light waves

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35.1 Interference and coherent sources of light Waves, Optics & Modern Physics 203-NYC-05 Greg Mulcair

Light as learned in High School n  In high school you learned about light and

how it behaves when reflecting off surfaces and passing through lenses.

n This branch of physics (geometric optics) treats light as “rays” (straight lines).

n Although this works very accurately, light is actually a wave when you look close enough. Wave optics is the topic of Unit 2.

Electromagnetic (EM) waves

n  Light is the term used for the visible portion of a wide range of waves around us known as electromagnetic (EM) radiation.

n  Light is EM radiation with a wavelength between approx. λ = 400nm and 700nm.

EM waves are transverse waves.

Electromagnetic (EM) waves

n  Just like for mechanical waves, EM waves transmit energy at a certain wave speed. ¨ But, EM waves don’t need a medium! They can

transmit this energy through vacuum. ¨ EM waves travel at speed v = 3E8 m/s (which

we will call the speed of light, “c”)

n  We will say “light” during most of our discussions, but what we will learn applies to all EM waves.

Note for the curious: The nature of EM radiation is explained in chapter 32 but is not part of the NYC curriculum. The energy transmitted in EM radiation is both electrical and magnetic. The waves alternate rapidly, from positive to negative in electrical terms, and from North to South pole in

magnetic terms. A nice visualization of this can be seen here: http://www.ephysics.ca/Abbott/ActiveFigures/Ch.23-46/SWFs/AF_3403.html

Interference of light waves

n  Light waves interfere with each other the same way we’ve seen: by superposition. ¨ Recall: This is a fancy way of saying that the

waves add up. The resultant displacement is the displacement of all the waves combined.


n  Note: We use the term “displacement” in a general sense here. ¨ With waves on the surface of a liquid, we mean

the actual displacement of the surface above or below its normal level, as we learned.

¨ With sound waves, the term refers to the excess or deficiency of pressure, as we learned.

¨ For electromagnetic waves, we usually mean a specific component of electric or magnetic field.

Note: Waves in 2D and 3D

n Waves on a string travel in one dimension along the length of the string) and ripples on a pond go in two dimensions (along the plane of the surface).

n For light we will work with waves traveling in two or three dimensions.

Visualizing waves as wavefronts

n  A wavefront is just a line joining common points on the wave. E.g.: All the peaks (crests) of the wave.

n  These move outward with time as the wave travels, and the distance between each is the wavelength (which makes sense).

Note: Waves in 2D and 3D

n  For example, the figure shows a succession of wave crests (the peaks) from a sound source. ¨ If the waves are in 2D, like

waves on the surface of a liquid, the circles represent circular wave fronts.

¨ If the waves are in 3D, the circles represent spherical wave fronts spreading away from the source.

Note: Monochromatic light

n Most light sources (light bulbs, fire…) emit a continuous distribution of wavelengths.

n But for our study, it will be easier to consider light that is made to shine at only one frequency and wavelength (colour): monochromatic light. ¨ We will assume monochromatic light for this

chapter (interference) and the next (diffraction).

n An excellent example of this is the laser

Note: Coherent sources

n Furthermore, sources of monochromatic waves that have a constant phase difference (not necessarily zero) are coherent.

n We often refer to waves from such sources as coherent waves.

n  If the waves are light, we usually just say coherent light.

Note: Polarization

n And the last assumption we will make is that the wave sources have the same polarization (the transverse wave disturbances lie along the same line). ¨ You don’t need to know more than that,

although it’s an interesting topic!


n Ok, with all those assumptions and simplifications to our scenario, we can now take a look at interference of light waves!

n Good news, it’s the same rule as we learned for sound wave interference.

Example of two ideal sources

n Consider two sources (S1 and S2) of equal amplitude, equal wavelength (monochromatic) and same polarization (in the case of transverse waves such as light).

