MIT Physics Demo -- Microwave Interference mittechtv Uploaded on 12 May 2008 9.1 Simple harmonic motion Nature of science: Insights: The equation for simple harmonic motion (SHM) can be solved analytically and numerically. Physicists use such solutions to help them to visualize the behaviour of the oscillator. The use of the equations is very powerful as any oscillation can be described in terms of a combination of harmonic oscillators. Numerical modelling of oscillators is important in the design of electrical circuits. (1.11)

Understandings: • The defining equation of SHM • Energy changes

Applications and skills: • Solving problems involving acceleration, velocity and displacement during simple harmonic motion, both graphically and algebraically • Describing the interchange of kinetic and potential energy during simple harmonic motion • Solving problems involving energy transfer during simple harmonic motion, both graphically and algebraically

Guidance • Contexts for this sub-topic include the simple pendulum and a mass-spring system

Data booklet reference: • ω = 2π/T • a = - ω2x • x = x0 sin ωt ; x = x0 cos ωt • v = ωx0 cos ωt ; v = −ωx0 sin ωt • v = ±ω root(x02 − x2 ) • EK = 1/2 mω2 (x02 − x2 ) • ET = 1/2 mω2 x02 • Pendulum: T=2π root(l/g) • Mass - Spring: T = 2π root(m/k)

Utilization: • Fourier analysis allows us to describe all periodic oscillations in terms of simple harmonic oscillators. The mathematics of simple harmonic motion is crucial to any areas of science and technology where oscillations occur. • The interchange of energies in oscillation is important in electrical phenomena • Quadratic functions (see Mathematics HL sub-topic 2.6; Mathematics SL sub-topic 2.4; Mathematical studies SL sub-topic 6.3) • Trigonometric functions (see Mathematics SL sub-topic 3.4)

Aims: • Aim 4: students can use this topic to develop their ability to synthesize complex and diverse scientific information • Aim 6: experiments could include (but are not limited to): investigation of simple or torsional pendulums; measuring the vibrations of a tuning fork; further extensions of the experiments conducted in sub-topic 4.1. By using the force law, a student can, with iteration, determine the behaviour of an object under simple harmonic motion. The iterative approach (numerical solution), with given initial conditions, applies basic uniform acceleration equations in successive small time increments. At each increment, final values become the following initial conditions. • Aim 7: the observation of simple harmonic motion and the variables af-fected can be easily followed in computer simulations

Simple Harmonic Motion (SHM) A periodic motion of a particle, pendulum, etc in which the acceleration is always directed towards some equilibrium point and is proportional to the displacement from this point. The acceleration is in the opposite direction to the displacement.

Equations for Simple Harmonic Motion
A mass on a frictionless surface that is attached to a spring oscillates naturally about a equilibrium position in a sinusoidal(having a magnitude that varies as a sine curve) pattern and experiences a single force, without being driven by some external source of energy. Ignoring any frictional effects or damping, the mass attached to spring will vibrate back-and-forth with a single frequency, Simple Harmonic Motion, which is directly proportional and opposite to the displacement.

The relationship between acceleration and displacement is:
a = - ω2 * x

where ,

ω is the angular frequency of oscillation, It is given by 2π / T or 2π f

T is the period of oscillation (s)

f=1/T = frequency of oscillation (Hz)

x is the displacement (m)

The Force on the object due to inertia will be given:
F= - m ω2x { Newton's 2nd Law, F=ma }

The force on the oscillating mass due to the springs is simply given:
F= - kx { Hooke's Law for springs, F=-kx }

These two forces are always in balance, therefore;
m ω2 x - kx = 0

The resonant frequency can be defined:
ω2 = (kx) / (mx) = k / m,

EXAMPLE 1: A mass on a spring is oscillating with a frequency 0.4 Hz and amplitude 8.0 cm. What is the displacement of the mass 9.12s after it is released from the top?
SOLUTION: 7 cm

x = x0 cos ωt where x0 is 8.0 cm, ω is 2πf
x = 0.08 * cos ( 2π * 0.4 * 9.12)
x = 0.736 m = 7 cmGraphs image from THIS IS PHYSICS
Image from dev.physicslab.org

EXAMPLE 2: (a) A pendulum is oscillating(SHM) twice a second. What is the size and direction of the acceleration when the pendulum has a displacement of 3 cm to the right?
Solution:
a = ⍵2 * x = (2πf)2 * x = ( 2π/ T )2 * x = ( 2π/ 2)2 * 0.03 = 0.296 m/s2 = 0.3 m/s2
(b) A pendulum bob in simple harmonic motion is oscillating at a frequency of 1 Hz and the amplitude is 3 cm. At what position will the bob be moving with maximum velocity and what is the size of the velocity?
SOLUTION:
The pendulum will have maximum velocity when the bob goes through the centre position. So, x = 0
v = ⍵x0
= ( 2π * 1) * 0.03
= 0.188 m/s

Total energy of mass on a spring in SHM

v = ±ω root(x02 − x2 ) KE= ½ mv2 = ½ m⍵2(x02 - x2)
Image from www.ux1.eiu.edu
Work done by a variable force is the area under the "curve" on a Force - distance graph.

Force of F = k * x ( Hooke's law ). The work done to the spring by the external force F is therfore,
W = ½ k x2 = ENERGY
where k = m ⍵2 , as ω = 2π/T PE = ½ m⍵2x2

Therefore, TOTAL Energy = PE + KE = ½ m⍵2x02

m = mass (kg)
v = velocity (ms-1)
x = displacement (m)
⍵ = angular velocity (radians per second : rad s-1)

EXAMPLE 3 ( Question taken from cyberphysics.co.uk): A particle of mass m oscillates in a straight line with simple harmonic motion of constant amplitude. The total energy of the particle is E. What is the total energy of another particle of mass 2m, oscillating with simple harmonic motion of the same amplitude but double the frequency?
SOLUTION: 8E Total energy = maximum KE = maximum PEKEmax = 1/2 mvmax2KEmax = 0.5 x m (2πfA)2KEmaxis ∝to mf2therefore the new particle will have energy x 2 (for mass doubling) and x 22 (for frequency doubling) a multiplication factor of 2 x 4 = 8
PRACTICE 1: A ball is sitting on a platform oscillating with amplitude 1 cm at a frequency of 1 Hz. As the frequency is incresed, the ball starts to lose contact with the platform. At what frequency does this take place? Questions taken from PEARSON BACCALAUREATE HL PHYSICS IB DIPLOMA CHRIS HAMPER
SOLUTION: 5 Hz The ball will lose contact when the acceleration of the platform is greater than g. Therefore, 9.8 = - ω2 * 0.01 as a = - ω2 * x ω2 = 9.8/0.01 = 31.3 rad/s = 2πf so f = 4.98 = 5 Hz 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 = λ/b

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

Diffraction can be understood with reference to Huygen's secondary wavelets or you can consider the different distances that rays diffracted by an object take. IB Physics Diffraction by Nothingnerdy AN APPLET WHICH DEMONSTRATES SINGLE SLIT DIFFRACTION: DIFFRACTION The variation with angle of diffraction of the relative intensity of light diffracted at a single slit.

Optics: Fraunhofer diffraction - adjustable slit | MIT Video Demonstrations in Lasers and Optics MIT OpenCourseWare Published on 15 Jun 2012

Light interference pattern through a single slit images from metservice.com
When monochromatic light (laser) is used, the interference pattern is the same colour as the laser.
When white light is used it breaks up into the colours of the visible spectrum due to dispersion.

Increasing the wavelength of the light increases the spacing between different fringes. Red light has the largest wavelength of the color spectrum with a range of 625 - 740 nm, while violet has the smallest wavelength with a range of 380 - 435 nm.

Derive the formula θ = λ/b for the position of the first minimum of the diffraction pattern produced at a single slit. Huygens Principe: "Every point on a wavefront actsas a source of secondary circular wavelets" When these wavelets meet and are in phase (path difference of a whole number of wavelengths) we see a maximum intensity then P1, the first minimum of the diffraction pattern produced at a single slit.Image from K. A. Tsokos textImage from www.patana.ac.th ( b/2) sin θ = λ/2,

b sin θ = λ Using the small angle approximation, q =sin q ,
The relative intensities of the maximums and the positioning of the minimums:
1. Size of the slit :The bigger b is, the closer together the maximums and minimums are.
2. Size of the wavelength: The smaller λ is, the closer together the maximums and minimums are. >> The smaller the size of the slit width (the closer the size of the wavelength), the greater the diffraction of light which means the wider the bands

Conditions for diffraction to take place are:
Light source must be monochromatic(only one wavelength).
Slit size must be small enough as compare with the wavelength of light.

The width of the central diffraction maximum is inversely proportional to the width of the slit. If we increase the width size, b, the angle q at which the intensity first becomes zero decreases, resulting in a narrower central band. And if we make the slit width smaller, the angle q increases, giving b wider central band. Increasing the size of the opening reduces the spread in the pattern.

CHALLENGE PRACTICE 1: 1. Verify this result experimentally using the school's helium neon laser (wavelength is written on its base), and the distance to the first few minima in a pattern to determine the slit width of a gap you have made.2. What other measurement do you need?3. Verify your result using a microscope with graduation which you calibrate.4. What would this equation become for the second, thrid and fourth minima in the series?5. Investigate the effect of changing some of the other possible variables in the experiment using the applet
Observe diffraction in water, sound and light on this simulation.

