Cambridge Pre-U

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Physics

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Coursebook

Cambridge Elevate edition Original material ÂŠ Cambridge University Press 2016

Cambridge Pre-U Physics

S13: Waves

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describe sound waves in terms of the displacement of molecules or changes in pressure explain what is meant by a plane-polarised wave, and use Malus’ law to calculate the amplitude and intensity of transmission through a polarising filter understand refraction of waves at the interface between two media, and relate the refractive index to the wave speeds in those media derive the equation for the critical angle and use it to solve problems recall that total internal reflection occurs when a wave is incident at an angle greater than the critical angle, and that optical fibres use total internal reflection to transmit signals recall that, in general, waves are partially transmitted and partially reflected at an interface between media

S13.1 Terminology

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Learning Outcomes

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Another name for a progressive wave is a travelling wave. We can use the terms frequency and period to describe other periodic (or cyclic) phenomena too, as you will see in later chapters on oscillations and rotation. The period, T is the time for one cycle, and the frequency, f is the number of cycles per unit time. They 1 are always related by the reciprocal relationship f = . T Figure 13.8 in the Coursebook (Chapter 13) shows how we can represent longitudinal and transverse waves. The high pressure regions of a longitudinal wave are called compressions, and the low pressure regions are called rarefactions. The sine graph used to represent a longitudinal wave may be plotted as pressure change against distance, with zero on the pressure axis corresponding to the equilibrium pressure. Alternatively it may be plotted as displacement against distance, where the displacement refers to the displacement of the particles from their equilibrium position. The maximum displacement does not correspond to the maximum pressure, though! At the centre of a compression (maximum pressure) or rarefaction (minimum pressure), the displacement is zero. The largest displacements correspond to the points that are between and equidistant (equal distances) from the compressions and rarefactions. We can describe the displacement and pressure in a sound wave as being 90° out of phase with each other. Phase difference is discussed further in the next section.

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after time t = T/2, the wave has advanced half a wavelength

wave at time t = 0

+ Distance

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Displacement

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So far, we have described waves in terms of how their displacement varies with distance along the direction of travel of the wave. The graphs we have been plotting are a ‘snapshot’ of what the wave looks like at a particular instant in time. If we were to take a second ‘snapshot’ half a period later, we would see that the wave had moved half a wavelength to the right (along the distance axis). This is shown in Figure S13.1a. Instead of plotting displacement against position (at a given time), we could plot displacement against time, at a given position. This produces the graph shown in Figure S13.1b, on which we can identify the period of the wave. We measured this period in Box 13.1 and the accompanying worked example.

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Displacement

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Figure 13.1  a A progressive wave travels along the direction of propagation, so at later times the graph of displacement against distance will be shifted along the distance axis.  b A graph of displacement against time, for a fixed point along the direction of travel of the wave. The time for one complete oscillation to pass that point is known as the period, T.

Note that phase difference can be measured in radians, where a complete cycle of 360° = 2π radians. (See also Chapter 17.) For example, this means that a phase difference of 90° is π radians. 2

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13.2 Waves at boundaries

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When waves meet a boundary between two different materials, they may be reflected, absorbed or transmitted. We will not deal specifically with the case where they are absorbed here – when this happens, energy is transferred from the wave into the medium, heating it up.

Reflection

You will be very familiar with the phenomenon of reflection from your everyday life. You probably see your own reflection in a mirror or a reflective surface several times daily, and often you will hear the reflection of sound as echoes. The reflection of seismic waves can be used to investigate the structure of rocks beneath the surface and search for oil. Police radar detectors reflect radio waves off vehicles. If the vehicle is moving, the reflected wave undergoes a Doppler shift, which can be used to calculate the speed of the vehicle. All types of waves can be reflected, although the properties of the surface required to reflect them vary depending on the type of wave. When waves are reflected, they obey the law of reflection, illustrated in Figure S13.2: The incident and reflected rays are at equal angles to the normal at the reflection point.

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a The law of reflection normal incident ray

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mirror

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b A parabolic mirror

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Figure S13.2  a The law of reflection. The normal is a line drawn at right angles to the surface. For curved surfaces, at any point the normal is a line at right angles to the tangent to the curve at that point.  b Reflections from a curved surface – a parabolic mirror.

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We have already met the idea of wavefronts, which ‘join up’ points of equal phase on the wave. A ray is a line that is at right angles to the wavefront. If we start a ray from the wave source in a given direction, it will follow a path that is at right angles to all the wavefronts it crosses. You will already be familiar with the idea of light rays from your earlier physics courses, but we can extend the use of rays to any other types of wave. We can use a ray diagram to analyse the properties of the reflection. Figure S13.3 is a ray diagram showing how an image is formed from a reflection in a plane mirror. This image is known as a virtual image since no real rays of light actually cross (or converge) at the image location. To find the image, we have to project the reflected rays backwards behind the mirror to the point where they meet (the dotted lines in the diagram). The reflected light rays are said to be diverging (spreading apart) in front of the mirror. They diverge in the same way as light would if it travelled directly from an object placed at the image location (if the mirror were not there).

blue arrow is object, placed in front of mirror

dotted lines are the continuation of the reflected light rays to the position where they appear to come from (as if they had come directly from a light source rather than being reflected)

mirror

the image, in the position where it is seen

Figure S13.3  Image formation in a mirror. The image can be seen when viewed from the position marked by the eye symbol.

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questionS 13.1 Use a ray diagram to prove that the image of a point in a plane mirror is the same distance behind the mirror as the object point is in front.

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13.2 Using the result in question 1, explain what the image looks like when a threedimensional object is placed in front of the mirror – use diagrams to help you.

Change of phase on reflection

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Imagine you stretch out a Slinky spring between you and a friend. Then, while they firmly hold one end, you send a transverse wave pulse down the spring, by quickly moving your end of the Slinky up and down. What happens to the pulse when it meets the other end? You should see that it is inverted or ‘flips’ as it is reflected – an ‘upwards’ pulse is returned as a ‘downwards’ pulse. This is a phase change of π radians (180°). We call this phase change inversion. Then hold the Slinky vertically, so that it is extended but only held at one end. If you send a transverse wave pulse down the Slinky like this, you will see that it is again reflected when it gets to the bottom, but in this case there is no inversion on reflection – there is no phase change on reflection. Figure S13.4 illustrates this effect.

Figure S13.4  A wave pulse passing along a string. In a the end of the string is fixed, and the pulse undergoes a phase change of π radians (180°) on reflection. In b the end of the string is free and the pulse is not inverted on reflection.

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We can explain the phase change on reflection using our knowledge of Newton’s laws of motion. Think about the case where the Slinky was held fixed at one end (Figure S13.4a), and imagine that an upward pulse is arriving at the fixed point. The upward movement of the Slinky exerts an upward force on the fixed point as it arrives. Therefore, by Newton’s third law, the fixed point must exert an equal downwards force on the Slinky. This accelerates this part of the Slinky downwards, and so the pulse is inverted. The same phase change on reflection can happen with light. A light ray travelling through air and reflecting off the surface of a piece of glass undergoes a phase shift of π radians on reflection. However, a ray travelling through glass and reflecting off the interface between the glass and air, does not undergo a phase shift on reflection. The general principle can be summarised as: • when a wave travels through a more dense medium and reflects off a less dense medium, there is no phase shift • when a wave travels through a less dense medium and reflects off a more dense medium, the wave is inverted. For light, we say that the medium with the higher refractive index (see below) is more optically dense. In the case of a mechanical wave on a spring or a rope, we are referring to density in the usual sense of mass per unit volume, assuming that the tension in the spring or rope remains the same across the boundary.

Refraction

You may have noticed that when you put a straw in a glass of water, the straw appears bent (Figure S13.5). Of course, the straw itself is not bent, but light rays travelling from the straw change direction as they leave the water. This phenomenon is called refraction, and occurs Original material © Cambridge University Press 2016

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whenever a wave travels through a boundary between two different materials and changes speed. The human eye uses refraction to form an image of the world around us on the retina. If you wear spectacles or use contact lenses, the refraction of light in the lens provides the correction necessary for an image to be formed in focus on the retina.

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Figure S13.5  A straw placed in a glass of water appears bent because the light rays reflected from the bottom of the straw are refracted when they leave the water.

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Imagine a car driving along a straight road with a hard surface. At the edge of the road there is soft mud. If the wheels on the left side of the car roll off the road into the mud, then they will be slowed down compared to the wheels that remain on the road. The car will turn to the left: its velocity vector will change from being almost parallel to the road, to pointing to the left of the road. This models what happens when waves are refracted.

The laws of refraction

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The diagram in Figure S13.6 shows what happens when a wave is refracted at a boundary between two materials. We can also use the diagram to derive the law of refraction, which is also known as Snell’s law. ray wavefront wavefront

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medium medium 11 (speed (speed vv1 )1 )

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medium 22 medium (speed vv2 ) ) (speed 2 v1 > v2 λ1 > λ22

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Figure 13.6  The speed of a wave depends on the medium (material) through which it travels. When a wave is transmitted across a boundary between two media that have a different wave speed, it is refracted.  a shows how the ray and wavefronts are refracted.  b shows a ­close-up view of a, and allows us to derive the law of refraction (see text).