Constructive interference

n  Consider a point a on the x-axis.

n  From symmetry the two distances from S1 to a and from S2 to a are equal.

n  So waves from the two sources take equal times to travel to a.


n So for any point on the x-axis, waves that leave in phase arrive in phase and there is constructive interference.

n The total amplitude is the sum of the two source amplitudes.


n Now consider point b, where the distance from S2 to b is exactly two wavelengths greater than the distance from S1 to b.


n So a wave crest from S1 arrives at b exactly two cycles earlier than a crest emitted at the same time from S2 and again the two waves arrive in phase.


n So the waves arrive in phase and again there is constructive interference.

n The total amplitude is the sum of the two source amplitudes.


n  In general, waves interfere constructively if their path lengths differ by an integral number of wavelengths:

r2 − r1 =mλm = 0, ±1, ±2, ±3,…

Destructive interference

n Now consider point c, where the path difference for the distance from S2 to c compared to distance from S1 to c is exactly 2.5 wavelengths.


n So the wave crests from the two sources arrive exactly one half cycle out of phase.

n  In this case, there is destructive interference.


n Waves interfere destructively if their path lengths differ by a half-integral number of wavelengths

m = 0, ±1, ±2, ±3,…

r2 − r1 = m+ 12

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Summary: interference of light

n  Waves interfere constructively if their path lengths differ by an integral number of wavelengths:

n  Waves interfere destructively if their path lengths differ by a half-integral number of wavelengths:

r2 − r1 =mλ

r2 − r1 = m+ 12

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&'λ

m = 0, ±1, ±2, ±3,…

m = 0, ±1, ±2, ±3,…

Question

n  Two sources S1 and S2 oscillating in phase emit sinusoidal waves. Point P is 7.3 wavelengths from source S1 and 4.3 wavelengths from source S2. As a result, at point P there is ¨ A. constructive interference.

¨ B. destructive interference.

¨ C. neither constructive nor destructive interference.

¨ D. not enough information given to decide.

Answer





We just subtract the two and find that the path difference is: r2 – r1 = 7.3λ – 4.3λ = 3λ As seen, this gives constructive interference.

Question





Answer





We just subtract the two and find that the path difference is: r2 – r1 = 7.3λ – 4.6λ = 2.7λ This gives neither constructive or destructive interference.

Interference of light n  If we draw a curve

through all points where constructive interference occurs, we get the shapes shown.

n We call these antinodal curves.

Interference of light n  In ch.15 and ch.16,

antinodes were points of maximum amplitude in

n  But be careful: There was no net flow of energy in those cases, but in this case there is (it is just channeled such that it’s greatest along the antinodal curves).

Interference of light n  For 3D sources, try to

picture rotating the red antinodal curves about the y-axis.

n  The result would be antinodal surfaces that follow the same idea.

35.2 Two source interference of light

Waves, Optics & Modern Physics 203-NYC-05 Greg Mulcair

Two-source interference of light

n Studying the interference of two sources of light is not as easy two wave ripples since we can’t see the light as easily.

n  In 1800, the English scientist Thomas Young devised an experiment that will be important for us now and in later chapters: Young’s double-slit experiment

Young’s double-slit experiment

n First, monochromatic light is sent through a small (1µm) slit S0 to obtain a source that is an idealized point source.

n  Reason: Emissions from different parts of an ordinary source are not synchronized, whereas if they all pass through a small slit they are very nearly synchronized.

Recall: 1µm = 1E-6m


n Next, this light encounters two other slits, S1 and S2 (each ≈1µm wide, ≈1mm apart)

n  Note: Because the cylindrical wave fronts travel an equal distance from the source S0 to the slits S1 and S2 , they reach the slits in phase with one another (i.e. when a crest is at S1 a crest is at S2 , and so on).


n The waves emerging from S1 and S2 are therefore always in phase, so S1 and S2 are coherent sources of monochromatic light.

n  Recall: Monochromatic light is light that is made to shine at only one frequency and wavelength (colour)

n  Recall: Coherent sources emit waves which have a constant phase difference


n  Lastly, we observe how these two monochromatic coherent sources of light interfere with each other on a screen.