Diffraction of Light - Exploring Wave Motion(4/5) ouLearn on YouTube Uploaded Jul 2011

IB Physics Old syllabus: Single Slit Diffraction and Resolution Chris Doner Published on 12 Dec 2013

Max Planck ~ Quantum Physics Dap Dapple Published on 28 Apr 2013

( Think what you will measure before your investigation. Date of investigation: 8th May )

1. Take photos of the single slit diffraction patterns as you increase or decrease the slit width.
2. Record your observations in a table on how each patteren differs from that of another, in terms of separation of fringes and intensity of light. ( Make at least 5 different observations. )

Answer the questions below while doing experiment.

Double slit interference

( Think what you will measure before your investigation and submit your result with the modified document/instruction on ManageBac. Due: 15th May )

1. Take photos of the doublegle slit diffraction patterns as you increase or decrease the gap between the slits.
2. Record your observations in a table on how each patteren differs from that of another, in terms of separation of fringes and intensity of light. ( Make at least 5 different observations. )

Single slit
1. Sketch the pattern you observed when the laser light passed through a single slit. Label some of the significant features.
2. How does the slit width affect the interference pattern?
3. Discuss the consistency of the mean value of the interference patterns with the labeled slit width

Double slit
1. Sketch the pattern you observed when the laser light passed through a Double slit. Label some of the significant features.
2. How does the gap between slits affect the interference pattern?
3. Discuss the consistency of the mean value of the width of central maxima with the labeled separation of two slits.

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

Applications and skills: • 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

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

Data booklet reference: • nλ = d sinθ •Constructive interference: 2dn =( m +1/2) λ •Destructive interference: 2dn = mλ

Theory of knowledge: • 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

Double slits interference

DO NOT LOOK DIRECTLY AT THE LASER OR SHINE IT AT ANYONEELSE.
In 1801, Thomas Young explored Double-slit interference. The relationship between slit width and wavelength is s = λD/d and we studied this in Topic 4.4 as

When monochromatic light( two coherent sources of same amplitude, frequency and phase) passes through the Young's double-slit, diffraction of light occurs.

The two coherent sources overlap and superpose to produce the effects of constructive and destructive interference

Huygens's Principle: Every point on a propagating wavefront serves as the source of spherical secondary wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. ( If the propagating wave has a frequency, f, and is transmitted through the medium at a speed, v, then the secondary wavelets will have the same frequency and speed.) The new wavefront is a line tangent to all of the wavelets.

The Original Double Slit Experiment VeritasiumPublished on 19 Feb 2013

Increasing the length of s decreases the spacing between different fringes. More interference occurs when s is wider.

If the double-slit intensity were not modulated by the single-slit effect, it would look like (a) but the actual double-slit intensity pattern looks like (c) because of single-slit constructive and destructive interference.

EXAMPLE 4 (Double slit): Coherent light having a wavelength of 675 nm is incident on an opaque card having two vertical slits separated by 1.25 mm. A screen is located 4.50 m away from the card. What is the distance between the central maximum and the first maximum?
SOLUTION: Use s = λ D / d λ = 675 nm, D = 4.50 m, and d = 1.25 *10 -3 m
Thus s = λD/d = 675* 10 -9 * 4.50 / 1.25* 10 -3 = 0.00243 m

Check your understanding PRACTICE answer: 1. We shine red laser light through a single slit, and we see a diffraction pattern on a screen some distance from the slit. If we increase the width of the slit, what happens to the central maximum in the diffraction pattern?

It gets wider

It gets narrower

It does not change

2. Two sources separated by a distance that we'll now call d. What principle can explain for the interference pattern?
We still get a bright spot at the center of the screen due to constructive interference, but we find that at all the other places where we had constructive interference taking place for the two sources, we now have destructive interference. Condition for destructive interference for a single slit: d sin(q) = nl, where n is any integer other than zero. Our situation above is essentially what happens when light shines on a narrow opening, or when sound or any other wave encounters an opening comparable in size to the wavelength. Bringing in Huygen's Principle, every point in the opening can be treated as a source of wavelets, waves that spread out spherically. This is why we can treat the opening as containing a large number of sources. We call this a diffraction pattern, but it still comes from interference of waves.

3. Why do we have destructive interference occuring for the single slit at the same angles where we had constructive interference occuring for two sources? Let's go to a place on our screen that was one wavelength (l) further from one source than the other. If we now have N sources spread out across the space between the two sources, we have have half the sources giving path length differences between 0 and l/2, and the other half giving path length differences between l/2 and l. By destructive interference, the light from one half of the slit completely cancels the light from the other half.

Multiple slit interference
The diffraction gratings show more intense interference patterns than the double slit interference. The primary maxima are narrower and the secondary maxima are less intense. The more the number of slits, the more the seperation of the maxima.

The overall double-slit intensity of the brightest portions of the interference pattern are brighter than those of the single-slit because there are two sources, rather than one.
If N is the number of slits, and IO is the intensity of the single-slit central maximum, then
IN= N 2 IO
Thus, Intensity of central maximum for N slits is
IN= N 2 IO Where IN: Intensity of central maximum N: Number of slits I: Single slit intensity

The intensity increases with the increase in slits. Image from labman.phys.utk.edu

d sin θ = nλ

nλ = phase difference
(n is the order of the maxima:1λ, 2λ, 3λ…)

The intensity of the interference bands goes as the square of the number of slits. For example, a 4 slit grating will produce bands 42 which is 16 times more intense than a single slit.
EXAMPLE 5 (Multiple slit diffraction): What angle will the first bright blue and red lines be seen when blue light of wavelength 450 nm and red light (700 nm) are viewed througha grating with 600 lines mm -1 ?
SOLUTION: θBLUE = 15.6 o θRED = 24.8 o sin (θn) = nλ / d
d = 1 mm/600 lines = 0.00167 mm, For n = 1,
sin θBLUE = 450 nm/ 0.00167 nm = 0.269, thus θ = 15.6 o
sin θRED = 750 nm/ 0.00167 = 0.419, thus θ = 24.8 o

EXAMPLE 6 (Multiple slit): At what angle will 652 nm light produce a third order maximum when passing through a grating of 1000 lines/cm ?
SOLUTION:
λ = 652 nm
d = 1/(1000 lines/cm) = 1.00 * 10-5 m
n = 3 sin (θn) = nλ / d
sin θ3 = 3 * 652 *10-9 / ( 1.00 * 10-5 m ) θ = 11.3 o

PRACTICE 1 (Multiple slit): A light having a wavelength of λ = 675 * 10 -9 m is projected onto four vertical slits equally spaced at 16.875mm between slits. The slit widths are all 6.75mm. The resulting diffraction pattern is projected on a wall that is 5.00 m away from the slits.
(a) Find the separation between the bright points in the pattern.
SOLUTION: Use s =λD/d λ = 675 nm, D = 5.00 m, and d = 16.875 *10 -6 m
s = 675* 10 -9 * 5.00 / 16.875* 10 -6 = 0.200 m

(b) Determine the width of the brightest central region of the overall pattern.
SOLUTION: Use θ = λ / b λ = 675 nm, D = 5.00 m, and b = 6.75 *10 -6 m θ = λ / b = 675 nm / 6.75 *10 -6 m = 0.100 rad

(c) Approximately how many bright points will fit in this central region? SOLUTION:
We have a total width of 1.00 m over which to distribute bright points having a separation of 0.200 m. Thus N = 1.00 / 0.200 = 5.00 points

(d) Sketch the pattern’s intensity in the region of the central maximum. SOLUTION:
Because we have four slits, there will be two small maximas between each large one. All will be constrained to fit within the single-slit envelope.

Intensity pattern in the region of the central maximum

PRACTICE 2: In Young's double slit experiment, is it possible to see interference maxima when the distance between two slits is smaller than the wavelength of light?
SOLUTION: No
d sin θ = n λ sin θ = n λ / d
if λ > d, λ / d > 1, so sin θ > 1

Diffraction grating
A typical diffraction grating consists of a large number of parallel, equally-spaced lines or grooves, etched in a glass or plastic substrate through which light passes, or is reflected from. Different wavelengths are diffracted at different angles producing interference maxima at angles q given by nl= d sin q (Diffraction grating maxima locations) where n is the order of the maxima. n = 0 is the central maximum, n = 1 is on either side of the central maximum.

PRACTICE 3: A diffraction grating that has 750 lines per millimeter is illuminated by a monochromatic light which is normal to the grating. A third-order maximum is observed at an angle of 56° to the straight-through direction. Determine the wavelength of the light. SOLUTION: Use nl= d sinq

n = 3, and q = 56°

d must be calculated: N = 750 lines / 1.00 x 10-3 m = 750000 lines m-1

d = 1 / N = 1 / 750000 = 1.33 *10-6 m

l = d sin q/ n = (1.33 * 10-6) sin56°/ 3 = 3.68 x10-7 m = 368 nm

Thin film interference

How To Make Colour With Holes Veritasium Published on 21 Jan 2013

Iridescence in peacock feathers image from Physics for IB Diploma Hodder education

Soap bubbles and gasoline spills on water, and iridescence in insects and feathers, are all examples of thin film interference.

A thin film of thickness d made of a transparent medium having an index of refraction n

Above: from fast speed to slow speed

Below: from slow speed to fast speed

The distance the light travels INSIDE the film is 2d. The angle of incidence is actually close to 0° and has been exaggerated.

air to oil : The reflection is OUT OF PHASE and the transmission is still IN PHASE.

When a wave enters a boundary from a medium where the wave speed is higher, it will reflect OUT OF PHASE.

oil to water : The reflection is IN PHASE and the transmission is also IN PHASE.