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AC =

v1 v2 = f sin θ1 f sin θ 2

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then rearranging:

λ1 λ = 2 sin θ1 sin θ 2

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Using v = f λ :

λ2 sin θ 2

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Equating the two expressions:

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In Figure S13.6, the wavefronts are continuous across the boundary between the two materials – that is, although the wavefronts change direction, each is an unbroken line as it crosses the boundary. The wavefronts are continuous because the frequency of the waves is the same on either side of the boundary. The frequency of the wave is set at the moment it leaves the source. The frequency cannot change as the wave crossed the boundary (otherwise a number of wavefronts would disappear completely). However, the wavelength does change as the wave crosses the boundary, because the speed of the wave is different in the two materials. Earlier in Chapter 13 of the Coursebook, we used the equation v = f λ to relate the wavespeed, frequency and wavelength. If the frequency remains constant but v decreases as the wave moves from medium 1 to medium 2, the wavelength λ must decrease. To allow this while keeping the wavefronts continuous across the boundary, the wavefronts have to change direction as they cross the boundary: they are refracted and the ray appears to bend. We can use the geometry of the two right-angled triangles shown on the diagram to produce two different expressions for the length of line AC, in terms of the wavelengths on each side of the boundary: λ AC = 1 sin θ1

v1 sin θ1 = = n v 2 sin θ 2

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Here n is called the refractive index of medium 2 with respect to medium 1, and is the v ratio of the wavespeeds in the two media, 1 . We could also call this the boundary refractive v2 index when travelling from medium 1 to medium 2. The line at right angles to the boundary,

at the point at which the ray crosses the boundary, is called the normal. The angles θ1 and θ 2 are the angles the ray makes with the normal on either side of the boundary. Notice that the incident ray, refracted ray and normal are all in the same plane. Because we measure the angles to the normal, we can use the same law for curved surfaces.

Absolute refractive index

The definition of refractive index above depends on both media – it is the refractive index of one medium relative to another medium. We can instead define the absolute refractive index of a medium as: nabs =

c v

where nabs is the (absolute) refractive index, c is the speed of light in a vacuum (3.00 × 108 m s−1), and v is the speed of light in the medium. You will see in most cases the absolute refractive Original material © Cambridge University Press 2016

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v2 =

c n2

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So the equation we derived above can be re-written as v1 sinθ1 n2 = = v 2 sinθ 2 n1

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index is simply called the refractive index of a medium. The speed of light in any medium can never be greater than the speed of light in a vacuum (empty space), so the refractive index is always greater than 1. For light travelling in air, the speed of light is very close to the speed of light in a vacuum, so we often approximate the refractive index of air to be 1. Table S13.1 gives some refractive indices for common media. We can use this new definition to write a new expression for Snell’s law. If the refractive index of medium 1 is n1 and that of medium 2 is n2, we can write:

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When re-arranged this gives us a simpler form of Snell’s law:

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n1 sinθ1 = n2 sinθ 2

Note that we can write the refractive index of medium 2 with respect to medium 1, n, as n n = n2 1 The use of the absolute refractive indices makes it easier to solve problems. To see how this works in practice, look at Worked example S13.1.

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Material Vacuum Air at 0 °C and 1 atm pressure Water at 20 °C Water ice Crown glass Flint glass Pyrex Perspex (acrylic glass) Sapphire Diamond

Refractive Index 1 (by definition) 1.000293 (usually taken to be 1) 1.3330 1.31 1.50 – 1.54 1.60 – 1.62 1.47 1.49 1.76 – 1.78 2.42

Table S13.1  Refractive indices of different media, for yellow light with a wavelength of 589 nm.

Worked Example S13.1 A ray of light falls on a glass block at an angle of incidence (angle to the normal) of 45°. The angle of refraction inside the block is measured to be 30°. What is the refractive index of the glass? Step 1 Decide which material is medium 1 and which is medium 2. You may wish to draw a labelled diagram to show the materials and the angles. In this case, we are going from

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Worked Example S13.1 (continued) air into glass, and so medium 1 is air and medium 2 is glass. The refractive index of air is 1.00 (to 2 s.f.). air (n = 1.00)

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glass Step 2 We need to find the refractive index of medium 2 (n2). Rearrange Snell’s law to find this quantity, then substitute in the values given in the question.

1.00 sin 45 = 1.41 sin 30

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n2 =

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n1 sin θ1 = n2 sin θ 2

Worked Example S13.2

A diver is working underwater and a ray of light from his lamp strikes the surface of the water at an angle of 55° to the horizontal. At what angle to the horizontal will the ray travel after it leaves the water? The refractive index of water is 1.33. Step 1 Read the question carefully! Here we are given and asked for angles to the horizontal, but remember that Snell’s law works with angles to the normal. Draw a labelled diagram with the given quantities and angles marked, and work out the angles to the normal.

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Step 2 Rearrange Snell’s law to find θ2. n1 sin θ1 = n2 sin θ 2 −1  n1 sin θ 1  ⇒ θ1 = sin  n  2

 1.33 sin 35  = sin −1    1.00 = 49.7° = 50° (2 s.f.) Step 3 Give the answer in the form asked for in the question: The ray leaves the water at an angle of 90° − 50° = 40° to the horizontal.

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SUMMARY OF THE LAWS OF REFRACTION 1 The incident ray, refracted ray and normal to the point of incidence are all in the same plane.

n1 sinθ1 = n2 sinθ 2

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θ1is the angle between the ray and the normal in the medium of refractive index n1and θ2is the angle between the ray and the normal in the medium of refractive index n2 (see Figure S13.7).

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2 If light travels from a medium of refractive index n1 into a medium of refractive index n2, then the angles that the rays make to the normal to the boundary are given by the relationship

Figure S13.7  Snell’s law

Apparent depth

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You may have noticed that when you look down into a pool of water, it appears to be less deep than it actually is. This effect is due to refraction – if you look back at the photograph of the straw at the start of this section (Figure S13.5) you will notice that the straw looks bent upwards in the water. In fact, if we look directly down into the water (at right angles to the surface), the refractive index of the water is given by the ratio n=

real depth apparent depth

If you look at the water at a smaller angle than a right angle, you will find that the apparent depth is reduced, so this formula only applies if you are looking directly down. We can explain this using our knowledge of refraction (see Figure S13.8).

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Figure S13.8  Refraction means that water appears to be less deep than it really is.

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A ray of light coming from the bottom of the container of water at an angle to the normal is refracted away from the normal as it leaves the water and passes into air (ray CO). A ray of light that comes from the bottom of the container but is normal to the surface passes through without changing angle (ray CA). If we trace the first ray back into the water (dotted line OB), then it meets the ray that came out along the normal at point B. The distance AB is the apparent depth (the real depth is the distance AC). If you redraw Figure S13.8 with a larger angle θ1, then you will notice that the rays cross higher up in the water and the apparent depth is reduced. So to find the maximum possible apparent depth, we need to work out what happens as θ1 tends to zero. Snell’s law tells us that: sin θ1 = n sin θ 2

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Trigonometry tells us that in triangle AOB

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sin θ 2 =

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and that in triangle AOC

Combining these two equations with Snell’s law tells us that: OA OA =n OB OC

However, as we make θ1 smaller, length OB tends to length AB and OC tends to AC (it doesn’t make sense to say what they are when θ1 = 0, but just before it becomes zero, these pairs of lengths are nearly equal). Applying this to our equation above tells us that OA OA =n AB AC real depth AC h= = AB apparent depth

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This equation can be used to measure the refractive index of a rectangular block of solid or a liquid, using a travelling microscope (a microscope that moves up and down on a scale). 1. Focus the microscope on a mark on a piece of paper laid on the bench. Call this measurement on the microscope scale a. 2. Put the block or liquid in place and refocus the microscope so it is again focused on the mark. Call this measurement on the microscope scale b. 3. Focus the microscope on the top of the block. Call this measurement on the microscope scale c. 4. The real depth is (c – a), and the apparent depth is (c – b). 5. You can use these measurements in the formula above to calculate the refractive index.

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question

Dispersion and the prism

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13.3 If you stand at one end of a swimming pool of constant depth, as you look to the far end it looks like the swimming pool gets shallower. Explain this effect using a ray diagram and your knowledge of refraction.

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So far, we have assumed that the refractive index is the same for all wavelengths (colours) of light. For many materials, however, this is not the case – the refractive index, and therefore the wavespeed, varies depending on the wavelength. This property is known as dispersion.

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Figure S13.9  White light entering a prism. Glass has a different refractive index for different colours (wavelengths), so the colours are refracted differently.

White light was known to be split into colours by a prism before Isaac Newton’s experiments with light, but the colours were thought to originate from the prism in some way. To test this idea, Newton took the coloured light from the prism and tried to split it further. Since no further colours were produced, he deduced that the white light was made up of a mixture of colours.

Total internal reflection

Our studies of refraction and Snell’s law have shown us that when light passes from a more optically dense medium (high refractive index) into a less optically dense medium (lower refractive index), the light bends away from the normal. However, once the angle of incidence in the high refractive index medium reaches a value where the angle of refraction would be greater than 90°, refraction can no longer happen. If you tried to solve Snell’s law for such a case, you would be trying to find the inverse sine of a number greater than 1, so there is no solution.

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The angle of incidence required for the angle of refraction to be 90° is known as the critical angle. A critical angle only exists for a ray going from one medium into another medium with a lower refractive index. (Think about the opposite situation: if the ray were going into a medium with a higher refractive index, it would be bent towards the normal, and we would not reach an angle of refraction of 90° before the angle of incidence reached 90°). For a ray of light going from a medium of refractive index n into air (which we will take to have a refractive index of 1), the critical angle can be found by using Snell’s law, with the angle of refraction set to 90°. We will call the critical angle c. n sin c = 1.0 sin 90° 1 n

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c = sin −1

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More generally, if light travels from a medium with refractive index n1 into a medium with refractive index n2, where n1 > n2, then the critical angle is:

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Once the angle of incidence becomes greater than or equal to the critical angle, no refraction takes place and the ray undergoes total internal reflection (see Figure S13.10), which obeys the laws of reflection discussed earlier in the chapter.

Figure S13.10  a Total internal reflection occurs when the angle of incidence is greater than the critical angle.  b Photograph showing total internal reflection in an acrylic block.

