• Recall from our work with mechanical waves that interference can occur between two wave fronts.
– And since light behaves like a wave, we can have interference with light.
• Although, observing interference in light waves is difficult because of the very short wavelengths.
• In order to have sustained interference so that it is observable, we must meet the following:
1.
The light sources must be coherent .
1.
They must maintain constant phase with respect to each other.
2.
The sources must have identical wavelengths .
2.
So blue light will not interfere with red light!
3.
The superposition principle must apply.
3.
Meaning the waves must share the same medium so their positions can be added.
d
r
1
L
r
2
• A ray emitted from source B will have to travel a greater distance to get to point C than a ray emitted from source
A.
– The extra distance can be found by r
2
–r
1
• The path difference ,
, between ray paths from the two slits is found by
= d sin
• The interference pattern that results from Young’s double-slit experiment creates a series of alternating high-low intensity regions.
• Each region of intensity (high or low) gets progressively more dim as the distance from the principal axis increases.
– Remember that waves intensity is proportional to r 2 .
• Each region of high intensity determines the order number of the wavefront we are analyzing.
– The numbering system begins at m = 0 when
= 0 o .
• That would be along the principal axis .
• This is also called the zeroth-order maximum .
– The numbering can be positive or negative as you move left/right or up/down.
• For instance, the first-order maximum occurs when m = + 1 .
m = 1 m = 0 m = -1
•
The equation to find the path difference assumes that the rays emitted by each slit travel parallel to each other.
– While not entirely true, they are close enough we can treat them as parallel.
•
This is because the distance from the slits to the screen, L , is much larger than d .
•
Remember, constructive interference occurs when waves are in phase with each other.
– So when the path difference is either zero or a whole integer multiple then the waves will be in phase.
•
An area of high intensity will occur at points of constructive interference .
= d sin
= m
• At any time waves are out of phase with each other, they will create a region of lower intensity.
• However, the areas that we are more concerned with are regions of near complete destructive interference .
– This will occur when waves are 180 o out of phase.
• Which occurs with a path difference of
=
/
2 or any integer multiple thereof.
• An area of low intensity will occur during destructive interference .
– Notice it is an area of low intensity and not total cancellation.
• That is due to the dual nature of light. The wave property may be cancelled out, but the particle still exists.
= d sin
= (m+ 1 /
2
)
• Identifying the path difference for several different interference patterns we assume the following:
– That L >> d
• Distance from slit to screen is much larger than distance from slit to slit
– d >>
• Distance between slits is much larger than the wavelength.
– x m
=
L m / d
• The fringe position of any high intensity order is directly proportional to:
– wavelength of light source
– screen distance
– order number
• and inversely proportional to:
– slit separation
• Interference is best observed in the light patterns from light interacting with thin films.
• The color pattern is determined by the wavelength of light as it travels inside the film.
– Remember that the speed of light changes when traveling through a dense material and therefore changes the wavelength of the light.
• n
=
/ n
– is the wavelength of light in a vacuum
• The “pattern” that you see created by the light rays is due to the destructive interference of light rays.
– Remember that some of the ray is reflected when it encounters the film layer.
• This ray is reflected with a 180 o phase change .
– Then some of the light is reflected when it encounters the other boundary of the film layer.
• This ray is reflected with no phase change .
• And this is the ray that determines the color change.
– This is how we get destructive interference from the same light source!
Examples of Thin Film Interference
Baylor University Explanation of Thin-Film Interference
• Light travels in a straight-line path.
– That is why we have shadows!
• Otherwise we would see that light can turn corners on its own, and that would defy the particle theory of light.
• We have seen that when light encounters a small opening, it ripples out much like a wave from a stone dropped in a pond.
• This is due to the wave nature of light.
• This divergence of light from its natural line of travel is called diffraction .
– Diffraction results in a central region of high intensity flanked by a series of narrower regions of diminishing intensity.
• In general, diffraction will occur when waves pass through small openings and try to bend around obstacles or sharp corners.
• The pattern that results consists of a central region of high intensity called the central maximum .
– Then there are repeating bands of destructive interference called minima .
– Those are followed by regions of visible yet diminishing intensity light called secondary maxima .
• In order to analyze the diffraction pattern produced from a single slit we can first split the slit in half.
– This is done to understand how the light waves interact with each edge of the slit.
– Each edge will create a path difference equal to a /
2 sin
• By dividing the slit in half, this creates a
180 o phase difference .
• So the light waves from one side of the slit will cancel out the waves from the other side of the slit.
• If we continue this division process of the slit into thirds, fourths, etc. we will begin to see where the diffraction pattern comes from.
– The division of the slit is our way of explaining how a continuous light source can be turned into a finite number of light sources as it passes through the slit.
• If it stays as one continuous light source, then the diffraction pattern would not result, which we have seen will always result.
• The location of the dark regions is much easier to calculate than the visible bright regions because they are all the same width and all the same distance from the principal axis .
– The secondary maxima are also the same distance from the principal axis.
• But the central maximum is located at the principal axis and is twice as large as any other maximum or minimum
This recognizes dark regions sin
= m
/ a mimima number width of slit
Bright regions would then occur at
/
2
.
• Due to the uncertainty of when an electron will emit a light carrying photon, most light sources emit noncoherent light .
–
Noncoherent light is any light in which the wave fronts are out of phase and do not lie in the same plane.
• Incandescent light bulbs, candles, and the sun are a few examples of noncoherent light sources.
• A laser is a light source in which its resulting light waves are coherent .
–
Coherent light is any light in which all of its wave fronts lie in phase, in the same plane, and parallel to one another.
• That way the interference pattern is nothing but constructive, creating a very intense stream of light.
L
A
S
E
R