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Physics 42S IBTopic 9 – Wave Phenomena

Physics 42S IBMiles Macdonell Collegiate

9.2 – Single-slit diffraction

Nature of science:

Development of theories: When light passes through an aperture the summation of all parts of the wave leads to an intensity pattern that is far removed from the geometrical shadow that simple theory predicts. (1.9)

Understandings:

The nature of single-slit diffraction

Applications and skills:

Describing the effect of slit width on the diffraction pattern

Determining the position of first interference minimum

Qualitatively describing single-slit diffraction patterns produced from white light and from a range of monochromatic light frequencies

Guidance:

Only rectangular slits need to be considered

Diffraction around an object (rather than through a slit) does not need to be considered in this sub-topic (see Physics sub-topic 4.4)

Students will be expected to be aware of the approximate ratios of successive intensity maxima for single-slit interference patterns

Calculations will be limited to a determination of the position of the first minimum for single-slit interference patterns using the approximation equation

Data booklet reference:

θ= λb

Theory of knowledge:

Are explanations in science different from explanations in other areas of knowledge such as history?

Utilization:

X-ray diffraction is an important tool of the crystallographer and the material scientist

Aims:

Aim 2: this topic provides a body of knowledge that characterizes the way that science is subject to modification with time

Aim 6: experiments can be combined with those from sub-topics 4.4 and 9.3

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Topic 9Wave Phenomena

Single-slit Diffraction

Diffraction at a Single Slit

When light passes through a narrow slit, it spreads out due to diffraction. The light does not spread out uniformly, but forms patterns

Huygens Construction

Dutch Physicist Christian Huygens proposed the model for the propagation of light. He suggested that light propagated as if the wave front was made of an infinite number of small wavelet sources. In diffraction, if the slit is small, only one wavelet passes through. However, if the slit is wider more wavelets pass through causing interference.

The Central Maximum

The central maximum occurs directly ahead of the slit. At a point ahead of the wave, all wavelets would have traveled the same distance and as a result they will be in phase.

The superposition of the wave will add up and give a region of high intensity (i.e. it will be bright).

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Figure 2 – a wave travelling through a single slit divided into wavelets

Figure 1 – an interference pattern produced by a wave travelling through a single slit


The First Minimum

The waves are now traveling at an angle, so they will not all be travelling the same distance. They will be out of phase and will cancel each other out.

Distance to the first minimum

For waves to be out of phase, the path

difference has to beλ2 .

sin θ=

λ2b2

= λb

The angle is very small, so we can use what is called the small angle

approximation. If θ is measured in radians andθ is small, sin θ=θ. So:

sin θ= λb

becomes:

θ= λb

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B5B4B3B2B1A5A4

A3A2A1

Figure 4 – angle and path difference of wavelets

θ

θ

path difference = λ2

b

Figure 3 – wavelets travelling to the first minima


Example 1

A diffraction pattern is formed on a wall 2.0m from a slit 0.10mm wide illuminated by light of wavelength of600nm. Determine the width of the central maxima.

Example 2

The central maxima of a diffraction pattern formed by a laser passing through a single slit on a wall is15.2mm. The wall is 4.75m from the single slit. The width of the slit 0.250mm. Determine the wavelength of the laser.

Homework: Tsokos – Page 364; Problems 14-16

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9.3 – Interference

Nature of science:

Curiosity: Observed patterns of iridescence in animals, such as the shimmer of peacock feathers, led scientists to develop the theory of thin film interference. (1.5)

Serendipity: The first laboratory production of thin films was accidental. (1.5)

Understandings:

Young’s double-slit experiment

Modulation of two-slit interference pattern by one-slit diffraction effect

Multiple slit and diffraction grating interference patterns

Thin film interference


Qualitatively describing two-slit interference patterns, including modulation by one-slit diffraction effect

Investigating Young’s double-slit experimentally

Sketching and interpreting intensity graphs of double-slit interference patterns

Solving problems involving the diffraction grating equation

Describing conditions necessary for constructive and destructive interference from thin films, including phase change at interface and effect of refractive index

Solving problems involving interference from thin films


Most two-slit interference descriptions can be made without reference to the one-slit modulation effect. To what level can scientists ignore parts of a model for simplicity and clarity?