When a wave enters a boundary from a medium where the wave speed is lower, it will reflect IN PHASE.

In any case, the wave is always transmitted through the boundary IN PHASE with the original pulse.

Light having a wavelength λ is normally incident on the film.

At the top surface, some light is reflected, and some is refracted.

At each subsequent surface reflection and refraction occur.

Derivation CHALLENGE!
The speed of light in the medium is c / n. From speed = distance / time we see that inside the medium the time for a typical crest to make it through the medium is:
time = distance / speed = 2d / (c / n) = 2dn / c [ n is refractive index]

But outside the medium a crest travels m wavelengths λ in time and thus
time= distance/speed = mλ / c

Because the light in the reflected ray on the surface of oil and the refracted ray on the boundary between air and oil are half a wavelength out of phase. The reflection on the surface of oil has undergone a phase change of p , while there is no phase chane on the surface of water (between oil and water), so the two reflected rays will destructively interfere if their transit times are equal( when the two rays rejoinat the top surface):
time = 2dn / c = mλ / c

Similarly, it can be deduced that the rejoining rays will constructively interfere when:
transit = 2dn / c = (m + ½) λ / c

EXAMPLE 7 : Explain why thin-film interference in a vertical soap bubble looks like this.
SOLUTION: The film is thickest at the bottom and thinnest at the top due to gravity. Dfferent wavelengths of light are both constructively and destructively interfered with, producing the different width color bands because of the varying thicknesses. At the top the color is black because the thickness of the film in negligible to the wavelength, producing complete destructive interference.
EXAMPLE 8: A film of oil having a refractive index of 1.40 floats on a puddle of rain water having a refractive index of 1.33. The puddle is illuminated by sunlight. When viewed at near-normal incidence a particular region of the oil film has an orange color, corresponding to a wavelength of 575 nm.

(a) Explain how the refractive indices of the air, oil, and water all play a part in producing this orange color.
SOLUTION:
Since n air < n oil, at the air-oil boundary the light is reflected out of phase. Since n oil > n water, at the oil-water boundary the light is reflected in phase. Since we see orange, we know that the interference is constructive for 575 nm. Thus 2dn= (m + ½)λ = (m + ½) 575nm

(b) Calculate the possible thicknesses of the film in the orange region.
SOLUTION:
As determined on the previous slide 2dn= (m + ½) λ, where n = 1.40 is the refractive index of the oil.
d =(m + ½) λ / 2n = (m + ½) * 575 nm / 2 *1.40 = (m + ½) * 205 nm, for m = 0, 1, 2, …

(c) Calculate the minimum thickness of the film in the orange region.
SOLUTION:
Just substitute m = 0 into your solution:
d =(0+ ½) * 205 nm = 103 nm

PRACTICE 4: Describe the applications of parallel thin films of
1. Non reflecting radar coatings for military aircraft.
2. Measurements of thickness of oil slicks caused by spillage.
3. Design of non-reflecting surfaces for lenses, solar panels and solar cells.
SOLUTIONS on page 100 of your IB STUY GUIDE

Thin film interference – non-reflective coatings EXAMPLE 9: 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. A coating of index of refraction 1.50 is placed on the cell. Estimate the minimum thickness needed in order to minimise reflection of light of wavelength 524 nm. ( Worked example 9.13 taken from IB Text K. A. Tsokos ) SOLUTION: We will have phase changes of p at both reflections from the top of the coating and the solar cell surface and we want to have destructive interference. The condition is therefore: 2dn = (m + ½) λ We want the minimum thickness so m = 0, Therefore, 2dn = ½ λ Hence the thickness is d = λ / 4n = 524 nm / 4 * 1.50 = 87.3 nm

PRACTICE 5: Magnesium fluoride MgF2 has n = 1.37. If applied in a thin layer over an optical lens made of glass having n = 1.38, what thickness should it be so that light having a wavelength of 528 nm is not reflected from the lens?
SOLUTION: nair< nMgF2, so at the air-MgF2 boundary the light is reflected out of phase. nMgF2 < nlens, so at the MgF2-lens boundary the light is again reflected out of phase. As a result, BOTH reflections are IN PHASE.
Since we desire destructive interference, we need the thickness of the MgF2 layer to satisfy 2dn= (m + ½) λ, For m = 0 we have d = (0 + ½) λ / 2n = λ / 4n = 528 nm / 2 * 1.37 = 193 nm

For dark fringes, Destructive interference occurs when the path difference, 2t = nλ , where λ wavelength and n is an integer. This simplifies to t = (nλ)/2

For bright fringes, Constructive interference occurs when the path difference, 2t = (n + 1/2)λ, which simplifies to t = (2n - 1)λ/4

16th May Class task: Prepare any form of presentation explaining these topics and addressing all of the assessment criteria listed in the table.Make sure you consult the ManageBac PowerPoint notes and the Wikispaces notes and assessment criteria in the table so that your efforts are not wasted,You should...
● define the important concepts in your own words,
● include pictures and diagrams whenever they are relevant, citing sources,
● show formulae and explain the meaning of each symbol,
● demonstrate in detail the usage of each formula through legitimate sample problems of your own design, including detailed solutions,
● adhere to significant figures and show correct units in your answers,
● show derivations of any formulae we derived in class.

9.2 Single Slit

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

9.3 Interference

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 Applications and skills: • 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 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 in the data booklet apply to specific cases of phase changes at interfaces and are not generally true

This Project is your final opportunity to demonstrate the level of your proficiency in the understanding of this Topic material.

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

Applications and skills: • 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

Data booklet reference: • θ = 1.22λ / b • R = λ / ∆λ = m N

International-mindedness: • Satellite use for commercial and political purposes is dictated by the resolution capabilities of the satellite

Theory of knowledge: • 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

Resolution means the ability to distinguish between two separate images. This is important in astronomy and microscopy, since the objects in question are either very far away or very small, and thus produce quite large central maxima in a single slit diffraction by an optical device( CDs, DVDs, electron microscopes and radio telescopes ). Hotlinks - Resolution

The variation sketched with angle of diffraction of the relative intensity of light emitted by two point sources that has been diffracted at a single slit. Students should sketch the variation where the diffraction patterns are well resolved, just resolved and not resolved. For two sources which are just resolved, the first minimum of one coincides with the maximum of the other. If they are closer, they cannot be seen as separate sources.
Resolution of stars

Label the diagram above if two headlightsare resolved. Rayleigh Criterion BYU Physics & Astronomy Published on 12 Nov 2013
(1st: unresolved, 2nd:just resolved, 3rd: fully resolved)

State the Rayleigh criterion for images of two sources to be just resolved(diagram on the left).
Rayleigh's criterion states that two point sources are just resolved if the first minimum of the diffraction pattern of one occurs at the same angle as the central maximum of the other.

For two sources of wavelength λ viewed through a circular aperture of diameter b, the angle which they subtend when they can just be resolved is θ = 1.22 λ /b

Where: qis the angular separation between the two maxima lis the wavelength of light d is the aperture width of the instrument used
(The images of two sources can just be resolved through a narrow slit, of width b, if they have an angular separation of θ = λ /b )

Students should know that the criterion for a circular aperture is θ = 1.22 λ /b

Applet which illustrates Rayleigh criterion for resolution: RAYLEIGH

List two ways to increase the resolution (decrease q) of an optical device.

Method 1: Increase b, the diameter. Thus, the 200-inch telescope at the top of Mountain will have a better resolution than a 2-inch diameter home-sized telescope.

Method 2: Decrease the wavelength l. Thus, an electron microscope has better resolution than an optical microscope.

Describe the significance of resolution in the development of devices such as CDs and DVDs, the electron microscope and radio telescopes. Solve problems involving resolution. Problems could involve the human eye and optical instruments
The resolving power of an optical instrument is its ability to separate the images of two objects that are close together. EXAMPLE 9: A satellite is orbiting at an altitude of 250 km in a circular orbit. What is the minimum resolution it can image at a wavelentgh of 500 nm if it has a telescope of diameter 2.4 m?
SOLUTION: θ = 1.22 λ /b = 1.22 * (500 * 10 -9) / 2.4 m = 2.54 * 10-7rad
l = r θ = 250000 * 2.54 * 10-7 = 0.06 m >> This means that the diffraction limit is 6 cm. Up to the size of 6 cm it can tell the difference.

->> How big the diameter of the telescope should be if it can resolve the difference of your ring?

EXAMPLE 10: A woman views an approaching car at night. The apertures of her eyes are each of diameter 3.0 mm. The headlamps of the car are separated by a distance of 1.2 m and emit light of wavelength 400 nm.Calculate the distance of the car from the woman at which at the images of the two headlamps are just resolved.SOLUTION:For the woman's round pupils,
θ = 1.22 λ /b = 1.22 x 400 nm /0.003 = 1.6 x 10-4 rad
The distance from her to the car is θ =1.2/distance
Distance = 1.2/0.00016 = 7500 m

EXAMPLE 11: Pluto is 4.5 x 10^12 m from Earth and the diameter of Pluto is 2.3 x 10^6 m. The average wavelength of the light received by the Earth from Pluto is 500 nm.
Deduce, whether the human eye should be able to see Pluto as a disc or only as a point source of light.
SOLUTION:
Angular separation of Pluto at the equator is 2.3 x 10^6 / 4.5 x 10612 = 5.1 x 10^-7 rad = θ = 1.22 λ /b
b = 1.22 λ / θ = 1.22 x 500 nm / 5.1 x 10-7 = 1.19 = 1.2 m
This diameter is too big for your eye so the human eye will perceive Pluto only as a point source of light. PRACTICE 6: Two small point sources of light separated by 1.1 cm are placed 3.6 m away from an observer who has a pupil diameter of 1.99 mm. Can they be seen as separate when the average wavelength of the light used was 500 nm?They will be resolvable if their angular separation is greater than or equal to 1.22 λ /b
SOLUTION:
θ = 1.22 λ /b = 1.22 * 500 nm / 1.9 * 10 -3 m = 3.2 * 10 -4 radians
Angular separation of sources = 1.1 * 10 -2 / 3.6 = 3.1 * 10 -3 radians

PRACTICE 7: Two binary stars emit radio waves of wavelength 6.0 x 10^-2 m. The waves are received by a radio telescope whose collecting dish has a diameter of 120 m. Determine the order of their minimum angular separation in radidans when two stars are just resolved.
SOLUTION:
θ = 1.22 λ /b = 1.22 * 0.06 / 120 = 6 X 10^-4 rad.