The critical angle is defined as the angle of incidence for a ray crossing the boundary from a medium of higher refractive index to one of lower refractive index for which the law of refraction predicts an angle of refraction of 90°. No refracted ray can form and the incident ray undergoes total internal reflection at all angles greater than or equal to the critical angle. Diamond has a very high refractive index and therefore a small critical angle. Diamonds used for jewellery are cut so that light entering through the top surface is totally internally Original material © Cambridge University Press 2016

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reflected and comes back out of the top, so it looks like light is streaming out of the diamond. Getting the cut right is critical to this – if one cut is not correct, then the light will exit through the sides of the diamond after being internally reflected. The small critical angle means that most light entering the diamond is totally internally reflected, and a small movement of the diamond can cause the light to illuminate a different facet – the diamond appears to sparkle. Many optical instruments such as binoculars and periscopes use total internal reflections in 45° prisms. Since the critical angle for glass with a refractive index of 1.5 is around 42°, the light is incident on the internal face of the prism at an angle greater than the critical angle, and is totally internally reflected (see Figure S13.11). periscope

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Figure S13.11  Light is totally internally reflected inside 45° prisms in a periscope.

Fibre optics

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Transparent glass fibres (often called optical fibres) guide light along them by total internal reflection. Light rays that pass into one end of a fibre meet the inner surface at an angle greater than the critical angle, and are therefore totally internally reflected. This continues to be occur even when the fibre is bent, as long as the radius of the bend in the fibre is much greater than the radius of the fibre. Most optical fibres produced have a diameter less than a millimetre, so the condition for total internal reflection is easy to achieve. The fibre used to transmit the light is usually clad (coated) in a layer of glass with a lower refractive index. This means that the critical angle is quite large, so the rays travel very close to the axis of the fibre. Optical fibres are mainly used for communication. In some cities, optical fibre is used instead of copper wire for high-speed internet communications. It is possible to send information down an optical fibre much more quickly, and with less signal loss, than sending electrical pulses down a copper cable (see also Chapter 20 on communications systems). This is because the high frequency of light (>1014 Hz) means that very short pulses can be used and detected. Light with a single frequency (monochromatic light) is used since the glass is dispersive, and light with a mix of different frequencies (colours) would travel at different speeds. If the fibres were used to communicate over long distances, then the different frequency components in non-monochromatic light would spread out and cause the signal to degrade. Optical fibres are also used in medicine. A device called an endoscope can be inserted into the body and used to see inside. An endoscope contains one bundle of optical fibres to transmit light inside the body to illuminate the area under investigation, and another bundle of fibres to transmit the image back to the physician. Endoscopes are used for diagnosis – determining the nature of a medical condition. They are also used in operations with special surgical instruments that can be inserted through a small incision in the patient’s tissue. Original material © Cambridge University Press 2016

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This minimises the need to cut through large amounts of tissue to perform an operation and helps to reduce the patient’s recovery time.

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cladding: n2 core: n1   Figure S13.12  a An optical fibre used for digital audio connections between devices.  b Diagram showing transmission of light through an optical fibre. This shows the maximum possible angle to the axis at which light can be incident, as it meets the fibre boundaries at the critical angle.

Partial reflection

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θ refracted     Figure S13.13  Partial reflection. a In this photograph, you can see a reflection from the buildings on the surface of the water – however, if you were viewing from under the surface of the water, you would be able to see a refracted image of the buildings, too. We can also see a refracted image of the bottom of the lake, but from within the water, you would be able to see a reflection of the pebbles on the upper surface of the water.  b When a ray of light is incident on a boundary between media, some of the light is transmitted (refracted) and some is reflected.

So far, we have discussed refraction and total internal reflection. When light is incident on an interface between two media at less than the critical angle, most of the light is refracted and transmitted, but some is reflected too (see Figure S13.13). The amount of light that is reflected depends on the angle of incidence in a complicated way, but once we get beyond the critical angle, we know that no light is transmitted – it is all reflected, hence the name total internal reflection. Original material © Cambridge University Press 2016

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You will have experienced partial reflection on a daily basis, but may not have thought about it. If you look out of a window when it is dark outside, you will see your reflection in the window. From the outside, though, a passer-by will be able to see you clearly. That’s one of the reasons we usually draw curtains or blinds across windows at night (although of course blinds or curtains also have other uses, such as thermal insulation). If there are streetlights nearby, you may be able to see the reflection and the view outside superimposed in the window (Figure S13.14). In fact, there will always be some reflection, but we usually do not notice the reflection so much when it is bright outside. This is because only a small fraction (<10%) of the light is reflected – so that when it is bright outside, the transmitted light is much brighter than the reflection. The same effect is used in so-called ‘two-way glass’. If you set up one side of the glass with much brighter lighting than the other, it appears to be mirror-like on that side, while allowing an observer on the dimly lit side to see through. This effect can be enhanced by partly silvering the side you wish to be reflective (coating the glass with a thin layer of reflective paint). We will look again at partial reflection when we discuss thin-film interference.

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Figure S13.14  Partial reflection in a window. The inside of the room is clearly visible in the righthand half of the window, but it is also possible to see outside through the left-hand half.

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S13.3 Polarisation of waves

In Figure 13.16 on Chapter 13 of the Coursebook, the electric field of an electromagnetic wave is shown as oscillating in the vertical plane. This wave can be described as being planepolarised in the vertical plane. If the electric field oscillated in the horizontal plane instead, the wave would be described as being plane-polarised in the horizontal plane (the magnetic field would now be now be in the vertical plane). We could also have a case where the electric field oscillated in some intermediate direction between vertical and horizontal. This is another plane polarisation of the wave. You may see plane polarisation referred to as linear polarisation. Figure S13.15 shows us how the concept of plane polarisation applies to a transverse wave on a string. Only transverse waves can be polarised – longitudinal waves cannot be polarised.

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a

vertically polarised wave

y

wave passes through vertical slit x

b

y

horizontally polarised wave

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z

wave unable to pass through vertical slit

c

wave with intermediate linear polarisation

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a component of the wave passes throughâ&#x20AC;&#x201C; this component is vertically polarised x

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z

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x

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question

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Figure S13.15â&#x20AC;&#x201A; Transverse waves on a string can be polarised. The polarisation plane of each wave is shown by the orange line. In a, we see a vertically plane-polarised wave, which passes through a vertical slit. In b, a horizontally plane-polarised wave is unable to pass through the vertical slit, so there is no transmission (the wave would be absorbed or reflected). In c, a wave with a polarisation between horizontal and vertical is partially transmitted, as it has a component which is in the vertical plane.

13.4 Explain why longitudinal waves cannot be polarised.

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Since electromagnetic waves are transverse, they can have a polarisation. For example, light from the Sun or from a light bulb is described as unpolarised, since it consists of light in all possible polarisation states superposed. This light can be polarised by a linear polarising filter, which is often called a Polaroid sheet. This filter only allows light in one plane of polarisation through. If two Polaroid sheets are placed so that the directions of polarisation are at right angles to each other, then no light is transmitted (all polarisations of light are blocked). A Polaroid sheet consists of a transparent polymer in which all of the long-chain molecules have been aligned in the same direction. The action of this is similar to the slit in Figure S13.15, except that the waves are absorbed if they are polarised parallel to the chains of molecules, and transmitted if they are at right angles to the chains. Figure S13.16 shows the effect of Polaroid sheets on unpolarised light.

Original material ÂŠ Cambridge University Press 2016

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N

Figure S13.16  No light is transmitted in the region where these polarising filters overlap, because their directions of polarisation are at right angles to each other. Where they are not overlapped, the light passing through is plane-polarised. There is a reduction in the intensity of light because not all the light incident on the filter is able to pass through. When unpolarised light of intensity I is incident on a polarising filter, the transmitted intensity is I 2

unpolarised light

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Light reflected from the surface of a still lake is partially plane-polarised. This means more of the reflected light is plane-polarised parallel to the lake’s surface than would be expected in unpolarised light. At an angle of approximately 37° to the lake’s surface, the reflected light is fully plane-polarised parallel to the surface. Reflections from other transparent media, such as glass, are also partially polarised (the angle at which the reflection is fully polarised is different: it depends on the refractive index). Why this happens is explained in Figure S13.17. If we take a photograph of a lake or a window through a polarising filter, we can reduce the intensity of the reflected light compared to the intensity of the transmitted light. If you have polarising sunglasses you may have noticed this effect: you may be able to see through a car window when without the polarising sunglasses you would have just seen your reflection in the window (see Figure S13.18). polarised reflected ray

53°

air

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water (refractive index n = 1.33)

53°

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partially polarised refracted ray

indicates polarisation direction in plane of paper indicates polarisation direction out of plane of paper

Figure S13.17  Light polarisation at the surface of a lake.

When light is incident on water at 53° to the normal, the fraction of the light that is reflected from the water is fully plane-polarised. When light enters the surface, it causes electrons in the surface to oscillate. These oscillations are in the two directions that are perpendicular to the refracted ray, and are the source of both the refracted and the reflected rays. However, the oscillations that are in the plane of the diagram (represented by the bars across the ray) are parallel to the reflected ray. They cannot therefore contribute to it, since the oscillations making up the reflected ray must be perpendicular to the ray. For this reason, the reflected ray is polarised and only consists of the oscillations out of the plane of the diagram (represented by the circles on the ray). The full polarisation of the reflected ray only occurs when the reflected and refracted ray are at right angles to each other. At other angles, the oscillations in the plane of the diagram have a component that is perpendicular to the reflected ray, and so can contribute to it. This gives rise to a partially polarised reflected ray. Original material © Cambridge University Press 2016

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worked example S13.3 Show that the angle of incidence required for the reflected and refracted ray to be at right angles to each other in Figure S13.17 is 53°.

