PART 3Wave Optics

1. Huygen’s Principle2. Young’s Two Slits Experiment3. Air Wedge4. Interference of Thin Film5. Diffraction by a Single Slit6. Diffraction Grating7. Spectrometer and Spectroscopy8. Polarization

Christian Huygens

• 1629 – 1695• Best known for

contributions to fields of optics and dynamics

Huygen’s Principle

Huygen’s Principle, Cont.

• All points on a given wave front are taken as point sources for the production of spherical secondary waves, called wavelets, which propagate in the forward direction with speeds characteristic of waves in that medium.

Wave Optics

• The wave nature of light is needed to explain various phenomena.– Interference– Diffraction– Polarization

Conditions for Interference

• For sustained interference between two sources of light to be observed, there are two conditions which must be met.– The sources must be coherent.

• The waves they emit must maintain a constant phase with respect to each other.

– The waves must have identical wavelengths.

Resulting Interference Pattern

• The light from the two slits form 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 appears.

• Destructive interference results in a dark fringe.

Fringe Pattern

• The bright areas represent constructive interference.

• The dark areas represent destructive interference.

Young’s Double Experiment

Interference Patterns

• Constructive interference occurs at the center point.

• The two waves travel the same distance.– Therefore, they arrive in

phase.

Interference Patterns, 3

• The upper wave travels one-half of a wavelength farther than the lower wave.

• The trough of the bottom wave overlaps the crest of the upper wave.

• This is destructive interference.– A dark fringe occurs.

Geometry of Young’s Double Slit Experiment

x

Interference Equations, 2

• For a bright fringe• d sin θbright = m λ

– m = 0, 1, 2, … – m is called the order number.

• When m = 0, it is the zeroth order maximum.• When m = 1, it is called the first order maximum.

Interference Equations, 3

• When destructive interference occurs, a dark fringe is observed.

• This needs a path difference of an odd half wavelength.

• d sin θdark = (m + ½) λm = 0, 1, 2, …

Interference Equations, Final

• For bright fringes

• For dark fringes

0,1,2...m ,mdL

xbright

0,1,2...m ),21

m(dL

xdark

Interference Equations, Final

• For fringe separation

dL

yseparation

Interference Equations, Final

Example 1

A screen is separated from a double-slit source by 1.2m. The distance between the two slits is 0.030 mm. The second order bright fringe is measure to be 4.50 cm from the centerline. Determine:a) the wavelength of the lightb) the fringe separation

Example 2

In a young’s double slits experiment, an interference pattern is formed on a screen 1.0 m away from the double slits. The double slits are separated by 0.25 mm and the wavelength of light used is 550 nm. Determine the distance from the central bright fringe the distance ofa) the 5th bright fringeb) the 3rd dark fringe

Air Wedge

Air Wedge

Air Wedge

nyL

HL

21

y

Separation between bright fringes

Angular size of air wedge

y21

LH

tan

n is number of bright fringe

Example 3

Air wedge is formed by placing a piece of thin paper at the edges of a pair of glass plates. Light of wavelength 600 nm is incident normally onto the plates. Fringes are observed with fringe separation of 0.25 mm. The length of the air wedge is 5.0 cm.a)Determine of the thickness of the piece of paper.b)Estimate the number of bright fringes that are formed on the plate.

Example 4

Figure below shows how a piece of material of thickness 5μm forms an air wedge between a pair of a glass plates A and B. Light of wavelength 500 nm is incident normally onto the plates. Estimate the number of bright fringes produced.

Interference in Thin Films

• Interference effects are commonly observed in thin films.– Examples are soap bubbles and oil on water

• Varied colours observed when incoherent light is incident on the water.

• The interference is due to the interaction of the waves reflected from both surfaces of the film.

Interference in Thin Films

1 Phase Reversal 0 or 2 Phase Reversal

Problem Solving with Thin Films

Equationm = 0, 1, 2, …

1 phase reversal

0 or 2 phase reversals

2nt = (m + ½) constructive destructive

2nt = m destructive constructive

Interference in Thin Films

• Identify interference in the film(constructive/destructive)

• Starts the number of order (m) with m=0 (otherwise)• Find the refraction index of thin film.• Determine the number of phase reversals. (0,1 or 2)• Using table, use correct formula and column

Example 5

a) Calculate the minimum thickness of a soap-bubble film (n=1.33) that will result in constructive interference in the reflected light if the film is illuminated by light with wavelength 602 nm.

b) In (a), if the soap-bubble is on top of a glass slide with n=1.50, find the minimum thickness (nonzero thickness) for constructive interference?

Example 6

Nearly normal incident light of wavelength 600 nm falls onto a thin uniform transparent film in air. The refractive index of the film is 1.34. A very bright ray reflected by the surface of the film is observed. Determine the minimum thickness of the film.

Refer the figure, find the minimum thickness of the film that will produce at least reflection at a wavelength of 552 nm.

Hint: minimum thickness is refer to destructive interference

Example 7

Diffraction• This spreading out of light

from its initial line of travel is called diffraction.– In general, diffraction occurs

when waves pass through small openings, around obstacles or by sharp edges.

Fraunhofer Diffraction• Fraunhofer Diffraction

occurs when the rays leave the diffracting object in parallel directions.– Screen very far from the slit– Converging lens (shown)

• A bright fringe is seen along the axis (θ = 0) with alternating bright and dark fringes on each side.

Single Slit Diffraction, 3

• In general, destructive interference occurs for a single slit of width a is given by: a sin θdark = mλ m = 1, 2, 3, …

Example 8

The angular separation between the two first order minima diffraction pattern produced by single slit is 200. The incident light has wavelength 600 nm. Determine the width of the slit.

Example 9

Parallel light of wavelength 550 nm is incident normally upon a single slit, thus forming a diffraction pattern on a screen. The first order minimum is at angular position of 10. Determine the width of the slit.

Diffraction Grating, Cont.• The condition for maxima is

– d sin θbright = m λ• m = 0, ±1, ±2, …

• The integer m is the order number of the diffraction pattern.

• Number per lines per cm = 1/d

Diffraction Grating, Final• All the wavelengths are

focused at m = 0– This is called the zeroth order

maximum• The first order maximum

corresponds to m = 1

Example 10

A parallel light beam of wavelength 600 nm is incident normally onto a diffraction. The angle between the two first order maxima is 300. Determine the number of lines per cm on the diffraction grating.

Example 11

A parallel light beam of wavelength 540 nm is incident normally onto a diffraction grating which has 5000 lines per cm. Determine the maximum number of orders of maxima which can be formed on both sides of the central maximum.

Polarization of Light Waves

• This is an unpolarized wave.

Polarization

Selective Absorption

• The intensity of the polarized beam transmitted through the second polarizing sheet (the analyzer) varies as– I = Io cos2 θ

• Io is the intensity of the polarized wave incident on the analyzer.

• This is known as Malus’ Law and applies to any two polarizing materials whose transmission axes are at an angle of θ to each other.

Polarization by Reflection

• Brewster’s Law relates the polarizing angle to the index of refraction for the material.

• θp may also be called Brewster’s Angle.

Example 12

A plane-polarised light beam has intensity I0. It is incident normally onto a polaroid. If the angle between the transmission axis of the polaroid and the plane of polarisation is 600, determine the intensity of the light beam after it has passed through the polaroid. (0.25I0)

Example 13

The index of refraction of a glass plate is 1.52. What is the Brewster’s angle when the plate is in air.