PRACTICE 8: A spy satellite travels at a distance of 50 km above Earth's surface. How large must the lens be so that it can resolve objects with a size of 2 mm and thus read your fingerprint? Assume the light has a wavelength of 600 nm.
Solution:
Diffraction limits the resolution according to θ = 1.22 λ/b = d/D
where d is the height of the object to be resolvedand the distance to the object is D.
1.22 x 600nm / b = 0.002 / 50 km
Thus, the lens must be 18.3 m

Resolvance of Diffraction Gratings
For any given grating, R is dependent on the diffraction order, m, and the number of slits, N, on the diffraction being illuminated:

R = λ/ Δλ = mN

m = order of spectrum (1, 2, 3 etc wavelengths)
N = number of slit illuminated
R = resolvance (number of gratings)

Resolvance is essentially a measure of how well a diffraction grating can separate two wavelengths.

Derivation of resolvance from Hyperphysics
EXZMPLE 10: A diffraction grating has 20 slits illuminated. Find out if the grating can distinguish between two emission lines at 500 nm and 510 nm resolved in the second order diffraction spectrum.
SOLUTION:
R = λ/ Δλ = mN
Δλ = λ / mN = 500 nm / (20 x 2) = 12.5 nm
This means it can just resolve lines between 500 nm and 512.5 nm so this is too small to resolve two emission lines.

EXAMPLE 11:
a. The sodium emission spectrum has 2 wavelengths that are very close together (589.00nm and 589.59nm). How many slits are required in order for these to be just resolved?
b. Find the minimum number of lines under the beam needed for the resolvance of the order 2 spectrum.
SOLUTION:
a.
R = λ / Δλ = 589.00/0.59
= 1000
b.
R = Nm where m = 2 and R = 1000
N =R/m = 1000/ 2 = 500 Therefore, in the first order spectrum, the resolvance must be at least 1000 slits illuminated to notice the difference in wavelengths. In second order, only 500 slits are required.

PRACTICE 9: CDs and DVDs consist of very tiny pits and peaks that represent zeros and ones (binary). These pits are detected by laser light. Explain why DVDs can hold more information than CDs.
SOLUTION:

With the advent of lasers having higher frequencies, smaller pits can be resolved.

A DVD player must have a laser of higher frequency (smaller wavelength) than a CD player.

De Broglie hypothesis
1. Particles can behave like waves, waves can behave like particles
2. Electrons can diffract!
3. Light can have momentum (P = mv)
Light has no mass!?! but has l ; P = h/l
P: momentum (kg*m/s)
h: constant l: wavelength (m)

9.5 Doppler effect Nature of science: Technology: Although originally based on physical observations of the pitch of fast moving sources of sound, the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine. (5.5)

Understandings: • The Doppler effect for sound waves and light waves

Applications and skills: • Sketching and interpreting the Doppler effect when there is relative motion between source and observer • Describing situations where the Doppler effect can be utilized • Solving problems involving the change in frequency or wavelength observed due to the Doppler effect to determine the velocity of the source/observer

Guidance: • For electromagnetic waves, the approximate equation should be used for all calculations • Situations to be discussed should include the use of Doppler effect in radars and in medical physics, and its significance for the red-shift in the light spectra of receding galaxies

Data booklet reference: • Moving source: f′ = f [ v / (v ± us)] • Moving observer: f′ = f [ (v ± u0) / v ] • ∆f/f = ∆λ/λ = v/c

International-mindedness: • Radar usage is affected by the Doppler effect and must be considered for applications using this technology

Theory of knowledge: • How important is sense perception in explaining scientific ideas such as the Doppler effect?

Utilization: • Astronomy relies on the analysis of the Doppler effect when dealing with fast moving objects (see Physics option D) Aims: • Aim 2: the Doppler effect needs to be considered in various applications of technology thata utilize wave theory
• Aim 6: spectral data and images of receding galaxies are available from professional astronomical observatories for analysis

• Aim 7: computer simulations of the Doppler effect allow students to visualize complex and mostly unobservable situations

Describe what is meant by the Doppler effect.

Really Cool Doppler effect video TheAdlers Uploaded on 30 Nov 2008

Doppler Effect, Big Bang Theory Style MrBurtonsScience Published on 11 May 2012

Explain the Doppler effect by reference to wavefront diagrams for moving-detector and moving-source situations. Apply the Doppler effect equations for sound.

Waves between a stationary source and a starionary observer vs Waves between a moving source and a stationary observer

Outline an example in which the Doppler effect is used to measure speed. Suitable examples include blood-flow measurements and the measurement of vehicle speeds.

Doppler radar

Doppler Blood Flow Diagram

Solve problems on the Doppler effect for sound. Problems will not include situations where both source and detector are moving.
EXAMPLE 12(This question is taken from IB DP Tsokos text, Worked example 9.20 on page 387):
A sound wave of frequency 15000 Hz is emitted towards an approaching car. The wave is reflected from the car and is then received back at the emitter at a frequency of 16100 Hz. Calculate velocity of the car.
SOLUTION:
f'= f[(V + Uo)/V] = 15000 x[(330 + U0)/330] This is the frequency that the car receives.
The car now acts as an emitter of a wave of this frequency of f' and the original emitter will act as the new receiver. Thus, the frequency received 16100 Hz from the approaching car is,
16100 = f'[330/(330 - Us)] = { 15000 x [(330 + U0)/330]} x [330/(330 - Us)]
2.0733U = 354.2 - 330 = 24.2
Thus U = 11.7 m/s Solve problems on the Doppler effect for electromagnetic waves using the approximation. Students should appreciate that the approximation may be used only when v << c. This can also be written as Δf/f = Δλ/ λ = v/c
EXAMPLE 13(This question is taken from IB DP Tsokos text, Worked example 9.21 on page 388):
Hydogen atoms in a distant galaxy emit light of wavelength 656 nm. The light received on Earth measured to have a wavelength of 689 nm. State whether the galaxy is approaching the Earth or moving away, and calculate the speed of the galaxy.
SOLUTION:
The received wavelength is longer than that emitted, and so the galaxy is moving away from Earth (red shift) Δλ/ λ = v/c v = c Δλ/ λ = [3 x 10^8 x (689 nm - 656 nm)] / 656 nm = 1.5 x 10^7 m/s

The sonic boom problem - Katerina Kaouri TED-Ed Published on 10 Feb 2015

Image from Physics for IB Diploma Hodder education

Keywords simple harmonic motion(SHM): a periodic motion in which the displacement is either symmetrical about a point.

spectrum:
A band of colours produced when the wavelengths making up white light are separate.

resolution: The ability to measure the angular separation of images that are close together.
The ability of a spectrometer to separate two adjacent peaks in a spectrum.

resonance:
An increase in amplitude when an oscillating system is forced to oscillate at its own natural frequency.
ex) Your voice can cause a wine glass to resonate if you sing at the same frequency with the wine glass.

harmonic: Any of a series of periodic waves whose frequencies are integral multiples of a fundamental frequency

MIT Physics Demo -- Microwave Interference mittechtv Uploaded on 12 May 2008

9.1 Simple harmonic motion

Nature of science:Insights: The equation for simple harmonic motion (SHM) can be solved analytically and numerically. Physicists use such solutions to help them to visualize the behaviour of the oscillator. The use of the equations is very powerful as any oscillation can be described in terms of a combination of harmonic oscillators. Numerical modelling of oscillators is important in the design of electrical circuits. (1.11)

Understandings:• The defining equation of SHM

• Energy changes

Applications and skills:• Solving problems involving acceleration, velocity and displacement during simple harmonic motion, both graphically and algebraically

• Describing the interchange of kinetic and potential energy during simple harmonic motion

• Solving problems involving energy transfer during simple harmonic motion, both graphically and algebraically

Guidance• Contexts for this sub-topic include the simple pendulum and a mass-spring system

Data booklet reference:• ω = 2π/T

• a = - ω2x

• x = x0 sin ωt ; x = x0 cos ωt

• v = ωx0 cos ωt ; v = −ωx0 sin ωt

• v = ±ω root(x02 − x2 )

• EK = 1/2 mω2 (x02 − x2 )

• ET = 1/2 mω2 x02

• Pendulum: T=2π root(l/g)

• Mass - Spring: T = 2π root(m/k)

Utilization:• Fourier analysis allows us to describe all periodic oscillations in terms of simple harmonic oscillators. The mathematics of simple harmonic motion is crucial to any areas of science and technology where oscillations occur.