Let the angle of incidence (and therefore angle of reflection) be θ. The diagram below shows all the angles in our problem: polarised reflected ray

unpolarised light 53°

53°

N

air

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Step 1 Set up the problem and use a diagram.

partially polarised refracted ray

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water (refractive index n = 1.33)

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indicates polarisation direction in plane of paper indicates polarisation direction out of plane of paper

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We know that the angles along one side of the normal must add up to 180°, hence we can work out that the angle of refraction is 90° – θ. Step 2 Use Snell’s law to write down an equation to solve for θ. n1 sin θ1 = n2 sin θ 2

1.00 sin θ = 1.33 sin (90° – θ)

Step 3 Recall that sin ( 90° − θ ) = cosθ and tanθ =

sin θ = 1.33 cos θ

R

sinθ , and hence solve the equation for θ. cosθ

tan θ = 1.33

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θ = tan−1(1.33)= 53.1°

Figure S13.18  An example of photographs taken a without and b with a polarising filter. With the filter, most of the light reflected from the surface of the water is absorbed, allowing the refracted light from beneath the water to be seen.

Original material © Cambridge University Press 2016

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Polarisation of microwaves

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Microwaves are another form of electromagnetic radiation, so they can also be polarised. You can use the equipment shown in Figure S13.19 to investigate the polarisation of microwaves. The microwaves used have a wavelength of a few centimetres, much larger than the wavelengths of visible light. Microwaves therefore need a different kind of polarising filter. The metal grids shown in Figure S13.19 are used as polarising filters for microwaves. When the electric field of the electromagnetic wave oscillates parallel to the metal bars, it makes electrons in the metal move up and down the bars. This absorbs the wave energy and the microwaves are not transmitted. However, when the electric field of the waves oscillates at right angles to the bars, then the electrons are not moved up and down the bars. (The electrons do not move across the bars provided the bars are thin compared to the wavelength of the microwaves.) The wave energy is not absorbed and hence the wave is transmitted.

20 cm

T

polarisation grid

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R

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If the grids are placed so that the metal bars cross each other at right angles, then no microwaves will be transmitted through the combination of grids. This is the same effect as in Figure S13.16, where we observed that no light was transmitted through two crossed polarising filters.

Figure S13.19  Microwave transmitter, receiver and polarising grids.

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Another source of light that is partially polarised is the light from the daytime sky. Sunlight is scattered by molecules in the atmosphere and this scattered light is partially planepolarised. It is completely plane-polarised when the scattered light is at 90° to the incident light, as the oscillations in the scattered ray cannot have a component in the direction of the incident ray. This means that there is only one possible oscillation direction for the scattered ray in this case. Light reflected from the clouds is unpolarised, so by using a polarising filter on a camera, we are able to reduce the intensity of the light from the sky and increase the contrast with the clouds.

Malus’ law

When plane-polarised light falls on a linear polarising filter at an angle θ to the polarisation direction of the filter, the component of the light that is parallel to the filter’s polarisation direction is transmitted. The remainder of the light is absorbed by the filter – this energy must be transferred to thermal energy in the filter. If the incident polarised light has an amplitude A0 (see Figure S13.20), then the component of light that is transmitted must have an amplitude given by: At = A0 cos θ

Remember that intensity is proportional to (amplitude)2 (see Chapter 13 of the Coursebook), so the transmitted intensity is: It = I0 cos 2 θ

Original material © Cambridge University Press 2016

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This result is called Malus’ law, and the graph of the function It is plotted in Figure S13.21. polarisation direction of filter

incident amplitude A0

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transmitted θ amplitude A = A0 cos θ

absorbed amplitude

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polarisation of incident light

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Figure S13.20  When plane-polarised light is incident on a linear polarising filter, the component in the direction of polarisation of the filter is transmitted.

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We can use this idea to show that the intensity of polarised light transmitted through a polaroid sheet is half the incident intensity of unpolarised light on the sheet: • Let the intensity of the unpolarised light be I0 . Since the light is unpolarised, this intensity is equally distributed over all polarisation angles. • To work out how much light is transmitted, we need to add up the contributions transmitted at each possible polarisation angle in the incident light. The size of the contribution at a particular angle is the value read from the graph, and we have already said that the intensity is equally distributed over all angles. If we work out the area under the graph from 0 to π radians, this will be p times the total transmitted intensity. So, considering the graph of It in Figure S13.21a, if we find the area under the curve between 0 Iπ and π radians, we get a value of 0 see Figure S13.21b), so the transmitted intensity is I0 2 2 • We do not need to calculate this over the full circle from 0 to 2π because polarisation angles of, say, π and 3π refer to the same polarisation. However, we would get the same 2 2 result if we did the calculation between 0 and 2π (and divided by 2π rather than π – look at the graph and convince yourself of this). We can also approach this problem more formally using calculus: • The intensity is distributed evenly over all angles, so the incident intensity over a small range of polarisation angles θ to (θ + dθ) is: dI =

I0dθ π

(where the possible range of polarisation angles is 0 to π). • The transmitted intensity over a small range of polarisation angles θ to (θ + dθ) is: dIt =

I0 cos 2 θ dθ π

• Therefore the total transmitted intensity is the integral (the sum) of this over all possible polarisation angles, that is: π

It = dIt = 0

π

∫ 0

I0 cos 2 θ dθ π

Original material © Cambridge University Press 2016

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2 • We use the trigonometric identity cos θ =

π

It =

∫ 0

1 (1 + cos2θ ) to do this integral, so: 2 π

I0 (1 + cos2θ ) I I  dθ =  0  = 0 2 2π π  0 2

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Note that the cos 2θ term gives zero when integrated over this range: you can either do the calculation to show this or use symmetry considerations).

a

0

π/2

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Intensity

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I0

π

3 π/2

θ (angle to polarisation axis of filter)

b I0

I0 cos2 θ

Intensity

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π

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π

0

π/2

π

I0 /2

I0 sin2 θ 3 π/2

Since cos2 θ + sin2 θ = 1, at any point the sum of these graphs is I0. So the area under both graphs from 0 to π is is I0π. Since the area under each graph is equal in this range, the area under the I0 cos2 θ graph is I0π/2. The average transmitted intensity through the polaroid filter is therefore I0/2.

2 Figure S13.21  a Graph of the function It = I 0 cos θ (Malus’ law).  b Calculating the area under the graph.

Original material © Cambridge University Press 2016

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N

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You can investigate Malus’ law in the laboratory using some Polaroid sheets and a light level meter. Take two Polaroid sheets and place them so that their directions of polarisation are at 90° to each other (‘crossed Polaroids’). Place another sheet between them and vary the angle of this middle sheet’s polarisation direction to the polarisation direction of the top sheet (see Figure S13.22). Use the light meter to investigate how the intensity of transmitted light changes as you change the angle.

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Figure S13.22  Two crossed Polaroid sheets with a third polariser inserted between them at an angle.

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worked example S13.4

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Two pieces of polaroid sheet are placed with their directions of polarisation at 90° to each other, so that no light passes through. A third piece is placed between them so that its polarisation direction is at 45° to the polarisation direction of the top sheet. Unpolarised light of intensity I 0 is incident on the stack of sheets. What fraction of this light is transmitted through the stack? Step 1 Determine how much light passes through the first sheet.

We know that the intensity of the plane-polarised light transmitted by the polarised sheet is half the intensity of the unpolarised light incident on it. So after the first sheet, we are left with an intensity: I1 =

I0 2

R

Step 2 Determine how much of that light passes through the second sheet.

FO R

Malus’ law tells us that if polarised light is incident on a polarising filter, the transmitted intensity is given by: It = I0 cos 2 θ

The second sheet has a polarisation direction at 45° to the polarisation direction of the light transmitted through the first sheet. So the intensity after the second sheet is given by: I 2 = I1 cos 2 45° =

I1 I0 = 2 4

Step 3 Determine how much of that light passes through the third sheet. The polarisation of light passing through the second sheet is also at 45° to the polarisation direction of the third sheet, so the same reasoning as in Step 2 applies, and the intensity passing through the third sheet is: I3 = I 2 cos 2 45° =

I 2 I1 I0 = = 2 4 8

So one-eighth of the incident light intensity is transmitted through the stack of sheets.

Original material © Cambridge University Press 2016

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Applications of polarisation

b

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Certain complex molecules have a ‘handedness’. This means that they exist in two forms which are mirror images of each other, called enantiomers. These mirror images cannot be superimposed by rotating the molecules, in the same way as you cannot lay one hand on top of the other and match up the fingers. If the substance is dissolved to form a solution: • a solution of one of the two enantiomers of a molecule will rotate the plane of polarisation of plane-polarised light clockwise • a solution of the other enantiomer will rotate plane-polarised light anticlockwise. Molecules that do this are said to be optically active. A solution containing a 1:1 mixture of the two enantiomers will not rotate the plane of polarisation of the light, since the two contributions to the rotation from each enantiomer cancel out. The amount of rotation of the light that passes through a solution can be measured. If we have a solution consisting of only one of the enantiomers, then this measurement can be used to determine its concentration. Glucose (a sugar) is an optically active molecule, and its concentration in solution can be determined in this way. This is used in the food industry. The polymer molecules in certain plastics such as Perspex can also rotate the plane of polarisation of light (see Figure S13.23a). The extent to which a plastic rotates the plane of polarisation depends on the stress exerted on the polymer sample and the colour of the light. This leads to concentrations of stress showing up as colourful patterns under plane-polarised light. This is used by engineers to investigate stresses in structures, by building and loading a Perspex model and examining it under plane-polarised light. Thin slices of rocks, thin enough for light to be transmitted through the mineral crystals, allow us to investigate the optical properties of those crystals (see Figure S13.23b). Many mineral crystals rotate the plane of polarised light in a similar way to Perspex, and the extent of that rotation is frequency (colour) dependent. This property is known as birefringence. By examining the crystals under polarised light, it is possible to identify the minerals present in the rock. Usually, polarised light is shone through the crystals and a second polarising filter, placed at 90° to the plane of polarisation of the first one, is placed at the eyepiece or detector of a microscope. Only light that has had its plane of polarisation rotated can be seen at the eyepiece or detector.