Utilization:

Compact discs are a commercial example of the use of diffraction gratings

Thin films are used to produce anti-reflection coatings

Aims:

Aim 4: two scientific concepts (diffraction and interference) come together in this sub-topic, allowing students to analyse and synthesize a wider range of scientific information

Aim 6: experiments could include (but are not limited to): observing the use of diffraction gratings in spectroscopes; analysis of thin soap films; sound wave and microwave interference pattern analysis

Aim 9: the ray approach to the description of thin film interference is only an approximation. Students should recognize the limitations of such a visualization

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Guidance:

Students should be introduced to interference patterns from a variety of coherent sources such as (but not limited to) electromagnetic waves, sound and simulated demonstrations

Diffraction grating patterns are restricted to those formed at normal incidence

The treatment of thin film interference is confined to parallel-sided films at normal incidence

The constructive interference and destructive interference formulae listed below and in the data booklet apply to specific cases of phase changes at interfaces and are not generally true


nλ=d sin θ

Constructive Interference: 2dn=(m+ 12 ) λ

Destructive Interference: 2dn=mλ

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Interference

Young’s Double Slit Experiment

When light (or any other type of wave) passes through a double slit two thing occur:

1) The wave passes through each slit. Single slit diffraction occurs at both slits.

2) The diffraction patterns of each slit interfere with each other.

We can predict the interference pattern with the following equation.

Double slit Diffraction

s= λDd

Where,

s – is the separation between adjacent maxima.

λ – is the wavelength.

D – is the separation between the double slit and the screen.

d – is the separation of the two slits.

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Figure 5 – Young’s double slit experiment


Example 1

Use figure 6 for this question. In a double-slit interference experiment the two slits are separated by a distance of4.2×10−4mand the screen is3.8mfrom the slits.

(a) Determine the wavelength of light used in this experiment.

(b) Suggest the effect on the separation of the fringes of decreasing the wavelength of light.

(c) State the feature of the graph that enables you to deduce that the slit width is negligible.

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Figure 6

2

4

3

1

−2 −1 0 1 2x /cm

Intensity


Example 2

In a Young’s two-slit experiment, a source of light of unknown wavelength is used to illuminate two very narrow slits a distance of0.15mmapart. On a screen at a distance of1.30mfrom the slits, bright spots are observed separated by a distance of4.95mm. What is the wavelength of light being used?

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An IB Question

(a)Outline what is meant by the principle of superposition of waves. [2]

(b)Red laser light is incident on a double slit with a slit separation of 0.35 mm.A double-slit interference pattern is observed on a screen 2.4 m from the slits.The distance between successive maxima on the screen is 4.7 mm.

Calculate the wavelength of the light. Give your answer to an appropriate number of significant figures. [3]

(c) Explain the change to the appearance of the interference pattern when the red-light laser is replaced by one that emits green light. [2]

(d)One of the slits is now covered. Describe the appearance of the pattern on the screen.[2]

An IB Question

Yellow light from a sodium lamp of wavelength 590 nm is incident at normal

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incidence on a double slit. The resulting interference pattern is observed on a screen. The intensity of the pattern on the screen is shown.

The double slit is replaced by a diffraction grating that has 600 lines per millimetre. The resulting pattern on the screen is shown.

(a)i. Explain why zero intensity is observed at position A. [2]

The distance from the centre of the pattern to A is4.1×10–2m. The distance from the screen to the slits is7.0m.

ii. Calculate the width of each slit. [2]

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iii. Calculate the separation of the two slits. [2]

(b)i. State and explain the differences between the pattern on the screen due to

the grating and the pattern due to the double slit. [3]

ii. The yellow light is made from two very similar wavelengths that produce two lines in the spectrum of sodium. The wavelengths are 588.995 nm and 589.592 nm. These two lines can just be resolved in the second-order spectrum of this diffraction grating. Determine the beam width of the light incident on the diffraction grating. [3]

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Multiple Slit Diffraction

The number of secondary maxima between primary maxima isN−2.

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Figure 7 – diffraction patterns from multiple slits


Example

Look at figure which shows the intensity pattern for diffraction through multiple slits.

Verify that the slit width is twice the wavelength and determine the separation of two consecutive slits in terms of the wavelength.

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40−40 302010−30 −20

30

20

10

Figure 8

−10 0θ /degrees

Intensity


The Diffraction Grating

d sin θ=nλ

Example 1

Light of wavelength 680nm falls normally on a diffraction grating that has 600 rulings per mm. What is the angle separating the central maximum(n=0 )from the next(n=1 )? How many maxima can be seen?