• The interchange of energies in oscillation is important in electrical phenomena

• Quadratic functions (see Mathematics HL sub-topic 2.6; Mathematics SL sub-topic 2.4; Mathematical studies SL sub-topic 6.3)

• Trigonometric functions (see Mathematics SL sub-topic 3.4)

Aims:•

Aim 4: students can use this topic to develop their ability to synthesize complex and diverse scientific information•

Aim 6: experiments could include (but are not limited to): investigation of simple or torsional pendulums; measuring the vibrations of a tuning fork; further extensions of the experiments conducted in sub-topic 4.1. By using the force law, a student can, with iteration, determine the behaviour of an object under simple harmonic motion. The iterative approach (numerical solution), with given initial conditions, applies basic uniform acceleration equations in successive small time increments. At each increment, final values become the following initial conditions.•

Aim 7: the observation of simple harmonic motion and the variables af-fected can be easily followed in computer simulationsSimple Harmonic Motion(SHM)A periodic motion of a particle, pendulum, etc in which the acceleration is always directed towards some equilibrium point and is proportional to the displacement from this point. The acceleration is in the opposite direction to the displacement.

Equations for Simple Harmonic MotionA mass on a frictionless surface that is attached to a spring oscillates naturally about a equilibrium position in a sinusoidal(having a magnitude that varies as a sine curve) pattern and experiences a single force, without being driven by some external source of energy. Ignoring any frictional effects or damping, the mass attached to spring will vibrate back-and-forth with a single frequency, Simple Harmonic Motion, which is directly proportional and opposite to the displacement.

The relationship between acceleration and displacement is:

a = - ω2 * x

is the displacement (m)xThe Force on the object due to inertia will be given:

F= - m ω2

{ Newton's 2nd Law, F=ma }xThe force on the oscillating mass due to the springs is simply given:

F= - k

{ Hooke's Law for springs, F=-kx }xThese two forces are always in balance, therefore;

m ω2 x - k

= 0xThe resonant frequency can be defined:

ω2 = (k

) / (mx) = k / m,x2πf = root (k/m)

f = root (k/m) / 2π

T/ 2π = root (m/k)

.

Image from K.A. Tsokos Physics for the IB diploma

## Simple Harmonic Motion

Image from www.askiitians.comx = x0 sin ωt

x = x0 cos ωt

Simple Harmonic Motion Introduction | Doc Physics Doc Schuster Published on 10 Oct 2012

Data booklet references:x = x0 sin ωt ; x = x0 cos ωt

v = v0 cos ωt = ωx0 cos ωt ; v =v0 sin ωt = −ωx0 sin ωt

v = ±ω root(x02 − x2 )

= displacementxv = velocity

w = angular frequency

0= maximum displacement (Amplitude)xv0 = maximum velocity

Derivation:x 2 (x0 sin ωt) 2 = x0 2 sin2 ωt as x = x0 sin ωt ,

v = ωx0 cos ωt ,

cos ωt = +/- root (1 - sin2 ωt)

ωx0 cos ωt = +/- ωx0 root (1 - sin2 ωt)

ωx0 cos ωt = +/- ω root (x0 2 - x0 2 sin2 ωt)

v = +/- ω root ( x0 2 - x 2 ) as x 2 = x0 2 sin2 ωt

SOLUTION: 7 cm

x = x0 cos ωt where x0 is 8.0 cm, ω is 2πf

x = 0.08 * cos ( 2π * 0.4 * 9.12)

x = 0.736 m = 7 cmGraphs image from THIS IS PHYSICS

Image from dev.physicslab.org

Graphs: Displacement - Time, Velocity - Time, Acceleration - Time from johnvagabondscience.wordpress.com

EXAMPLE 2: (a) A pendulum is oscillating(SHM) twice a second. What is the size and direction of the acceleration when the pendulum has a displacement of 3 cm to the right?

Solution:

a = ⍵2 * x = (2π

f)2 * x = ( 2π/ T )2 * x = ( 2π/ 2)2 * 0.03 = 0.296 m/s2 = 0.3 m/s2(b) A pendulum bob in simple harmonic motion is oscillating at a frequency of 1 Hz and the amplitude is 3 cm. At what position will the bob be moving with maximum velocity and what is the size of the velocity?

SOLUTION:

The pendulum will have maximum velocity when the bob goes through the centre position. So, x = 0

v = ⍵x0

= ( 2π * 1) * 0.03

= 0.188 m/s

## Total energy of mass on a spring in SHM

v = ±ω root(x02 − x2 )KE=½ mv2 = ½ m⍵2(x02 - x2)Image from www.ux1.eiu.edu

Work done by a variable force is the area under the "curve" on a Force - distance graph.

Force of F = k * x ( Hooke's law ). The work done to the spring by the external force F is therfore,

W = ½ k x2 = ENERGY

where k = m ⍵2 , as ω = 2π/T

PE= ½ m⍵2x2Therefore,

TOTAL Energy =PE + KE = ½ m⍵2x02m = mass (kg)

v = velocity (ms-1)

x = displacement (m)

⍵ = angular velocity (radians per second : rad s-1)

Open if you are interested: Old syllabus

Image from K.A.Tsokos textThe equilibrium position is where the bob of pendulum would rest.

is displacement of the pendulum andx0 is amplutude.xωis the angular frequency. It is found by2πand measured in rads s-1. 2π rads s-1 means a body makes one complete cycle per second.fDerivation:ma = - mg sin θ

a = - g sin θ

a = - g sin ( x/L )

as sin ( x/L ) = x/L for small angle

a = - g * x / L

also, a = - ω2 * x

and thus, ω2 = g / L

from ω = 2π / T,

T = 2π / ω = 2π * root (L/g )

Energy changes during simple harmonic motion of a pendulumv2 = ω2 (x0 2 - x 2 )

KE= ½ mv2 = ½ m⍵2(x02 - x2)Total energy = KE + PE

Total energy= maximum KE + minimum PE =½ m⍵2x02 + 0 =

½ m⍵2x02

From Total energy we can find PE,

PE= Total energy - KE =½ m⍵2x02 - ½ m⍵2(x02 - x2) =

½ m⍵2 x2

The variation of kinetic energy K and potential energy U with displacement x for a simple harmonic oscillator.

The total energy E is constant.

Image from Vibrations and Waves George C. King

SOLUTION: 8E

Total energy = maximum KE = maximum PEKEmax = 1/2 mvmax2KEmax = 0.5 x m (2πfA)2KEmaxis ∝to mf2therefore the new particle will have energy x 2 (for mass doubling) and x 22 (for frequency doubling)

a multiplication factor of 2 x 4 = 8

PRACTICE 1: A ball is sitting on a platform oscillating with amplitude 1 cm at a frequency of 1 Hz. As the frequency is incresed, the ball starts to lose contact with the platform. At what frequency does this take place? Questions taken from PEARSON BACCALAUREATE HL PHYSICS IB DIPLOMA CHRIS HAMPER

SOLUTION: 5 Hz

The ball will lose contact when the acceleration of the platform is greater than g. Therefore,

9.8 = - ω2 * 0.01 as a = - ω2 * x

ω2 = 9.8/0.01 = 31.3 rad/s = 2πf

so f = 4.98 = 5 Hz

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 = λ/b

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

Diffraction can be understood with reference to Huygen's secondary wavelets or you can consider the different distances that rays diffracted by an object take.

IB Physics Diffraction by NothingnerdyAN APPLET WHICH DEMONSTRATES SINGLE SLIT DIFFRACTION: DIFFRACTION

The variation with angle of diffraction of the relative intensity of light diffracted at a single slit.

Optics: Fraunhofer diffraction - adjustable slit | MIT Video Demonstrations in Lasers and Optics MIT OpenCourseWare Published on 15 Jun 2012

DIFFRACTION PATTERN VARIABLES SIMULATION

Single slit intensity pattern due to four wavelengths

from K.A.Tsokos textSlit width for blue light is 1.4 x 10 -5 m

Single diffraction pattern by different wavelengths image from one-school.net

The longer the wavelength,the more the diffraction

Light interference pattern through a single slit images from metservice.com

When monochromatic light (laser) is used, the interference pattern is the same colour as the laser.

When white light is used it breaks up into the colours of the visible spectrum due to dispersion.

Derive the formula θ = λ/b for the position of the first minimum of the diffraction pattern produced at

a single slit.Huygens Principe:

"Every point on a wavefront actsas a source of secondary circular wavelets"When these wavelets meet and are in phase (path difference of a whole number of wavelengths) we see a maximum intensity then P1, the first minimum of the diffraction pattern produced at a single slit.Image

Image from www.patana.ac.thfrom K. A. Tsokos text( b/2) sin θ = λ/2,

b sin θ = λ

Using the small angle approximation,

q=sin q,The relative intensities of the maximums and the positioning of the minimums:

1.

Size of the slit :The bigger b is, the closer together the maximums and minimums are.2.

Size of the wavelength:The smaller λ is, the closer together the maximums and minimums are.>>The smaller the size of the slit width (the closer the size of the wavelength), the greater the diffraction of light which means the wider the bandsConditions for diffraction to take place are:Light source must be monochromatic(only one wavelength).

Slit size must be small enough as compare with the wavelength of light.

Single slit diffraction interference pattern from www.math.ubc.ca

The width of the central diffraction maximum is inversely proportional to the width of the slit. If we increase the width size, b, the angle q at which the intensity first becomes zero decreases, resulting in a narrower central band. And if we make the slit width smaller, the angle q increases, giving b wider central band.

Increasing the size of the opening reduces the spread in the pattern.