Figure S13.23  a A plastic protractor viewed in plane-polarised light. The patterns of stress that were locked in to the structure of the plastic as the shape was formed and cooled are visible. b The lower image shows a thin-section (a very fine slice) of the rock in the upper image, viewed in cross-polarised light. Birefringence colours are visible.

Original material © Cambridge University Press 2016

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The law of reflection: when rays are reflected from a surface, the incident and reflected rays are at equal angles to the normal at the reflection point.

When a wave travels through a less dense medium and is reflected off a more dense medium, there is a 180° (π radians) phase shift in the reflected wave (the wave is inverted). This also applies to light, where a more optically dense medium is one with a higher refractive index.

When a wave travels between two media with different wave speeds, it is refracted according to the law of refraction:

N

n1 sin θ1 = n2 sin θ 2

LY

Summary

Many transparent materials, such as glass or plastic, are dispersive, which means that different frequencies of light travel through them at different speeds – they have a different refractive index for different colours of light.

The critical angle is the angle of incidence for a ray crossing the boundary from a medium of higher refractive index to one of lower refractive index for which the law of refraction predicts an angle of refraction of 90°.

No refracted ray can form and the incident ray undergoes total internal reflection at all angles greater than or equal to the critical angle.

When a wave is refracted, not all of the wave is transmitted – some is reflected too. This is known as partial reflection.

Transverse waves may be polarised, which means that their oscillations are confined to a particular plane.

Polarising filters only allow one polarisation of light to pass through. If light is incident on them at an angle θ to the filter’s polarisation direction, then only a component of that light will pass through. If the incident wave has intensity I0 the transmitted wave has intensity:

EV IE

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It = I0 cos 2 θ

Original material © Cambridge University Press 2016

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End-of-chapter questions

S13: Waves and Optics S13.1

A semi-circular convex mirror is attached to the wall.

N

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mirror

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A

Using a ray diagram, explain why a viewer positioned at A has almost a 180° field of view in the mirror (that is, they can almost see directly along the wall).

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S13.2 a Define the term critical angle.

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b  Explain what happens when light is incident on an interface at an angle greater than the critical angle.

[4]

[2] [2]

A diamond has a refractive index of 2.4.

c Calculate the critical angle for diamond.

[2]

R

The diagrams below show three possible cuts for a diamond, to be used in a ring.

too shallow

perfect

too deep

[3]

e  A ray of light meets a flat surface of the diamond at an angle of 85° to the surface. Calculate the angle at which the light leaves the diamond into the air. Take the refractive index of the air to be 1.00.

[3]

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d  Using these diagrams, and your knowledge of total internal reflection, explain why the cut of a diamond is very important if you wish to have a ‘brilliant’ gemstone (one which appears to be illuminated from within).

S13.3

a  Explain how an optical fibre is able to transmit light efficiently over a long distance. Use a diagram in your answer.

[3]

b  An optical fibre consists of an inner cylindrical core, through which the light is transmitted, and a cladding with a lower refractive index. The core has a refractive index of 1.50, and the cladding a refractive index of 1.45. Calculate the critical angle within the fibre.

[2]

Original material © Cambridge University Press 2016

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c  Calculate the maximum angle θmax, relative to the axis of the fibre, at which the light may enter and be transmitted down the core of the fibre.

S13: Waves and Optics θc

θc

core: n1

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θc

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90 – θ c

θ max

S13.4

explosion

10 km direct wave

reflected wave

10°

wave speed 8 km/s

6 km/s

α

1 km

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wave speed

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In a seismic survey, an explosion is triggered. The P-waves (compressional waves, like sound) generated by the explosion travel through the Earth, and are detected by seismometers (sensors). The diagram below shows some possible paths the wave can take.

W

[3]

refracted wave

26

[3]

b  When a light wave travels from one medium to another, it obeys Snell’s law relating the angles and the refractive indices. Write down Snell’s law and then express it in terms of the speed of light in the two media, v1 and v2.

[3]

a  Seismic waves also obey a version of Snell’s law. Using the expression you derived in b, calculate the angle of refraction α in the diagram.

[2]

S13.5

An interview room in a police station is set up as shown in the diagram below.

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a  At a distance of 10 km from the source, calculate the difference in arrival time between the direct wave and the reflected wave.

dimly lit viewing room

brightly lit interview room

glass window

Explain carefully why someone in the viewing room is able to see into the interview room, but someone in the interview room cannot see into the viewing room and only sees their own reflection in the glass. [3]

Original material © Cambridge University Press 2016

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S13.6 a Explain why light can be Polarised but sound cannot. b  Explain why Polaroid sunglasses allow you to see beneath the surface of a lake, but without the sunglasses, you can only see a reflection.

S13: Waves and Optics

[3] [3]

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S13.7 A Polaroid sheet, when held in front of an unpolarised light source, reduces the intensity of the light by a factor of 2.

Two Polaroid sheets are placed ‘crossed’, so that their directions of polarisation are perpendicular to each other. No light passes through the two sheets.

A third Polaroid sheet is placed in between these two sheets at an angle θ.

N

Sketch how the light intensity varies with θ.

1 2

[4] [2]

R

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cos 2 ( 45° ) =

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b

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a  If the angle θ is set at 45°, what fraction of incident light does the combination of Polaroid sheets allow through?

Original material © Cambridge University Press 2016

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Syllabus Mapping Document Syllabus content

Present?

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Supplement

Chapter title

Section number

1

Start page

End page

To add to which chapter of Coursebook?

Section number

Title

4

S4.1

O

Part A

Section title

N

Section number

Mechanics

Content • scalars and vectors • moment of a force

W

• kinematics • Newton’s laws of motion • conservation of linear momentum

• pressure Candidates should be able to: (a) distinguish between scalar and vector quantities and give examples of each

Y

C

1

Kinematics – describing motion

Distance and displacement, scalar and vector

4

5

01b

(b) resolve a vector into two components at right angles to each other by drawing and by calculation

Y

C

4

Forces – vectors and moments

Components of vectors

56

58

01c

(c) combine any number of coplanar vectors at any angle to each other by drawing

Y

B

4

Forces – vectors and moments

Combining forces

54

56

01d

(d) calculate the moment of a force and use the conditions for equilibrium to solve problems (restricted to coplanar forces)

Y

C

4

Forces – vectors and moments

The turning effect of a force

59

62

01e

(e) construct displacement-time and velocity-time graphs for uniformly accelerated motion

Y

C

2

Accelerated motion

Deducing acceleration; Deducing displacement

17

18

01f

(f) identify and use the physical quantities derived from the gradients of displacement-time and areas and gradients of velocity-time graphs, including cases of non-uniform acceleration

C

1

Kinematics – describing motion

Deducing velocity from a displacement–time graph

6

8

C

2

Accelerated motion

Deducing acceleration; Deducing displacement

17

18

B

1

Kinematics – describing motion

Speed and velocity

5

6

1

S1.1

Using x instead of s for displacement

B

2

Accelerated motion

The meaning of accleration; Calculating acceleration

15

16

1

S2.1

Describing motion using calculus

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01g

R

01a

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• density

(g) recall and use: v = Δ x /Δt , a = Δv/Δt

Y

Y

Original material © Cambridge University Press 2016

01h

(h) recognise and use the kinematic equations for motion in one dimension with constant acceleration: s = ut + 1/2 at 2, v 2 = u2 + 2as, s = (u + v)t /2

01i

(i) recognise and make use of the independence of vertical and horizontal motion of a projectile moving freely under gravity

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Chapter title

Y

C

2

Accelerated motion

Y

C

2

Accelerated motion

C

2

Accelerated motion

(j) recognise that internal forces on a collection of objects sum to zero vectorially

Y

S

01k

(k) recall and interpret statements of Newton’s laws of motion

Y

C

01l

(l) recall and use F = ma in situations where mass is constant

3

Dynamics – explaining motion

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01j

Supplement

Section number

Y

C

3

Section title

Dynamics – explaining motion

Start page

End page

To add to which chapter of Coursebook?

Section number

Title

4

S4.1

N

Present?

The equations of motion

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22

Acceleration caused by gravity

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25

Motion in two dimensions – projectiles

28

31

O

Syllabus content

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Section number

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Cambridge Pre-U Physics

Calculating the acceleration; Defining the newton; Mass and inertia; Top speed; Identifying forces; Newton’s third law of motion

38

49

Defining the newton; The pull of gravity; Mass and inertia

40

43

(m) understand the effect of kinetic friction and static friction

Y

S

4

S4.2

Friction

(n) use Fk = µkN and Fs = µs N, where N is the normal contact force and µk and µs are the coefficients of kinetic friction and static friction, respectively

Y

S

4

S4.2

Friction

01o

(o) recall and use the independent effects of perpendicular components of a force

Y

C

4

Forces – vectors and moments

Solving problems by resolving forces

58

58

01p

(p) recall and use p = mv and apply the principle of conservation of linear momentum to problems in one dimension

Y

C

6

Momentum

The idea of momentum; Modelling collisions

86

88

01q

(q) distinguish between elastic and inelastic collisions

Y

C

6

Momentum

Understanding collisions; Two types of collision

89

93

01r

(r) relate resultant force to rate of change of momentum in situations where mass is constant and recall and use F = Δp/Δt

Y

C

6

Momentum

Momentum and Newton’s laws; Understanding motion

95

97

01s

(s) recall and use the relationship impulse = change in momentum

Y

S

6

S6.1

Impulse

01t

(t) recall and use the fact that the area under a force-time graph is equal to the impulse

Y

S

6

S6.2

Determining impulse from a force-time graph

01u

(u) apply the principle of conservation of linear momentum to problems in two dimensions

Y

C

6

Momentum

Collisions in two dimensions

93

95

01v

(v) recall and use density = mass/volume

Y

C

7

Matter and materials

Density

102

102

FO R

R

01m 01n

Original material © Cambridge University Press 2016

Syllabus content

01w

(w) recall and use pressure = normal force/area

01x

(x) recall and use p = ρgh for pressure due to a liquid. 2

Present?