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Example 2

A diffraction grating is used to resolve two lines in the spectrum of sodium in the second order. The two lines have wavelengths588.995nmand589.592nm.

Determine the minimum number of slits in the grating that will enable the two lines to be resolved.

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An IB Question

This question is about laser light.

(a)Laser light is monochromatic and coherent. Explain what is meant by [3 marks](i) monochromatic.

(ii) coherent.

(b) A beam of laser light is incident normally on a diffraction grating which has 600 lines per millimetre. A fringe pattern is formed on a screen 2.0 m from the diffraction grating.

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The fringe pattern formed on the screen is shown below.

Determine the wavelength of the laser light. [4]

(c) The number of lines per millimetre in the diffraction grating in (b) is reduced. Describe the effects of this change on the fringe pattern in (b).

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Thin Film Interference

When light is incident on a layer of film (oil, water, soap, varnish, etc.) with a different index of refraction, it will refract and reflect. The refracted ray will then reflect and refract of the other side of the film. The ray transmitted to the film will travel a distance that is the thickness of the film. As the wave propagates, it might be out of phase with the reflected ray. As a result, destructive interference will occur. If the transmitted ray is in phase with the reflected ray, constructive interference will occur. The following diagram show light exhibiting thin film interference of a soap bubble.

In the preceding diagram, the light is incident on soap bubble of thicknessd . At pointA, the light is reflected and transmitted. The transmitted ray is then reflected off the other side of the film and transmitted atB. The ray reflected atB, is then incident atC where it is refracted out to the air. If the ray reflected atA and the ray transmitted atC are in phase, constructive interference occurs. If the ray reflected atA and the ray transmitted atC are out of phase, destructive interference occurs.

The wave that is reflected atA is inverted as it enters a slower medium (see refraction). For the transmitted ray to be in phase, it needs travel just one-half wavelength from A to C.

pathdifference=(m+12 )λs

m is an integer number(m=0,1,2,3 ,… ), but in practice we only usem=0. λ s is the wavelength of light in the soap film. The path difference in twice the thickness of the film.

pathdifference=(m+12 )λs

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Figure 9 – Thin film interference with one phase change

d

n=1.00n=1.40

n=1.00C

B

A


2d=(m+ 12 ) λs

The wavelength of light in the film is related to the wavelength of light in air by:

λ s=λn

So:

2d=(m+ 12 ) λs

2d=(m+ 12 ) λ

n

2dn=(m+ 12 ) λ

For the thin film to be a minimum, m=0 for constructive interference.

Similarly, for the wave to be out of phase it needs to travel one full wavelength:

2dn=mλ

For the thin film to be a minimum, m=1 for destructive interference.

There are two scenarios in which thin film interference can occur. The previous example is where there is only one phase change (i.e. only one different index of refraction). The soap bubble has air on the outside and air on the inside. The film of soap has a different index of refraction to the air. If there is a second phase change, like in a transparent coating over glass, where the final index of refraction is higher than the previous indices of refraction, the reflected ray atB is also inverted. This requires a modification to the formula.

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Fortunately, the formulas are the same, just switched, as are their conditions form.

Type of interference 1 phase 2 phases

Constructive 2dn=(m+ 12 ) λ ;m=0 for min 2dn=mλ;m=1 formin

Destructive 2dn=mλ;m=1 formin 2dn=(m+ 12 ) λ ;m=0 for min

Example 1

Monochromatic light of wavelength600nm is incident normally on a thin film of soap that has a refractive index ofn=1.40. The film is surrounded by air. The intensity of the reflected light is a minimum. Determine the minimum thickness of the film.

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Figure 10 – Thin film interference with two phase changes

d

n=1.52n=1.40

n=1.00C

B

A


Example 2

A solar cell must be coated to ensure as little as possible of the light falling on it is reflected. A solar cell has a very high index of refraction (aboutn=3.50). A coating of index of refractionn=1.50is placed on the cell. Estimate the minimum thickness needed in order to minimize reflection of light of wavelength524nm.

Example 3

A thin film of oil lies on a puddle of water. White light from above shines on the film at normal incidence. The refractive index of oil is 1.45 and the refractive index of water is 1.33. The thickness of the oil film is250nm. Calculate the maximum wavelength of the incident light for which destructive interference occurs.