Questions:

What do you think is happening at these bright bands?

And what’s happening at these dark bands?


n  We see an interference pattern on the screen: ¨ Bright bands: there was constructive interference ¨ Dark bands: there was destructive interference

Question

n What did we learn earlier was the way to know if there is constructive or destructive interference between two waves?

Answer: Their path difference

n  Waves interfere constructively if their path lengths differ by an integral number of wavelengths:

n  Waves interfere destructively if their path lengths differ by a half-integral number of wavelengths:

r2 − r1 =mλ

r2 − r1 = m+ 12

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m = 0, ±1, ±2, ±3,…

m = 0, ±1, ±2, ±3,…


n  Consider a point P on the screen. We see that the wave coming from source S2 travels a bit further than the wave coming from source S1.

n  But how do we calculate the exact difference in the distances the waves travel (r2 – r1) ?


n  First we simplify: Since the distance R is much greater than the distance d between the two slits, we can consider the rays as being parallel.


n  And so the path difference is approximately equal to the extra distance travelled from S2 (which is dsinθ from trigonometry and similar triangles).

n  This is a very good approximation when R is large. Make sure that you understand the trigonometry (it is important now and will be important later too).

Similar triangles, so this θ equals this θ

θ

r2 − r1 = d sinθ


n  So the difference in path length is ¨ Where θ is the angle between a line from slits to screen

(the thick blue lines in the figure) and the normal to the plane of the slits (the thin black line in the figure).

r2 − r1 = d sinθ


n  And we’ve learned that we have: ¨ Constructive interference when

¨ Destructive interference when

r2 − r1 =mλ

r2 − r1 = m+ 12

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m = 0, ±1, ±2,…

m = 0, ±1, ±2,…


n  So we can conclude that we have: ¨ Constructive interference when

¨ Destructive interference when

d sinθ =mλ

d sinθ = m+ 12

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m = 0, ±1, ±2,…

m = 0, ±1, ±2,…


n  So we get a succession of bright and dark bands on the screen as the angle changes (i.e. for different locations on the screen). We call these interference fringes.

n  The central band (which corresponds to m = 0) is the brightest

Let’s be sure we understand n  We describe locations on the screen in

terms of the angle from horizontal. n  For any θ, if the value of dsinθ works

out to be an integer multiple of wavelengths ( )… …then you will have constructive interference (a bright band).

n  And if you get a dark band.

d sinθ =mλLight from two very close

coherent sources. We see interference fringes

on the screen at different angles, θ, from the

horizontal

+θ

-θ

d sinθ = m+ 12

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Measuring the y-position of fringes

n  If the angle θ is small (< 20º) we can describe each fringe’s position in terms of it’s distance from the middle fringe.

n  For example, for the m=3 fringe the angle θm = θ3

n  Corresponds to a height above horizontal of ym = y3. From trig, what is this? Note that the screen is a distance R from the slits

θm = θ3 ym = y3


n  From trig, the y-position of the third bright fringe (m=3) is: y3 = Rtanθ3

n  R is actually very large compared to y3 so this equation is very nearly equal to: y3 = Rsinθ3

n  And since we also know that we can say dsinθ3 = 3λ è sinθ3 = 3λ/d which gives us: y3 = R3λ/d

θm = θ3 ym = y3

R

d sinθ =mλ


n  In general, the y-position of any bright (constructive interference) fringe is:

n  So the distance between adjacent bright bands in the pattern is inversely proportional to the distance d between the slits. The closer together the slits are, the more the pattern spreads out. When the slits are far apart, the bands in the pattern are closer together.

θm

ym

ym = Rmλd

Measuring the wavelength of light

n  This equation can also be used to find the wavelength of incoming light since we can measure R, d and ym.

n  Young's experiment was actually the first direct measurement of wavelengths of light.

θm

ym ym = R

mλd

http://www.awakensolutions.com/Abbott/ActiveFigures/NYB/SWFs/AF_3702.html

This is a great simulation of this scenario. Start by watching it with max slit separation and minimum wavelength. The red/green dots where the wave crests (maximus) meet follow antinodal curves on their path to the screen. When they reach the screen, this constructive interference (large amplitude) gives a bright band on the screen.