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Chapter title

Y

C

7

Matter and materials

Y

C

7

Matter and materials

Dynamics – explaining motion

Section title

Start page

End page

Pressure

102

102

Pressure in a fluid

103

103

The pull of gravity

41

42

Forces – vectors and moments

Centre of gravity

59

59

Gravitational fields

Representing a gravitational field

272

273

Gravitational fields

To add to which chapter of Coursebook?

Section number

Title

7

S7.1

Describing materials

7

S7.1

Describing materials

O

Content • gravitational field strength • centre of gravity

Y

C

3

02b

(b) recall that the weight of a body appears to act from its centre of gravity

Y

C

4

02c

(c) sketch the field lines for a uniform gravitational field (such as near the surface of the Earth)

Y

C

18

02d

(d) explain the distinction between gravitational field strength and force and explain the concept that a field has independent properties. Deformation of solids

Content • elastic and plastic behaviour • stress and strain Candidates should be able to:

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(a) recall and use the fact that the gravitational field strength g is equal to the force per unit mass and hence that weight W = mg

W

Candidates should be able to: 02a

3

Supplement

Section number

N

Section number

LY

Cambridge Pre-U Physics

Y

C

18

Gravitational fields

Representing a gravitational field; Gravitational field strength g

272

276

Y

B

7

Matter and materials

Compressive and tensile forces; Stretching materials; Elastic potential energy

104

108

(a) distinguish between elastic and plastic deformation of a material

03b

(b) recall the terms brittle, ductile, hard, malleable, stiff, strong and tough, explain their meaning and give examples of materials exhibiting such behaviour

Y

S

03c

(c) explain the meaning of, use and calculate tensile/ compressive stress, tensile/compressive strain, spring constant, strength, breaking stress, stiffness and Young modulus

Y

B

7

Matter and materials

Compressive and tensile forces; Stretching materials

104

107

7

S7.1

Describing materials

03d

(d) draw force-extension, force-compression and tensile/ compressive stress-strain graphs, and explain the meaning of the limit of proportionality, elastic limit, yield point, breaking force and breaking stress

Y

B

7

Matter and materials

Compressive and tensile forces; Stretching materials; Elastic potential energy

104

109

7

S7.2

Explaining stress– strain graphs

03e

(e) state Hooke’s law and identify situations in which it is obeyed

Y

B

7

Matter and materials

Compressive and tensile forces;

104

105

7

S7.1

Describing materials

FO R

R

03a

Original material © Cambridge University Press 2016

Syllabus content

03f

(f) account for the stress-strain graphs of metals and polymers in terms of the microstructure of the material. 4

Present?

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Y

S

C

Supplement

Chapter title

Section number

Section title

Start page

End page

N

Section number

LY

Cambridge Pre-U Physics

Energy concepts

Section number

Title

7

S7.2

Explaining stress– strain graphs

5

S5.1

Heat engines

5

S5.2

Gravitational potential

O

Content

To add to which chapter of Coursebook?

• work • power • potential and kinetic energy

• specific latent heat • specific heat capacity Candidates should be able to:

W

• energy conversion and conservation

04a

(a) understand and use the concept of work in terms of the product of a force and a displacement in the direction of that force, including situations where the force is not along the line of motion

Y

04b

(b) calculate the work done in situations where the force is a function of displacement using the area under a forcedisplacement graph

Y

C

04c

(c) understand that a heat engine is a device that is supplied with thermal energy and converts some of this energy into useful work

Y

S

04d

(d) calculate power from the rate at which work is done or energy is transferred

Y

C

5

Work, energy and power

Power

80

82

04e

(e) recall and use P = Fv

Y

C

5

Work, energy and power

Power

81

82

04f

(f) recall and use ΔE = mg Δh for the gravitational potential energy transferred near the Earth’s surface

Y

B

5

Work, energy and power

Gravitational potential energy

75

76

04g

(g) recall and use g Δh as change in gravitational potential

Y

S

04h

(h) recall and use E = 1/2 Fx for the elastic strain energy in a deformed material sample obeying Hooke’s law

Y

C

7

Matter and materials

Elastic potential energy

108

109

04i

(i) use the area under a force-extension graph to determine elastic strain energy

Y

C

7

Matter and materials

Elastic potential energy

108

108

04j

(j) derive, recall and use E = 1/2 kx 2

Y

C

7

Matter and materials

Elastic potential energy

109

109

04k

(k) derive, recall and use E = 1/2 mv 2 for the kinetic energy of a body

Y

C

5

Work, energy and power

Kinetic energy

76

76

04l

(l) apply the principle of conservation of energy to solve problems

Y

C

5

Work, energy and power

G.p.e.–k.e. transformations, Down, up, down – energy changes, Energy transfers

76

80

Work, energy and power

Doing work, transferring energy

71

73

Matter and materials

Elastic potential energy

108

109

FO R

R

EV IE

5

7

Original material © Cambridge University Press 2016

Section number

Syllabus content

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

04m

(m) recall and use % efficiency = useful energy (or power) out × 100 total energy (or power) in

Y

S

04n

(n) recognise and use ΔE = mc Δθ, where c is the specific heat capacity

Y

C

21

Thermal physics

04o

(o) recognise and use ΔE = mL , where L is the specific latent heat of fusion or of vaporisation.

Y

C

21

Thermal physics

Supplement

5 Electricity

Section number

Section title

Start page

End page

N

Chapter title

Calculating energy changes

336

338

Calculating energy changes

338

340

O

Present?

LY

Cambridge Pre-U Physics

Content

W

• electric current • potential difference and electromotive force (emf) • resistance and resistivity

Candidates should be able to: (a) discuss electrical phenomena in terms of electric charge

Y

C

9

Electric current, potential difference and resistance

Electric current

129

132

05b

(b) describe electric current as the rate of flow of charge and recall and use I = ΔQ /Δt

Y

C

9

Electric current, potential difference and resistance

Electric current

131

131

05c

(c) understand potential difference in terms of energy transfer and recall and use V = W/Q

Y

C

9

Electric current, potential difference and resistance

The meaning of voltage; Electrical power

134

136

05d

(d) recall and use the fact that resistance is defined by R = V/I and use this to calculate resistance variation for a variety of voltage-current characteristics

Y

C

9

Electric current, potential difference and resistance

Electrical resistance; Electrical power

135

138

C

11

Resistance and resistivity

The I–V characteristic for a metallic conductor; Ohm’s law; Resistance and temperature

157

162

C

9

Electric current, potential difference and resistance

The meaning of voltage

134

134

C

12

Practical circuits

Internal resistance

169

171

R

05a

EV IE

• conservation of charge and energy

(e) define and use the concepts of emf and internal resistance and distinguish between emf and terminal potential difference

FO R

05e

Y

05f

(f) derive, recall and use E = I(R + r ) and E = V + Ir

Y

C

12

Practical circuits

Internal resistance

170

170

05g

(g) recall and use P = VI and W = VI t , and derive and use P = I 2R

Y

C

9

Electric current, potential difference and resistance

Electrical power

136

138

05h

(h) recall and use R = ρℓ/A

Y

C

11

Resistance and resistivity

Resistivity

162

164

Original material © Cambridge University Press 2016

To add to which chapter of Coursebook?

Section number

Title

5

S5.3

Calculating efficiency using power

Coursebook Chapter number

Chapter title

05i

(i) recall the formula for the combined resistance of two or more resistors in series and use it to solve problems R T = R 1 + R 2 + …

Y

C

10

Kirchhoff’s laws

05j

(j) recall the formula for the combined resistance of two or more resistors in parallel and use it to solve problems 1/R total = 1/R 1 + 1/R 2 + …

Y

C

10

Kirchhoff’s laws

05k

(k) recall Kirchhoff’s first and second laws and apply them to circuits containing no more than two supply components and no more than two linked loops

Y

C

10

Kirchhoff’s laws

05l

(l) appreciate that Kirchhoff’s first and second laws are a consequence of the conservation of charge and energy, respectively

Y

C

10

Kirchhoff’s laws

05m

(m) use the idea of the potential divider to calculate potential differences and resistances.

Y

C

12

• progressive waves • longitudinal and transverse waves • electromagnetic spectrum • polarisation • refraction Candidates should be able to: (a) understand and use the terms displacement, amplitude, frequency, period and wavelength intensity wave speed 06b

(b) recall and apply f = 1/ T to a variety of situations not limited to waves

Section title

Start page

End page

Resistor combinations

148

149

Resistor combinations

149

151

Kirchhoff’s first law; Kirchhoff’s second law; Applying Kirchhoff’s laws

144

148

Kirchhoff’s first law; Kirchhoff’s second law

144

146

Potential dividers

172

172

180

Y

C

13

Waves

Describing waves

179

Y

C

13

Waves

Wave energy

182

183

Y

C

13

Waves

Wave speed

183

183

Y

C

13

Waves

Describing waves

182

183

R

06a

Practical circuits

EV IE

Waves

Content

Supplement

Section number

N

Location (C)oursebook or (S)upplement or (B)oth

6

Present?

O

Syllabus content

W

Section number

LY

Cambridge Pre-U Physics

(c) recall and use the wave equation v = f λ

Y

C

13

Waves

Wave speed

183

184

(d) recall that a sound wave is a Iongitudinal wave which can be described in terms of the displacement of molecules or changes in pressure

Y

C

13

Waves

Longitudinal and transverse waves

181

182

06e

(e) recall that light waves are transverse electromagnetic waves,

Y

C

13

Waves

Longitudinal and transverse waves

181

182

and that all electromagnetic waves travel at the same speed in a vacuum

Y

C

13

Waves

186

187

(f) recall the major divisions of the electromagnetic spectrum in order of wavelength, and the range of wavelengths of the visible spectrum

Y

C

13

Waves

Orders of magnitude

187

187

06f

FO R

06c 06d

Original material © Cambridge University Press 2016

To add to which chapter of Coursebook?