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Example 4

A mirror has a coating of glassn=1.62 over aluminium. Determine the minimum thickness of the glass so that the reflection of the mirror is at a maximum.

Homework: Giancoli – Page 650; Problems 4-10Tsokos – Page 374; Problems 17-26

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This question is about thin-film interference.

The anti-reflective coating of a lens consists of a thin layer of a suitable material placed between the air and the glassof the lens.

The following data are available.Refractive index of air ¿1.0Refractive index of coating ¿1.2Refractive index of glass ¿1.5

(a)State what phase change occurs on reflection at the air-coating

boundary and at the coating-glass boundary.

(b)The thickness d of the coating layer is 110 nm. Determine the wavelength for which there is no resultant reflection from the surface of the lens for light at normal incidence (θ=0 °) .

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9.4 – Resolution

Nature of science:

Improved technology: The Rayleigh criterion is the limit of resolution. Continuing advancement in technology such as large diameter dishes or lenses or the use of smaller wavelength lasers pushes the limits of what we can resolve. (1.8)

Understandings:

The size of a diffracting aperture

The resolution of simple monochromatic two-source systems


Solving problems involving the Rayleigh criterion for light emitted by two sources diffracted at a single slit

Resolvance of diffraction gratings

Guidance:

Proof of the diffraction grating resolvance equation is not required


θ=1.22 λb

R= λΔ λ

=mN

International-mindedness:

Satellite use for commercial and political purposes is dictated by the resolution capabilities of the satellite


The resolution limits set by Dawes and Rayleigh are capable of being surpassed by the construction of high-quality telescopes. Are we capable of breaking other limits of scientific knowledge with our advancing technology?

Utilization:

An optical or other reception system must be able to resolve the intended images. This has implications for satellite transmissions, radio astronomy and many other applications in physics and technology (see Physics option C)

Storage media such as compact discs (and their variants) and CCD sensors rely on resolution limits to store and reproduce media accurately

Aims:

Aim 3: this sub-topic helps bridge the gap between wave theory and real-life applications

Aim 8: the need for communication between national communities via satellites raises the awareness of the social and economic implications of technology

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Resolution

When Diffraction occurs, this may cause problems for specific devices, such as cameras and telescopes, and can even cause problems for our eyes.

The Rayleigh Criterion

The Rayleigh Criterion puts a limit on the resolvability of two points, based on the diffraction of light. It states that:

“Two points will just be resolved if the central maximum of the diffraction pattern of one point coincides with the first minimum of the second point.”

The diffraction patterns are not those formed by slits, but they are formed by a circular aperture, so the diffraction patterns are circular.

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(b) – two points, easily resolvedFigure 11 (a) – one point

one point

(d) – two points, not resolved(c) – two points, just resolved


To calculate the distances for resolution, it is similar to diffraction, given by the equation:

θ= λb

For circular openings (apertures), the central maximum is 1.22 times wider than the aperture, so the angular position of the first minimum is given by the formula:

θ=1.22 λb

where b is the diameter of the aperture.

Using this formula, we can calculate whether two spots can be resolved, as long as we know the wavelength of light and the diameter of the aperture.

If the angleθ formed by the rays of the observed images is equal to or greater than the angle θcalculated by the equation, then the images can be easily resolved as two separate points. If the angle θ formed by the rays of the observed images is less than the calculated value ofθ, the two points will not be resolved and will appear as one single point.

θ= sd

sd=1.22 λ

b

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s

d

Figure 12 – two stars just resolved through an aperture

θ θ


Example 1

Two small red spots 0.50mm apart reflect light of wavelength 650nm in to a camera that has an aperture of5.0mm. What is the maximum distance that the camera can resolve the two spots?

Example 2

A telescope with an aperture of 4.00 cm is used to view a collection of stars that are 4.00×1012m away that give out light of wavelength570nm. Determine if the telescope could resolve two stars separated by7.5×107m.

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Example 3

To read the headlines in a newspaper, you need to resolve points that are about 1.00 cm apart. If a spy camera is to be put in orbit 200 km above the Earth, could an aperture of 5.0mm be used to read the newspaper headlines (assume a wavelength of600nm)? What is the minimum aperture size required to read the headline?

