1.25) is located on a smooth wet pavement. When viewed perpendicular to the pavement, the film reflects most strongly red light at 640 nm and reflects no blue light at 512 nm. How thick is the oil film?

33. A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of 3.00 cm and the index of refraction of the polymer is n = 1.50, how thick would you make the coating?

37. A beam of 580-nm light passes through two closely spaced glass plates, as shown in Figure P37.37. For what minimum nonzero value of the plate separation d is the transmitted light bright?

Mach Zehnder Interferometer

• No compensating plate needed• Test cell easily inserted in one leg. • No factor of two as in the Michelson interferometer.• Difficult to align

Detector

Sagnac Interferometer

Detector

R

Ring laser gyro. The rotation effectively shortens the path

taken by one direction over the other.

4 2

2

cw

Rt

vc

4 2

2

ccw

Rt

vc

2

R

v

s rv r

2

R

2

v

cw ccwt t t

8 8

2 2

R Rt

c v c v

8 8

2 2

R Rt

c R c R

8 1 1 81 1

2 2 2 21 12 2

R R R Rt

R Rc c c cc c

2

2

8 2 8

2 2

R R Rt

cc c

28Rc t

c

Phase Angle

1E

2E2

1

1 2

1 2 01 02i t i tE E E E e E e

E

0

i tE E e

01 1 02 2

01 1 02 2

sin sintan

cos cos

E E

E E

2 2

01 02 01 02 2 12 cosE E E E E

01 1sinE

02 2sinE

01 1cosE 02 2cosE