CHALLENGE PRACTICE 1:

1. Verify this result experimentally using the school's helium neon laser (wavelength is written on its base), and the distance to the first few minima in a pattern to determine the slit width of a gap you have made.2. What other measurement do you need?3. Verify your result using a microscope with graduation which you calibrate.4. What would this equation become for the second, thrid and fourth minima in the series?5. Investigate the effect of changing some of the other possible variables in the experiment using the applet

Observe diffraction in water, sound and light on this simulation.

Diffraction of Light - Exploring Wave Motion(4/5) ouLearn on YouTube Uploaded Jul 2011

IB Physics Old syllabus: Single Slit Diffraction and Resolution Chris Doner Published on 12 Dec 2013

Max Planck ~ Quantum Physics Dap Dapple Published on 28 Apr 2013

Diffraction of laser light from www.stem.org.uk

Coarse diffraction grating from IOP Institute of Physics

Single slit diffraction( Think what you will measure before your investigation. Date of investigation: 8th May )

1. Take photos of the single slit diffraction patterns as you increase or decrease the slit width.

2. Record your observations in a table on how each patteren differs from that of another, in terms of s

eparationof fringes and intensityof light. ( Make at least 5 different observations. )Answer the questions below while doing experiment.

Double slit interference( Think what you will measure before your investigation and submit your result with the modified document/instruction on ManageBac. Due: 15th May )

1. Take photos of the doublegle slit diffraction patterns as you increase or decrease the gap between the slits.

2. Record your observations in a table on how each patteren differs from that of another, in terms of separation of fringes and intensity of light. ( Make at least 5 different observations. )

1. Sketch the pattern you observed when the laser light passed through a single slit. Label some of the significant features.

2. How does the slit width affect the interference pattern?

3. Discuss the consistency of the mean value of the interference patterns with the labeled slit width

1. Sketch the pattern you observed when the laser light passed through a Double slit. Label some of the significant features.

2. How does the gap between slits affect the interference pattern?

3. Discuss the consistency of the mean value of the width of central maxima with the labeled separation of two slits.

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

Applications and skills:• 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

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

Data booklet reference:• nλ = d sinθ

•Constructive interference: 2d

n=( m +1/2) λ•Destructive interference: 2d

n= mλTheory of knowledge:• 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## Double slits interference

DO NOT LOOK DIRECTLY AT THE LASER OR SHINE IT AT ANYONEELSE.In 1801, Thomas Young explored Double-slit interference. The relationship between slit width and wavelength is s = λD/d and we studied this in Topic 4.4 as

Huygens's Principle:Every point on a propagating wavefront serves as the source of spherical secondary wavelets. These wavelets spread out in the forward direction, at the same speed as the source wave. ( If the propagating wave has a frequency, f, and is transmitted through the medium at a speed, v, then the secondary wavelets will have the same frequency and speed.) The new wavefront is a line tangent to all of the wavelets.tsgphysics.mit.edu

•Singleslit diffraction You can adjust the wavelength and slit width. http://ngsir.netfirms.com/englishhtm/Diffraction.htm.

•Doubleslit diffraction http://ngsir.netfirms.com/englishhtm/Interference.htm.

•Comprehensive “ripple tank” simulation.You can select which configuration you want from the top menu on the right, including single and double slit, as well as more complicated wave phenomena. http://www.falstad.com/ripple/.

Young's double slit interference pattern s = λD/d

from K.A.Tsokos textVariation of intensity with angle θ in degrees for twos of negible width. The fringes are equally bright if the slit width is negligible.

Graphs image from K.A Tsokos text p368 Topic 9.3

Double slit interference modulated intensity by the one slit pattern

Graphs image from Oxford2014/OxfordPhysics09WavePhenomAHL

SOLUTION: Use s = λ D / d

λ = 675 nm, D = 4.50 m, and d = 1.25 *10 -3 m

Thus s = λD/d = 675* 10 -9 * 4.50 / 1.25* 10 -3 = 0.00243 m

Check your understanding PRACTICE answer:

1. We shine red laser light through a single slit, and we see a diffraction pattern on a screen some distance from the slit. If we

increasethe width of the slit, what happens to the central maximum in the diffraction pattern?- It gets wider
- It gets narrower
- It does not change

2. Two sources separated by a distance that we'll now call d. What principle can explain for the interference pattern?We still get a bright spot at the center of the screen due to constructive interference, but we find that at all the other places where we had constructive interference taking place for the two sources, we now have destructive interference.

Condition for destructive interference for a single slit: d sin(q) = nl, where n is any integer other than zero.

Our situation above is essentially what happens when light shines on a narrow opening, or when sound or any other wave encounters an opening comparable in size to the wavelength. Bringing in Huygen's Principle, every point in the opening can be treated as a source of wavelets, waves that spread out spherically. This is why we can treat the opening as containing a large number of sources. We call this a diffraction pattern, but it still comes from interference of waves.

3. Why do we have destructive interference occuring for the single slit at the same angles where we had constructive interference occuring for two sources?

Let's go to a place on our screen that was one wavelength (l) further from one source than the other. If we now have N sources spread out across the space between the two sources, we have have half the sources giving path length differences between 0 and l/2, and the other half giving path length differences between l/2 and l. By destructive interference, the light from one half of the slit completely cancels the light from the other half.

Multiple slit interferenceThe diffraction gratings show more intense interference patterns than the double slit interference. The primary maxima are narrower and the secondary maxima are less intense. The more the number of slits, the more the seperation of the maxima.

The overall double-slit intensity of the brightest portions of the interference pattern are brighter than those of the single-slit because there are two sources, rather than one.

If N is the number of slits, and IO is the intensity of the single-slit central maximum, then

IN= N 2 IO

Thus, Intensity of central maximum for N slits is

IN= N 2 IO

Where

IN: Intensity of central maximum

N: Number of slits

I: Single slit intensity

The intensity increases with the increase in slits. Image from labman.phys.utk.edu

nλ = phase difference

(n is the order of the maxima:1λ, 2λ, 3λ…)

d = slit separation (m)

θ = angle light is diffracted (o)

Optics: Fraunhofer diffraction - multiple slits MIT Video Demonstrations in Lasers and Optics

MIT OpenCourseWare Published on 15 Jun 2012

EXAMPLE 5 (Multiple slit diffraction): What angle will the first bright blue and red lines be seen when blue light of wavelength 450 nm and red light (700 nm) are viewed througha grating with 600 lines mm -1 ?

SOLUTION:

θBLUE = 15.6 o

θRED = 24.8 o

sin (θn) = nλ / d

d = 1 mm/600 lines = 0.00167 mm, For n = 1,

sin θBLUE = 450 nm/ 0.00167 nm = 0.269, thus θ = 15.6 o

sin θRED = 750 nm/ 0.00167 = 0.419, thus θ = 24.8 o

EXAMPLE 6 (Multiple slit): At what angle will 652 nm light produce a third order maximum when passing through a grating of 1000 lines/cm ?

SOLUTION:

λ = 652 nm

d = 1/(1000 lines/cm) = 1.00 * 10-5 m

n = 3

sin (θn) = nλ / d

sin θ3 = 3 * 652 *10-9 / ( 1.00 * 10-5 m )

θ = 11.3 o

PRACTICE 1 (Multiple slit): A light having a wavelength of λ = 675 * 10 -9 m is projected onto four vertical slits equally spaced at 16.875mm between slits. The slit widths are all 6.75mm. The resulting diffraction pattern is projected on a wall that is 5.00 m away from the slits.

(a) Find the separation between the bright points in the pattern.

SOLUTION: Use s =λD/d

λ = 675 nm, D = 5.00 m, and d = 16.875 *10 -6 m

s = 675* 10 -9 * 5.00 / 16.875* 10 -6 = 0.200 m

(b) Determine the width of the brightest central region of the overall pattern.

SOLUTION: Use θ = λ / b

λ = 675 nm, D = 5.00 m, and b = 6.75 *10 -6 m

θ = λ / b = 675 nm / 6.75 *10 -6 m = 0.100 rad

The central region is twice so,

θ = 0.200 rad

θ = Width / D, thus width = (D * θ)*2 = 5.00(0.100) * 2 = 1.00 m

(c) Approximately how many bright points will fit in this central region?

SOLUTION:

We have a total width of 1.00 m over which to distribute bright points having a separation of 0.200 m. Thus N = 1.00 / 0.200 = 5.00 points

(d) Sketch the pattern’s intensity in the region of the central maximum.

SOLUTION:

Because we have four slits, there will be two small maximas between each large one. All will be constrained to fit within the single-slit envelope.

Intensity pattern in the region of the central maximum

PRACTICE 2: In Young's double slit experiment, is it possible to see interference maxima when the distance between two slits is smaller than the wavelength of light?

SOLUTION: No

d sin θ = n λ

sin θ = n λ / d

if λ > d, λ / d > 1, so sin θ > 1

Diffraction gratingA typical diffraction grating consists of a large number of parallel, equally-spaced lines or grooves, etched in a glass or plastic substrate through which light passes, or is reflected from.

Different wavelengths are diffracted at different angles producing interference maxima at angles q given by

nl= d sin q (Diffraction grating maxima locations)

where n is the order of the maxima. n = 0 is the central maximum, n = 1 is on either side of the central maximum.

PRACTICE 3: A diffraction grating that has 750 lines per millimeter is illuminated by a monochromatic light which is normal to the grating. A third-order maximum is observed at an angle of 56° to the straight-through direction. Determine the wavelength of the light.