Section number

Title

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Chapter title

06g

(g) recall that the intensity of a wave is directly proportional to the square of its amplitude

Y

C

13

Waves

06h

(h) use graphs to represent transverse and longitudinal waves,

Y

C

13

Waves

including standing waves

Y

C

15

Stationary waves

(i) explain what is meant by a plane-polarised wave

Y

S

(j) recall Malus’ law (intensity ∝ cos 2 θ) and use it to calculate the amplitude and intensity of transmission through a polarising filter

Y

S

06k

(k) recognise and use the expression for refractive index n = sin θ1/sin θ2 = v1/v 2

Y

S

06l

(l) derive and recall sin c = 1/n and use it to solve problems

06m

(m) recall that optical fibres use total internal reflection to transmit signals

06n

(n) recall that, in general, waves are partially transmitted and partially reflected at an interface between media. 7

Superposition

• phase difference • diffraction • interference • standing waves

Start page

End page

Wave energy

182

183

Describing waves; Longitudinal and transverse waves

179

182

From moving to stationary; Nodes and antinodes; Formation of stationary waves

211

213

FO R

07b

Section number

Title

13

S13.3

Polarisation

13

S13.3

Malus’ Law; Applications of polarisation

13

S13.2

Waves at boundaries – refraction

S

13

S13.2

Waves at boundaries – total internal reflection

Y

S

13

S13.2

Waves at boundaries – total internal reflection

Y

S

13

S13.2

Waves at boundaries – partial reflection

Candidates should be able to: 07a

To add to which chapter of Coursebook?

Y

R

Content

Section title

EV IE

06i 06j

Supplement

Section number

N

Present?

O

Syllabus content

W

Section number

LY

Cambridge Pre-U Physics

(a) explain and use the concepts of superposition

Y

C

14

Superposition of waves

The principle of superposition of waves

193

194

coherence, path difference and phase

Y

C

14

Superposition of waves

Interference

196

200

(b) understand the origin of phase difference and path difference, and calculate phase differences from path differences

Y

C

14

Superposition of waves

Interference

196

200

Original material © Cambridge University Press 2016

Coursebook Chapter number

Chapter title

07c

(c) understand how the phase of a wave varies with time and position

Y

C

14

Superposition of waves

07d

(d) determine the resultant amplitude when two waves superpose, making use of phasor diagrams

Y

B

14

Superposition of waves

07e

(e) explain what is meant by a standing wave, how such a wave can be formed, and identify nodes and antinodes

Y

C

15

Stationary waves

07f

(f) understand that a complex wave may be regarded as a superposition of sinusoidal waves of appropriate amplitudes, frequencies and phases

Y

C

14

07g

(g) recall that waves can be diffracted and that substantial diffraction occurs when the size of the gap or obstacle is comparable to the wavelength

07h

(h) recall qualitatively the diffraction patterns for a slit, a circular hole and a straight edge

07i

Supplement

Section number

Section title

Start page

End page

N

Location (C)oursebook or (S)upplement or (B)oth

Interference

196

200

Interference

196

200

From moving to stationary; Nodes and antinodes; Formation of stationary waves

211

216

The principle of superposition of waves

193

194

O

Present?

W

Syllabus content

Superposition of waves

EV IE

Section number

LY

Cambridge Pre-U Physics

To add to which chapter of Coursebook?

Section number

Title

14

S14.1, S14.2, S14.3

Representing waves using phasor diagrams; Doubleslit interference revisited; Multipleslit interference and diffraction gratings

14

S14.4, S14.5, S14.6

Single-slit diffraction; The Rayleigh criterion; Diffraction at an edge

C

14

Superposition of waves

Diffraction of waves

194

196

Y

B

14

Superposition of waves

Diffraction of waves

194

196

(i) recognise and use the equation nλ = b sin θ to locate the positions of destructive superposition for single slit diffraction, where b is the width of the slit

Y

S

14

S14.4

Single-slit diffraction

07j

(j) recognise and use the Rayleigh criterion θ ≈ λ for resolving power of a single aperture, where b is the width of the aperture

Y

S

14

S14.5

The Rayleigh criterion

07k

(k) describe the superposition pattern for a diffraction grating and for a double slit and use the equation d sin θ = nλ to calculate the angles of the principal maxima

Y

B

14

Superposition of waves

The Young doubleslit experiment; Diffraction gratings

200

206

14

S14.3

Multiple-slit interference and diffraction gratings

07l

(l) use the equation λ = ax /D for double-slit interference using light.

Y

B

14

Superposition of waves

The Young doubleslit experiment

200

203

14

S14.2

Double-slit interference revisited

FO R

R

Y

8

Atomic and nuclear processes

Content

• the nucleus

• nuclear processes

• probability and radioactive decay • fission and fusion

Candidates should be able to:

Original material © Cambridge University Press 2016

Present?

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Chapter title

08a

(a) understand the importance of the α-particle scattering experiment in determining the nuclear model

Y

C

16

08b

(b) describe atomic structure using the nuclear model

Y

C

16

08c

(c) show an awareness of the existence and main sources of background radiation

Y

08d

(d) recognise nuclear radiations (α, β–, γ) from their penetrating power and ionising ability, and recall the nature of these radiations

Y

C

16

08e

(e) write and interpret balanced nuclear transformation equations using standard notation

Y

C

31

08f

(f) understand and use the terms nucleon number (mass number), proton number (atomic number), nuclide and isotope

Y

C

16

08g

(g) appreciate the spontaneous and random nature of nuclear decay

Y

C

08h

(h) define and use the concept of activity as the number of decays occurring per unit time

Y

08i

(i) understand qualitatively how a constant decay probability leads to the shape of a radioactive decay curve

08j

Supplement

Section number

Section title

Start page

End page

Looking inside the atom; Alpha-particle scattering and the nucleus

223

225

A simple model of the atom; Nucleons and electrons; Forces in the nucleus

229

231

235

Nuclear physics

Balanced equations

490

491

Nucleons and electrons

226

228

31

Nuclear physics

Randomness and decay

496

497

C

31

Nuclear physics

497

99

Y

C

31

Nuclear physics

Decay graphs and equations

499

500

(j) determine the number of nuclei remaining or the activity of a source after a time which is an integer number of half-lives

Y

C

31

Nuclear physics

Decay graphs and equations; Decay constant and half-life

499

501

08k

(k) understand the terms thermonuclear fusion, induced fission and chain reaction

Y

08l

(l) recall that thermonuclear fusion and the fission of uranium-235 and plutonium-239 release large amounts of energy.

Y

C

31

Nuclear physics

494

494

Y

C

30

Quantum physics

Particulate nature of light; The photoelectric effect

468

475

R

EV IE

W

225

Quantum ideas

FO R

9

Content

• the photoelectric effect • the photon

• wave-particle duality

Candidates should be able to: 09a

(a) recall that, for monochromatic light, the number of photoelectrons emitted per second is proportional to the light intensity and that emission occurs instantaneously

To add to which chapter of Coursebook?

Section number

Title

16

S16.1

31

S31.5

Fusion and fission

31

S31.5

Fusion and fission

N

Syllabus content

O

Section number

LY

Cambridge Pre-U Physics

Original material © Cambridge University Press 2016

Coursebook Chapter number

Chapter title

09b

(b) recall that the kinetic energy of photoelectrons varies from zero to a maximum, and that the maximum kinetic energy depends on the frequency of the light, but not on its intensity

Y

C

30

Quantum physics

09c

(c) recall that photoelectrons are not ejected when the light has a frequency lower than a certain threshold frequency which varies from metal to metal

Y

C

30

Quantum physics

09d

(d) understand how the wave description of light fails to account for the observed features of the photoelectric effect and that the photon description is needed

Y

C

30

Quantum physics

09e

(e) recall that the absorption of a photon of energy can result in the emission of a photoelectron

Y

C

30

09f

(f) recognise and use E = hf

Y

C

30

09g

(g) understand and use the terms threshold frequency and work function and recall and use hf = Φ + 1/2 mv 2max

Y

C

30

09h

(h) understand the use of stopping potential to find the maximum kinetic energy of photoelectrons and convert energies between joules and electron-volts (i) plot a graph of stopping potential against frequency to determine the Planck constant, work function and threshold frequency

09j

(j) understand the need for a wave model to explain electron diffraction

09k

(k) recognise and use λ = h/p for the de Broglie wavelength. Part B

Section title

Start page

End page

The photoelectric effect

471

475

The photoelectric effect

471

475

Modelling with particles and waves; Particulate nature of light; The photoelectric effect

467

475

Quantum physics

The photoelectric effect

472

473

Quantum physics

Particulate nature of light

468

470

Quantum physics

The photoelectric effect

472

475

Y

Y

C

30

Quantum physics

Particulate nature of light

468

Y

Y

C

30

Quantum physics

Electron waves

480

482

Y

C

30

Quantum physics

Electron waves

480

482

17

Circular motion

Describing circular motion; Angles in radians

259

261

Learning outcomes marked with an asterisk (*) will be assessed only in Section 2 of Paper 3 Written Paper. 10

Rotational mechanics

Content

FO R

• kinematics of uniform circular motion • centripetal acceleration • moment of inertia

• kinematics of rotational motion Candidates should be able to: 10a

(a) define and use the radian

Y

C

To add to which chapter of Coursebook?

Section number

Title

30

S30.3

Measuring the work function

30

S30.3

Measuring the work function

470

R

09i

Supplement

Section number

N

Location (C)oursebook or (S)upplement or (B)oth

O

Present?

