Chapter 3

Interference of Light Waves

To observe interference in light wave, conditions are:

The sources must be coherent that is, they must maintain a constant phase with respect to each other.

The sources must be monochromatic that is of a single wavelength.

The superposition principle must apply.

Huygen’s Principle, Wavefronts and Coherence

E Em sin(2

2f t)

k

E Em sin(kx t)

Examples of coherence are: Laser light Small spot on tungsten filament Wavefront

Most light is incoherent: Two separate light bulbs Two headlight beams on a car Sun is basically incoherent

Interference is the combination of two or more waves to form a composite wave, based on the principle of superposition

In Phase Out of Phase by 180 degrees or radians or/2

In between

Young’s Double Slit Experiment,

The narrow slits, S1 and S2 act as sources of waves

The waves emerging from the slits originate from the same wave front and therefore are always in phase

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 fringe pattern

formed from a Young’s Double Slit Experiment would look like this

The bright areas represent constructive interference

The dark areas represent destructive interference

Interference Patterns

Constructive interference occurs at the center point

The two waves travel the same distance Therefore, they

arrive in phase

The upper wave has to travel farther than the lower wave

The upper wave travels one wavelength farther Therefore, the

waves arrive in phase

A bright fringe occurs

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

Interference Equations The path difference, δ,

is found from the tan triangle

δ = r2 – r1 = d sin θ This assumes the

paths are parallel Not exactly parallel,

but a very good approximation since L is much greater than d

For a bright fringe, produced by constructive interference, the path difference must be either zero or some integral multiple of the wavelength

δ = 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

The positions of the fringes can be measured vertically from the zeroth order maximum

y = L tan θ L sin θ Assumptions

L>>d d>>λ

Approximation θ is small and therefore the approximation tan

θ sin θ can be used

When destructive interference occurs, a dark fringe is observed

This needs a path difference of an odd half wavelength

δ = d sin θdark = (m + ½) λ m = 0, ±1, ±2, …

Interference Equations, final

For bright fringes

For dark fringes

0, 1, 2bright

Ly m m

d

10, 1, 2

2dark

Ly m m

d

Intensity distribution of the double-slit interference pattern The total electric

field intensity at point P on the screen is the vector superposition of the two waves.

We suppose the two slits represent coherent sources of sinusoidal waves. Hence,

the waves have the same angular frequency ω and a constant phase difference φ .

Assuming the two waves have the same amplitude Eo , the electric field intensities at P due to each wave separately is :

E1 = Eo sin ω t ,and

E2 = Eo sin ( ω t + φ )  Although the waves are in phase at the

slits, their phase difference φ at P depends on the path difference

δ = r2 - r1 = d sin θ .

Because a path difference of λ (constructive interference) corresponds to a phase difference of 2 π rad,

a path difference of λ/2 (destructive interference) corresponds to a phase difference of π rad, we obtain the ratio:

Using the superposition principle we can obtain the resultant electric field at P Ep = E1 + E2 = Eo [sin ω t + sin (ω t + φ)] sin A+sin B = 2 sin[(A + B) / 2] cos[(A – B)/2] Taking A = ω t + φ , and B = ω t, Ep = 2 Eo cos (φ/2) sin (ω t + φ/2) the electric field at P has the same

frequency as the light at the slits, but its amplitude is multiplied by the factor:

2 cos (φ /2).

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

I α Ep2

= 4 Eo2 cos2 (φ /2) sin2 (ω t + φ/2)

the time average value of sin2 (ω t + φ/2) over one cycle is 1/2,

Iav = Io cos2 ( φ / 2)

where lo is the maximum possible time average light intensity

Iav = Io cos2 ( π d sin θ /λ )  

since sinθ ≈ y/ L for small values of θ Iav = Io cos2 ( π yd /λ L)   Constructive interference, which produces

intensity maxima, occurs when the quantity

( π yd / λ L) is an integral multiple of π , corresponding to :

y = ( λ L / d) m. This is consistent with the equation of ybright .

Intensity distribution versus d sin θ or the double-slit pattern when the screen is far from the two slits (L » d).