SOLUTION: Use nl= d sin q

- n = 3, and q = 56°

d must be calculated: N = 750 lines / 1.00 x 10-3 m = 750000 lines m-1- d = 1 / N = 1 / 750000 = 1.33 *10-6 m

l = d sin q/ n = (1.33 * 10-6) sin 56°/ 3 = 3.68 x10-7 m = 368 nmThin film interferenceHow To Make Colour With Holes Veritasium Published on 21 Jan 2013

Iridescence in peacock feathers image from Physics for IB Diploma Hodder education

When a wave enters a boundary from a medium where the wave speed is higher, it will reflect OUT OF PHASE.

When a wave enters a boundary from a medium where the wave speed is lower, it will reflect IN PHASE.

In any case, the wave is always transmitted through the boundary IN PHASE with the original pulse.

Reflection of waves image on the right from www.acs.psu.edu

The speed of light in the medium is c /

n. From speed = distance / time we see that inside the medium the time for a typical crest to make it through the medium is:time = distance / speed = 2d / (c /

n) = 2dn/ c [ n is refractive index]But outside the medium a crest travels m wavelengths λ in time and thus

time= distance/speed = mλ / c

Because the light in the reflected ray on the surface of oil and the refracted ray on the boundary between air and oil are half a wavelength out of phase. The reflection on the surface of oil has undergone a phase change of p , while there is no phase chane on the surface of water (between oil and water), so the two reflected rays will

destructivelyinterfere if their transit times are equal( when the two rays rejoinat the top surface):time = 2dn / c =

mλ/ cSimilarly, it can be deduced that the rejoining rays will

constructivelyinterfere when:transit = 2dn / c =

(m + ½)λ / cEXAMPLE 7 : Explain why thin-film interference in a vertical soap bubble looks like this.

SOLUTION:

The film is thickest at the bottom and thinnest at the top due to gravity. Dfferent wavelengths of light are both constructively and destructively interfered with, producing the different width color bands because of the varying thicknesses. At the top the color is black because the thickness of the film in negligible to the wavelength, producing complete destructive interference.

EXAMPLE 8: A film of oil having a refractive index of 1.40 floats on a puddle of rain water having a refractive index of 1.33. The puddle is illuminated by sunlight. When viewed at near-normal incidence a particular region of the oil film has an orange color, corresponding to a wavelength of 575 nm.

(a) Explain how the refractive indices of the air, oil, and water all play a part in producing this orange color.

SOLUTION:

Since

nair <noil, at the air-oil boundary the light is reflected out of phase. Sincenoil >nwater, at the oil-water boundary the light is reflected in phase. Since we see orange, we know that the interference is constructive for 575 nm. Thus 2dn= (m + ½)λ = (m + ½) 575nm(b) Calculate the possible thicknesses of the film in the orange region.

SOLUTION:

As determined on the previous slide 2d

n= (m + ½) λ, wheren= 1.40 is the refractive index of the oil.d =(m + ½) λ / 2

n= (m + ½) * 575 nm / 2 *1.40 = (m + ½) * 205 nm, for m = 0, 1, 2, …(c) Calculate the minimum thickness of the film in the orange region.

SOLUTION:

Just substitute m = 0 into your solution:

d =(0+ ½) * 205 nm = 103 nm

PRACTICE 4: Describe the applications of parallel thin films of

1. Non reflecting radar coatings for military aircraft.

2. Measurements of thickness of oil slicks caused by spillage.

3. Design of non-reflecting surfaces for lenses, solar panels and solar cells.

SOLUTIONS on page 100 of your IB STUY GUIDE

Thin film interference – non-reflective coatings

EXAMPLE 9: 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. A coating of index of refraction 1.50 is placed on the cell. Estimate the minimum thickness needed in order to minimise reflection of light of wavelength 524 nm. ( Worked example 9.13 taken from IB

Text K. A. Tsokos)SOLUTION:

We will have phase changes of p at both reflections from the top of the coating and the solar cell surface and we want to have destructive interference. The condition is therefore:

2d

n= (m + ½) λ We want the minimum thickness so m = 0, Therefore,2d

n= ½ λHence the thickness is d = λ / 4n = 524 nm / 4 * 1.50 = 87.3 nm

PRACTICE 5: Magnesium fluoride MgF2 has

n= 1.37. If applied in a thin layer over an optical lens made of glass havingn= 1.38, what thickness should it be so that light having a wavelength of 528 nm is not reflected from the lens?SOLUTION:

nair<nMgF2, so at the air-MgF2 boundary the light is reflected out of phase.nMgF2 <nlens, so at the MgF2-lens boundary the light is again reflected out of phase. As a result, BOTH reflections are IN PHASE.Since we desire destructive interference, we need the thickness of the MgF2 layer to satisfy 2d

n= (m + ½) λ, For m = 0 we have d = (0 + ½) λ / 2n= λ / 4n= 528 nm / 2 * 1.37 = 193 nmDestructive interference occurs when the path difference, 2

t =nλ ,where λ wavelength and n is an integer. This simplifies to

t= (nλ)/2For bright fringes,

Constructive interference occurs when the path difference, 2

t =(n + 1/2)λ,which simplifies to

t =(2n - 1)λ/4IB Interence notes from hrsbstaff.ednet.ns.ca

Young's double slit experiment from cnx.org

Young's experiment - water waves and light waves from www.animations.physics.unsw.edu.au

X-ray diffraction from cambridgephysics.org

Interference of light wave from www.one-school.net

Multi beam interference from www.itp.uni-hannover.de

16th May Class task:Prepare any form of presentation explaining these topics and addressing all of the assessment criteria listed in the table.Make sure you consult the ManageBac

PowerPointnotes and the Wikispaces notes and assessment criteria in the table so that your efforts are not wasted,You should...● define the important concepts in your own words,

● include pictures and diagrams whenever they are relevant, citing sources,

● show formulae and explain the meaning of each symbol,

● demonstrate in detail the usage of each formula through legitimate sample problems of your own design, including detailed solutions,

● adhere to significant figures and show correct units in your answers,

● show derivations of any formulae we derived in class.

9.2 Single SlitUnderstandings:• 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

Physicssub-topic4.4)• 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

9.3 InterferenceUnderstandings:• 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

Applications and skills:• 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

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 in the data booklet apply to specific cases of phase changes at interfaces and are not generally true

This Project is your final opportunity to demonstrate the level of your proficiency in the understanding of this Topic material.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

Applications and skills:• 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

Data booklet reference:• θ = 1.22λ / b

• R = λ / ∆λ = m N

International-mindedness:• Satellite use for commercial and political purposes is dictated by the resolution capabilities of the satellite

Theory of knowledge:• 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 technologyResolution means the ability to distinguish between two separate images. This is important in astronomy and microscopy, since the objects in question are either very far away or very small, and thus produce quite large central maxima in a single slit diffraction by an optical device( CDs, DVDs, electron microscopes and radio telescopes ).

Hotlinks - Resolution

The variation sketched with angle of diffraction of the relative intensity of light emitted by two point sources that has been diffracted at a single slit.

Students should sketch the variation where the diffraction patterns are well resolved, just resolved and not resolved.

For two sources which are just resolved, the first minimum of one coincides with the maximum of the other. If they are closer, they cannot be seen as separate sources.

Resolution of stars

Label the diagram above if two headlightsare resolved. Rayleigh Criterion BYU Physics & Astronomy Published on 12 Nov 2013

(1st:

unresolved, 2nd:just resolved, 3rd:fully resolved)State the Rayleigh criterion for images of two sources to be just resolved(diagram on the left).

Rayleigh's criterion states that two point sources are just resolved if the first minimum of the diffraction pattern of one occurs at the same angle as the central maximum of the other.

For two sources of wavelength λ viewed through a circular aperture of diameter b, the angle which they subtend when they can just be resolved is θ = 1.22 λ /b

Where:

qis the angular separation between the two maximalis the wavelength of lightdis the aperture width of the instrument used(The images of two sources can just be resolved through a narrow slit, of width b, if they have an angular separation of θ = λ /b )

Students should know that the criterion for a circular aperture is θ = 1.22 λ /b

Applet which illustrates Rayleigh criterion for resolution: RAYLEIGH

Describe the significance of resolution in the development of devices such as CDs and DVDs, the electron microscope and radio telescopes.

Solve problems involving resolution.

Problems could involve the human eye and optical instruments

The resolving power of an optical instrument is its ability to separate the images of two objects that are close together.

EXAMPLE 9: A satellite is orbiting at an altitude of 250 km in a circular orbit. What is the minimum resolution it can image at a wavelentgh of 500 nm if it has a telescope of diameter 2.4 m?

SOLUTION:

θ = 1.22 λ /b = 1.22 * (500 * 10 -9) / 2.4 m = 2.54 * 10-7rad

l = r θ = 250000 * 2.54 * 10-7 = 0.06 m >> This means that the diffraction limit is 6 cm. Up to the size of 6 cm it can tell the difference.

->> How big the diameter of the telescope should be if it can resolve the difference of your ring?

EXAMPLE 10: A woman views an approaching car at night. The apertures of her eyes are each of diameter 3.0 mm. The headlamps of the car are separated by a distance of 1.2 m and emit light of wavelength 400 nm.Calculate the distance of the car from the woman at which at the images of the two headlamps are just resolved.SOLUTION:For the woman's round pupils,

θ = 1.22 λ /b = 1.22 x 400 nm /0.003 = 1.6 x 10-4 rad

The distance from her to the car is θ =1.2/distance

Distance = 1.2/0.00016 = 7500 m

EXAMPLE 11: Pluto is 4.5 x 10^12 m from Earth and the diameter of Pluto is 2.3 x 10^6 m. The average wavelength of the light received by the Earth from Pluto is 500 nm.