Now try decreasing the slit separation, and see that ym increases (the pattern spreads out) as expected. And you can change the wavelength as well and see that ym increases (the pattern spreads out) as well.

ym = Rmλd

Summary for Young’s double-slit experiment

n For any angle, the position of the fringes can be described using these equations:

n And if the angle θ is small (< 20º) this equation works too:

d sinθ =mλ

d sinθ = m+ 12

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m = 0, ±1, ±2,…

m = 0, ±1, ±2,…

ym = Rmλd

Recall from a few slides back, this equation came from saying the height ym = Rtanθm is very close to ym = Rsinθm . However this isn’t the case if θ cannot be considered small.

Example 35.1 (p.1213)

n  In a two-slit interference experiment, the slits are 0.200 mm apart, and the screen is at a distance of 1.00 m. The third bright fringe (not counting the central bright fringe straight ahead from the slits) is found to be displaced 9.49 mm from the central fringe. Find the wavelength of the light used.

d sinθ =mλ

d sinθ = m+ 12

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ym = Rmλd

Solution using the ‘approximate’ equation

ym = Rmλd

ym = Rmλd

λ =ymdmR

=(9.49E −3)(0.2E −3)

(3)(1)λ = 633E − 9m = 633nm

n  The third bright fringe corresponds to m = 3. To determine the wavelength we may can use since R = 1.00 m is much greater than d = 0.200 mm or y3 = 9.49 mm

Note that this fringe could also have corresponded to m = -3. You would obtain the same result for the wavelength since y and m would be negative.

ym = Rmλd

Solution using the ‘pure’ equation

n  Using the pure equation we need to first find the angle to the screen:

n  And we use this to find the exact wavelength:

Note that this fringe could also have corresponded to m = -3. You would obtain the same result for the wavelength since y and m would be negative.

d sinθ =mλ

nmmEmE

Emd

63396337326.6

3)544.0sin()32.0(sin

=−=

−=

=−

=

λ

λ

λ

λθ

oEadjopp 544.0

1349.9tan =⇒

−== θθ

Question 35.11 (p.1230)

n  Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except light of a single wavelength. It then falls on two slits separated by 0.460 mm. In the resulting interference pattern on a screen 2.20 m away, adjacent bright fringes are separated by 2.82 mm. What is the wavelength?

Answer 35.11 (p.1230)

( )

λ

λ

λ λ λ λ

λ

+

+

=

+=

+Δ = − = − = + − =

Δ − −= = = − =

1

1

( 1)

( 1) ( 1)

(0.46 3)(2.82 3) 5.9 7 5902.2

m

m

m m

my Rdmy Rd

m my y y R R R m m Rd d d d

d y E E E m nmR

Question 35.13 (p.1230)

n  Two very narrow slits are spaced 1.80 µm apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with λ = 550 nm?

n  For this problem, do not use the approximation that tanθ ≅ sinθ (consider that the angle is too large for this approximation since the distance from the screen, R, is less than 1m).

Two very narrow slits are spaced 1.80 µm apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with λ = 550 nm?

Answer 35.13 (p.1230)

n  In the previous problem we were told that the angle wasn’t small enough for us to use the small angle approximation (which lets us say sinθ ≅ tanθ, thus giving us ).

n So we had to use the main equation from which the above is derived:

λ=m

my Rd

d sinθ = m+ 12

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Answer 35.13 (p.1230)

n  Let’s see how the small angle approximation (sinθ ≅ tanθ) breaks down as the angle gets bigger: ¨ For θ1 = 8.8º we get:

n  sinθ1 = 0.1528 tanθ1 = 0.1546

¨ For θ2 = 27.3º we have a larger angle and get: n  sinθ2 = 0.4583 tanθ2 = 0.5157

n As you can see, for larger angles the small angle approximation is less and less valid

light waves

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