Example 4

The pupil of the human eye has a diameter of 2.00mm. The Pleiades is a cluster of stars in the Taurus constellation. The stars are136.2 parsec or4.20×1018m away from Earth. Pleione and Atlas are6.20×1015mapart.

(a) Determine the angular separation of Pleione and Atlas for an observer on Earth.

(b) Determine if the human eye can resolve these two stars as separate images.

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Resolving Power

Diffraction gratings can be used to distinguish between two wavelengths, λ1 and λ2, that are very close to each other.

In figure 13, if the angular separation is too small, the two lines will not be seen as distinct.

The resolving power (or resolvance) of a diffraction grating is defined as:

R= λΔ λ

λ is the average between the two wavelengths andΔ λ is their difference. The higher the resolving power, the smaller the differences in wavelength that can be resolved. It can be shown thatR=mN where m is the order at which lines are observed and N is the total number of slits or rulings on the diffraction grating.

λΔ λ

=mN

Δ λ= λmN

This means the smallest difference in wavelength can be resolved with a large number of slits or rulings.

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Figure 13 – the angular separation between red and blue waves of light

angular separation

θ /degrees

Intensity


Example

A beam of light containing different wavelengths is incident on a diffraction grating. The grating has 600 lines per mm and is2.0cmwide. The average wavelength in the beam is620nm. Calculate the least difference in wavelength that can be resolved by this grating in the second order.

This question is about diffraction and interference.Light of wavelength 620 nm from a laser is incident on a single rectangular slit of width 0.45 mm.

(a) After passing through the slit, the light is incident on a screen that is a distance of3.4m from the slit. Calculate the distance between the centre and the first minimum of the diffraction pattern. [2]

(b) The laser in (a) is replaced by two identical lasers so that the light from both lasers illuminates the slit. The lasers are both 6.0 m from the slit. The two diffraction patterns on the screen are resolved according to the Rayleigh criterion. [3]

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(i) State what is meant by the Rayleigh criterion.

(ii) The minimum separation of the two laser beams isx. Determinex.

(c) Compare the appearance of a single-slit diffraction pattern formed by laser light to that formed by a source of white light. [2]

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Improving Resolution

Whenever electromagnetic radiation passes through an aperture, it is diffracted. This causes problems if you need to resolve the part of the radiation that come different places. We have seen how this affects optical instruments and how we can improve the situation by increasing the size of the aperture. An alternative way of increasing resolution is to use radiation of a different wavelength.

How is more information stored on a DVD than a CD?

A CD is a shiny disc covered in small bumps that are called ‘pits’. These pits are about 5×10−7m wide. As the CD rotates in the CD reader, a laser of wavelength 780 nm is reflected off the shiny surface into a detector. If a pit goes past, the detector senses a change and sends a signal to the computer. The computer decodes the signal converting it into music, pictures or something else. To get more information on the disc, you would need to make the pits smaller. However, if you do that they diffract the light so much that the detector could not resolve the difference between the pits and the gaps. To solve this problem, a laser of shorter wavelength (640 nm) is used.

Blu-ray

DVDs and CDs use a red laser (longer wavelength) which has its own limitations. With Blu-ray discs the pits are much smaller, and the use of a blue laser (shorter wavelength 405nm) allows more information to be stored on a Blu-ray disc.

CD 700MB DVD 4GB−8.5GB

Blu-ray 25GB−100GB

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The Electron Microscope

A microscope is used to give a large image of a small object, but there is no point of making a big image if you cannot resolve the details. This is a problem with microscopes that use light – their resolving power is limited by the aperture size and wavelength. The closest points that can be resolved by an optical microscope are about 200 nm apart. To increase the resolving power of a microscope you could use shorter wavelengths, but then you would not be able to see them because the light would not be in the visible spectrum. An electron microscope uses high energy electrons that have wavelengths as small as0.2nm, enabling them to resolve points as close as0.1nm (about the size of an atom). You can’t see the image directly, but the electrons are detected sensors and the image is produced electronically.

The Radio Telescope

A radio telescope detects the radio waves emitted by celestial bodies and converts them into images. These radio waves have wavelengths of about20cm, so to have a good resolution, the telescopes must be very big. The Lovell Radio telescope at Jodrell Bank has an aperture of76m.

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http://upload.wikimedia.org/wikipedia/commons/d/d3/Lovell_Telescope_5.jpg

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