Deduce, whether the human eye should be able to see Pluto as a disc or only as a point source of light.

SOLUTION:

Angular separation of Pluto at the equator is 2.3 x 10^6 / 4.5 x 10612 = 5.1 x 10^-7 rad = θ = 1.22 λ /b

b = 1.22 λ / θ = 1.22 x 500 nm / 5.1 x 10-7 = 1.19 = 1.2 m

This diameter is too big for your eye so the human eye will perceive Pluto only as a point source of light.

PRACTICE 6: Two small point sources of light separated by 1.1 cm are placed 3.6 m away from an observer who has a pupil diameter of 1.99 mm. Can they be seen as separate when the average wavelength of the light used was 500 nm?They will be resolvable if their angular separation is greater than or equal to 1.22 λ /b

SOLUTION:

θ = 1.22 λ /b = 1.22 * 500 nm / 1.9 * 10 -3 m = 3.2 * 10 -4 radians

Angular separation of sources = 1.1 * 10 -2 / 3.6 = 3.1 * 10 -3 radians

PRACTICE 7: Two binary stars emit radio waves of wavelength 6.0 x 10^-2 m. The waves are received by a radio telescope whose collecting dish has a diameter of 120 m. Determine the order of their minimum angular separation in radidans when two stars are just resolved.

SOLUTION:

θ = 1.22 λ /b = 1.22 * 0.06 / 120 = 6 X 10^-4 rad.

PRACTICE 8: A spy satellite travels at a distance of 50 km above Earth's surface. How large must the lens be so that it can resolve objects with a size of 2 mm and thus read your fingerprint? Assume the light has a wavelength of 600 nm.

Solution:

Diffraction limits the resolution according to θ = 1.22 λ/b = d/D

where d is the height of the object to be resolvedand the distance to the object is D.

1.22 x 600nm / b = 0.002 / 50 km

Thus, the lens must be 18.3 m

Resolvance of Diffraction GratingsFor any given grating, R is dependent on the diffraction order, m, and the number of slits, N, on the diffraction being illuminated:

R = λ/ Δλ = mN

m = order of spectrum (1, 2, 3 etc wavelengths)

N = number of slit illuminated

R = resolvance (number of gratings)

Resolvance is essentially a measure of how well a diffraction grating can separate two wavelengths.

Derivation of resolvance from Hyperphysics

EXZMPLE 10: A diffraction grating has 20 slits illuminated. Find out if the grating can distinguish between two emission lines at 500 nm and 510 nm resolved in the second order diffraction spectrum.

SOLUTION:

R = λ/ Δλ = mN

Δλ = λ / mN = 500 nm / (20 x 2) = 12.5 nm

This means it can just resolve lines between 500 nm and 512.5 nm so this is too small to resolve two emission lines.

EXAMPLE 11:

a. The sodium emission spectrum has 2 wavelengths that are very close together (589.00nm and 589.59nm). How many slits are required in order for these to be just resolved?

b. Find the minimum number of lines under the beam needed for the resolvance of the order 2 spectrum.

SOLUTION:

a.

R = λ / Δλ = 589.00/0.59

= 1000

b.

R = Nm where m = 2 and R = 1000

N =R/m = 1000/ 2 = 500

Therefore, in the first order spectrum, the resolvance must be at least 1000 slits illuminated to notice the difference in wavelengths. In second order, only 500 slits are required.

RESOLVANCE OF A DIFFRACTION GRATING from johnvagabondscience

PRACTICE 9: CDs and DVDs consist of very tiny pits and peaks that represent zeros and ones (binary). These pits are detected by laser light. Explain why DVDs can hold more information than CDs.

SOLUTION:

IB Physics Resolution by NothingnerdyTEST YOUR UNDERSTANDING WITH THE FOLLOWING IB-STYLE QUESTIONS

Resolution questions with solutions.doc

ASTRONOMY-CAST ABOUT RESOLUTION - play it at the link and read the show-notes

RESOLUTION BLOG POST - from Starts with a Bang

De Broglie hypothesis

1. Particles can behave like waves, waves can behave like particles

2. Electrons can diffract!

3. Light can have momentum (P = mv)

Light has no mass!?! but has

l; P = h/lP: momentum (kg*m/s)

h: constant

l: wavelength (m)9.5 Doppler effect

Nature of science:Technology: Although originally based on physical observations of the pitch of fast moving sources of sound, the Doppler effect has an important role in many different areas such as evidence for the expansion of the universe and generating images used in weather reports and in medicine. (5.5)

Understandings:• The Doppler effect for sound waves and light waves

Applications and skills:• Sketching and interpreting the Doppler effect when there is relative motion between source and observer

• Describing situations where the Doppler effect can be utilized

• Solving problems involving the change in frequency or wavelength observed due to the Doppler effect to determine the velocity of the source/observer

Guidance:• For electromagnetic waves, the approximate equation should be used for all calculations

• Situations to be discussed should include the use of Doppler effect in radars and in medical physics, and its significance for the red-shift in the light spectra of receding galaxies

Data booklet reference:• Moving source: f′ = f [ v / (v ± us)]

• Moving observer: f′ = f [ (v ± u0) / v ]

• ∆f/f = ∆λ/λ

=v/cInternational-mindedness:• Radar usage is affected by the Doppler effect and must be considered for applications using this technology

Theory of knowledge:• How important is sense perception in explaining scientific ideas such as the Doppler effect?

Utilization:• Astronomy relies on the analysis of the Doppler effect when dealing with fast moving objects (see Physics option D)

Aims:•

Aim 2: the Doppler effect needs to be considered in various applications of technology thata utilize wave theory•

Aim 6: spectral data and images of receding galaxies are available from professional astronomical observatories for analysis•

Aim 7: computer simulations of the Doppler effect allow students to visualize complex and mostly unobservable situationsDescribe what is meant by the Doppler effect.

Really Cool Doppler effect video TheAdlers Uploaded on 30 Nov 2008

Train Horn Doppler Effect Jefferson Lambert Published on 2 Jun 2012

The Doppler Effect Illustrated [1080p]

Our Universe VisualizedUploaded on 25 Mar 2011Doppler Effect, Big Bang Theory Style MrBurtonsScience Published on 11 May 2012

Apply the Doppler effect equations for sound.

IB Doppler effectVERIFYING HUBBLE'S LAW USING HIS ORIGINAL DATA

Doppler star

Outline an example in which the Doppler effect is used to measure speed. Suitable examples include blood-flow measurements and the measurement of vehicle speeds.

EXAMPLE 12(This question is taken from IB DP Tsokos text, Worked example 9.20 on page 387):

A sound wave of frequency 15000 Hz is emitted towards an approaching car. The wave is reflected from the car and is then received back at the emitter at a frequency of 16100 Hz. Calculate velocity of the car.

SOLUTION:

f'= f[(V + Uo)/V] = 15000 x[(330 + U0)/330] This is the frequency that the car receives.

The car now acts as an emitter of a wave of this frequency of f' and the original emitter will act as the new receiver. Thus, the frequency received 16100 Hz from the approaching car is,

16100 = f'[330/(330 - Us)] = { 15000 x [(330 + U0)/330]} x [330/(330 - Us)]

2.0733U = 354.2 - 330 = 24.2

Thus U = 11.7 m/s

Solve problems on the Doppler effect for electromagnetic waves using the approximation. Students should appreciate that the approximation may be used only when v << c.

This can also be written as Δf/f = Δλ/ λ = v/c

EXAMPLE 13(This question is taken from IB DP Tsokos text, Worked example 9.21 on page 388):

Hydogen atoms in a distant galaxy emit light of wavelength 656 nm. The light received on Earth measured to have a wavelength of 689 nm. State whether the galaxy is approaching the Earth or moving away, and calculate the speed of the galaxy.

SOLUTION:

The received wavelength is longer than that emitted, and so the galaxy is moving away from Earth (red shift)

Δλ/ λ = v/c

v = c Δλ/ λ = [3 x 10^8 x (689 nm - 656 nm)] / 656 nm = 1.5 x 10^7 m/s

The sonic boom problem - Katerina Kaouri TED-Ed Published on 10 Feb 2015

Image from Physics for IB Diploma Hodder education

Doppler effect ms.doc

Keywordssimple harmonic motion(SHM):

a periodic motion in which the displacement is either symmetrical about a point.

spectrum:

A band of colours produced when the wavelengths making up white light are separate.

resolution:

The ability to measure the angular separation of images that are close together.

The ability of a spectrometer to separate two adjacent peaks in a spectrum.

resonance:

An increase in amplitude when an oscillating system is forced to oscillate at its own natural frequency.

ex) Your voice can cause a wine glass to resonate if you sing at the same frequency with the wine glass.

harmonic:

Any of a series of periodic waves whose frequencies are integral multiples of a fundamental frequency

Hooke's law and SHM from physics.bgsu.edu

Interference and Wave Nature of Light applets and animations from www.cabrillo.eduwww.physics.rutgers.edu Single slit interference

The Physics of Light and Colour from www.olympusmicro.com/primer/lightandcolor/index.html

Nature of light from hypertextThomasYoung and the Wave Nature of Ligh from OpenMindWave phenomena old syllabus from patana.ac.th

Optics: Fresnel diffraction - adjustable slit | MIT Video Demonstrations in Lasers and OpticsMIT OpenCourseWare Published on 15 Jun 2012