W

Syllabus content

EV IE

Section number

LY

Cambridge Pre-U Physics

Original material © Cambridge University Press 2016

Section number

Syllabus content

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Chapter title

10b

(b) understand the concept of angular velocity, and recall and use the equations v = ωr

Y

C

17

Circular motion

and T = 2π/ωv 2

Y

10c

(c) derive, recall and use the equations for centripetal acceleration a = v 2/r and a = rω2

Y

C

17

Circular motion

10d

(d) recall that F = ma applied to circular motion gives resultant force mv 2/r

Y

C

17

Circular motion

10e

(e) describe qualitatively the motion of a rigid solid object under the influence of a single force in terms of linear acceleration and rotational acceleration

Y

S

10f

(f) *recall and use I = ∑mr2 to calculate the moment of inertia of a body consisting of three or fewer point particles fixed together

Y

S

10g

(g) *use integration to calculate the moment of inertia of a ring, a disk and a rod

Y

10h

(h) *deduce equations for rotational motion by analogy with Newton’s laws for linear motion, including E = 2 Iω2, L = Iω and C = I dt

10i

(i) *apply the laws of rotational motion to perform kinematic calculations regarding a rotating object when the moment of inertia is given.

• simple harmonic motion • energy in simple harmonic motion • forced oscillations, damping and resonance Candidates should be able to:

Start page

End page

To add to which chapter of Coursebook?

Section number

Title

17

S17.1

Rotational motion

17

S17.1

Rotational motion

17

S17.2

Moment of inertia, kinetic energy and torque

S

17

S17.3

Using moments of inertia

Y

S

17

S17.2

Moment of inertia, kinetic energy and torque

Y

S

17

S17.3

Using moments of inertia

19

S19.1

A more mathematical approach to s.h.m.

N

Section title

261

262

Centripetal forces; Calculating acceleration and force

262

265

Calculating acceleration and force

264

265

W

O

Steady speed, changing velocity; Angular velocity

EV IE

Oscillations

Content

Supplement

Section number

R

11

Present?

LY

Cambridge Pre-U Physics

(a) recall the condition for simple harmonic motion and hence identify situations in which simple harmonic motion will occur

Y

C

11b

(b) *show that the condition for simple harmonic motion leads to a differential equation of the form d2 x /dt 2 = – ω2 x and that x = A cos ωt is a solution to this equation

Y

S

FO R

11a

19

Oscillations

Free and forced oscillations; Observing oscillations; Describing oscillations; Simple harmonic motion

Original material © Cambridge University Press 2016

286

290

Present?

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

11c

(c) *use differential calculus to derive the expressions v = – Aω sin ωt and a = – Aω2 cos ωt for simple harmonic motion

Y

S

11d

(d) *recognise and use the expressions x = A cos ωt , v = – Aω sin ωt , a = – Aω2 cos ωt and F = –mω2 x to solve problems

Y

S

S Y

C

11f

(f) understand the phase differences between displacement, velocity and acceleration in simple harmonic motion

Y

S

11g

(g) *show that the total energy of an undamped simple harmonic system is given by E = 1/2 mA 2ω2 and recognise that this is a constant

Y

S

11h

(h) recognise and use E = 1/2 mA 2ω2 to solve problems

11i

(i) distinguish between free, damped and forced oscillations

11j

(j) recall how the amplitude of a forced oscillation changes at and around the natural frequency of a system and describe, qualitatively, how damping affects resonance. 12

Electric fields

• concept of an electric field • uniform electric fields • capacitance • electric potential

Start page

End page

19

Oscillations

Frequency and angular frequency

292

Y

S

Y

C

C

19

Oscillations

Damped oscillations

297

299

Y

C

19

Oscillations

Resonance

299

302

19

Oscillations

Free and forced oscillations

286

287

FO R

• electric field of a point charge Candidates should be able to: 12a

(a) explain what is meant by an electric field and recall and use E = F/Q for electric field strength

Y

C

8

Electric fields

Attraction and repulsion; The concept of an electric field; Electric field strength

117

121

12b

(b) recall that applying a potential difference to two parallel plates stores charge on the plates and produces a uniform electric field in the central region between them

Y

C

8

Electric fields

Electric field strength

119

121

Original material © Cambridge University Press 2016

To add to which chapter of Coursebook?

Section number

Title

19

S19.1

A more mathematical approach to s.h.m.

19

S19.1

A more mathematical approach to s.h.m.

19

S19.2

The simple pendulum

19

S19.1

A more mathematical approach to s.h.m.

19

S19.3

Energy of an undamped simple harmonic oscillator

19

S19.3

Energy of an undamped simple harmonic oscillator

293

R

Content

Section title

W

(e) recall and use T = 2π/ω as applied to a simple harmonic oscillator

Section number

EV IE

11e

Supplement

Chapter title

N

Syllabus content

O

Section number

LY

Cambridge Pre-U Physics

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

Chapter title

12c

(c) derive and use the equations Fd = QV and E = V/d for a charge moving through a potential difference in a uniform electric field

Y

C

8

Electric fields

12d

(d) recall that the charge stored on parallel plates is proportional to the potential difference between them

Y

C

24

Capacitance

12e

(e) recall and use C = Q / V for capacitance

Y

C

24

Capacitance

12f

(f) recognise and use W = 2QV for the energy stored by a capacitor, derive the equation from the area under a graph of charge stored against potential difference, and derive and use related equations such as W = 1/2CV 2

Y

C

24

Capacitance

12g

(g) analyse graphs of the variation with time of potential difference, charge and current for a capacitor discharging through a resistor

Y

S

Supplement

Section number

Section title

Start page

End page

Electric field strength; Force on a charge

119

123

Capacitors in use

373

375

Capacitors in use

373

375

Energy stored in a capacitor

375

377

(h) define and use the time constant of a discharging capacitor

Y

S

(i) analyse the discharge of a capacitor using equations of the form x = xo eRC

Y

S

12j

(j) understand that the direction and electric field strength of an electric field may be represented by field lines (lines of force), and recall the patterns of field lines that represent uniform and radial electric fields

Y

C

8

Electric fields

The concept of an electric field

118

120

Y

C

23

Coulomb’s law

Electric potential

363

365

Y

S

Y

C

23

Coulomb’s law

Coulomb’s law

360

361

Y

C

23

Coulomb’s law

Electric field strength for a radial field

362

362

Y

S

12k

(k) understand electric potential and equipotentials (l) understand the relationship between electric field and potential gradient, and recall and use E = − dV/dx

12m

(m) recognise and use F = Q1Q 2/4 πε0 r2 for point charges

12n

(n) derive and use E = Q /4 πε0 r for the electric field due to a point charge

12o

(o) *use integration to derive W = Q1Q 2/4 πε r from F = Q1Q 2/4 πε0 r2 for point charges

12p

Please choose techniques that have not been previously assessed with this group of students. Suggested topics are given on the right.

R

2

EV IE

12h 12i

12l

To add to which chapter of Coursebook?

Section number

Title

24

S24.1

Capacitor discharge

24

S24.1

Capacitor discharge

24

S24.1

Capacitor discharge

23

S23.1

More on electric potential

23

S23.1

More on electric potential

N

Present?

O

Syllabus content

W

Section number

LY

Cambridge Pre-U Physics

Y

S

23

S23.1

More on electric potential

(a) state Kepler’s laws of planetary motion:

Y

S

18

S18.1

Kepler’s laws

(i) planets move in elliptical orbits with the Sun at one focus

Y

S

18

S18.1

Kepler’s laws

13

Gravitation

FO R

Content

• Kepler’s laws

• Newton’s law of gravity • gravitational field

• gravitational potential energy Candidates should be able to: 13a

Original material © Cambridge University Press 2016

Present?

Location

Coursebook

(C)oursebook or (S)upplement or (B)oth

Chapter number

(ii) the Sun-planet line sweeps out equal areas in equal times

Y

S

(iii) the orbital period squared of a planet is proportional to its mean distance from the Sun cubed

Y

S

13b

(b) recognise and use F = − Gm1 m2/r2

Y

B

18

Gravitational fields

13c

(c) use Newton’s law of gravity and centripetal force to derive r3 ∝ T 2 for a circular orbit

Y

C

18

Gravitational fields

13d

(d) understand energy transfer by analysis of the area under a gravitational force–distance graph

Y

B

18

Gravitational fields

13e

(e) derive and use g = Gm/r2 for the magnitude of the gravitational field strength due to a point mass

Y

C

18

13f

(f) recall similarities and differences between electric and gravitational fields

Y

C

23

13g

(g) recognise and use the equation for gravitational potential energy for point masses E = – Gm1 m2/r

13h

(h) calculate escape velocity using the ideas of gravitational potential energy (or area under a force–distance graph) and energy transfer

13i

(i) calculate the distance from the centre of the Earth and the height above its surface required for a geostationary orbit.

Content • concept of a magnetic field • force on a current-carrying conductor • force on a moving charge

Section title

Start page

End page

Representing a gravitational field

272

274

Orbiting under gravity; The orbital period

277

279

Energy in a gravitational field

276

276

Gravitational fields

Gravitational field strength g

274

276

Coulomb’s law

Comparing gravitational and electric fields

366

367

W

EV IE

Electromagnetism

Section number

FO R

Title

18

S18.1

Kepler’s laws

18

S18.1

Kepler’s laws

18

S18.2

Gravity is always attractive

18

S18.3

Potential energy and gravitational force–distance graphs

18

S18.3

Potential energy and gravitational force–distance graphs

Y

S

18

S18.4

Escape velocity

Y

C

18

Gravitational fields

Orbiting the Earth

279

280

C

26

Magnetic fields and electromagnetism

Magnetic flux density; Measuring magnetic flux density

411

413

C

28

Electromagnetic induction

Explaining electromagnetic induction

437

441

• the Hall effect

Candidates should be able to:

(a) understand and use the terms magnetic flux density, flux and flux linkage

Section number

S

• electromagnetic induction

14a

To add to which chapter of Coursebook?

Y

R

14

Supplement

Chapter title

N

Syllabus content

O

Section number

LY

Cambridge Pre-U Physics

Y

Original material © Cambridge University Press 2016

Preview Cambridge Pre-U Physics Coursebook

Preview Cambridge Pre-U Physics Coursebook