Interference in Thin Films Interference effects are

commonly observed

in thin films Examples are soap

bubbles and oil on water The interference is due to

the interaction of the waves reflected from both surfaces of the film

An electromagnetic wave traveling from a medium of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1

There is no phase change in the reflected wave if n2 < n1

The wavelength of light λn in a medium with index of refraction n is λn = λ/n where λ is the wavelength of light in vacuum

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

Ray 2 also travels an additional distance of 2t before the waves recombine

For constructive interference 2nt = (m + ½ ) λ 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 destruction interference 2 n t = m λ m = 0, 1, 2 …

Two factors influence interference Possible phase reversals on reflection Differences in travel distance

The conditions are valid if the medium above the top surface is the same as the medium below the bottom surface

If the thin film is between two different media, one of lower index than the film and one of higher index, the conditions for constructive and destructive interference are reversed

Be sure to include two effects when analyzing the interference pattern from a thin film Path length Phase change

Newton’s Rings Another method for viewing interference is to

place a planoconvex lens on top of a flat glass surface

The air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness t

A pattern of light and dark rings is observed This rings are called Newton’s Rings The particle model of light could not explain

the origin of the rings Newton’s Rings can be used to test optical

lenses

The interference effect is due to the combination of ray l, reflected from the flat plate, with ray 2, reflected from the lower part of the lens.

Imperfection in lens

Ray 1 undergoes a

phase change of 180o

upon reflection,

because it is reflected from a medium of higher refractive index, whereas ray 2 undergoes no phase change

the conditions for constructive and destructive interference are given, with n = 1 because the film is air.

Point O is dark, because ray 1, reflected from the flat surface, undergoes a 180o phase change with respect to ray 2

Expressions for the radii of the bright and dark bands can be obtained in terms of the radius of curvature R and wavelength λ.

For example, the dark rings have radii given by: r ≈ √ m λ / n .

By measuring the radii of the rings, the wavelength can be obtained, provided R is known.

Conversely, if the wavelength is accurately known, this effect can be used to obtain R.

The Michelson Interferometer A beam of light provided by

a monochromatic source is split into two rays by a partially silvered mirror, M, inclined at 45o to the incident light beam. One ray is reflected vertically upward toward mirror M2, while the second ray is transmitted horizontally through M toward mirror M2.

Hence, the two rays travel separate paths L1 and L2.

After reflecting from M1 and M2, the two rays eventually recombine to produce an interference pattern, which can be viewed through a telescope.

The interference condition for the two rays is determined by their path length differences.

When the two mirrors are exactly perpendicular to each other, the interference

pattern is a target pattern of bright and dark circular fringes, similar to Newton’s rings.

As M1 is moved, the fringe pattern collapses or expands, depending on the direction in which M1 is moved.

For example, if a dark circle appears at the center of the target pattern (corresponding to destructive interference) and M1 is then moved a distance λ /4 toward M0, the path difference changes by λ /2. What was a dark circle at the center now becomes a bright circle.

As M1 is moved an additional distance &/4 toward M0, the bright circle becomes a dark

circle again. Thus, the fringe pattern shifts by one-half fringe each time M1 is

moved a distance λ /4. The wavelength of light is then measured by counting the number of fringe shifts for a given displacement of M1.

If the wavelength is accurately known, mirror displacements can be measured to within a fraction of the wavelength.

Anti-Reflecting Coatings We can apply a thin

film of a material with the proper thickness so that it will cause destructive interference of reflected light.

Thus all the light must be transmitted through the film into the lens.

Problem Solving Strategy with Thin Films Identify the thin film causing the interference Determine the indices of refraction in the film

and the media on either side of it Determine the number of phase reversals:

zero, one or two

The interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λ The conditions are reversed if one of the

waves undergoes a phase change on reflection

Problem Solving with Thin Films

Equation1 phase reversal

0 or 2 phase reversals

2nt = (m + ½) constructive destructive

2nt = m destructive constructive

Example An example of

different indices of refraction

A coating on a solar cell

There are two phase changes

CD’s and DVD’s Data is stored digitally

A series of ones and zeros read by laser light reflected from the disk

Strong reflections correspond to constructive interference These reflections are chosen to represent

zeros Weak reflections correspond to destructive

interference These reflections are chosen to represent

ones

CD’s and Thin Film Interference A CD has multiple tracks

The tracks consist of a sequence of pits of varying length formed in a reflecting information layer

The pits appear as bumps to the laser beam The laser beam shines on the metallic layer

through a clear plastic coating

Reading a CD

As the disk rotates, the laser reflects off the sequence of bumps and lower areas into a photodector The photodector converts

the fluctuating reflected light intensity into an electrical string of zeros and ones

The pit depth is made equal to one-quarter of the wavelength of the light

When the laser beam hits a rising or falling bump edge, part of the beam reflects from the top of the bump and part from the lower adjacent area This ensures destructive interference and

very low intensity when the reflected beams combine at the detector

The bump edges are read as ones The flat bump tops and intervening flat

plains are read as zeros

DVD’s DVD’s use shorter wavelength lasers

The track separation, pit depth and minimum pit length are all smaller

Therefore, the DVD can store about 30 times more information than a CD