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Edexcel-AS-Unit-2-Physics-at-work-Waves-Electricity-and-Nature-of-light

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AS-Unit-2-Physics-at-work ......................................................................................................................1 1 Waves .....................................................................................................................................................1 Chapter 1 Mechanical waves ....................................................................................................................1

1.1 Mechanical waves .......................................................................................................................1 1.2 Types of waves............................................................................................................................1 1.3 Calculating wave speed...............................................................................................................1 1.4 Periodic wave features ................................................................................................................2 1.5 Energy of a periodic wave...........................................................................................................3 1.6 The superposition principle.........................................................................................................4 1.7 standing waves ............................................................................................................................5

1.7 11 Worked examples ........................................................................................................................7 Chapter 2 Geometrical optics and Wave optics.......................................................................................10

2.1 Reflection and refraction...........................................................................................................10 2.2 Total internal reflection .............................................................................................................12 2.3 Optical fibers.............................................................................................................................13 2.4 Progressive waves & Wave features .........................................................................................13 2.5 Polarization & Intensity ............................................................................................................13 2.6 Superposition and interference..................................................................................................13 2.7 Diffraction .................................................................................................................................13

2.8 41 Worked examples ......................................................................................................................15 Chapter 3 Sound......................................................................................................................................29

3.1 The speed of sound....................................................................................................................29 3.2 vibrating air columns ................................................................................................................29 3.3 Audible sound waves ................................................................................................................29 3.4 the Doppler Effect .....................................................................................................................29

3.5 9 Worked examples ........................................................................................................................29 2 DC electricity .......................................................................................................................................30

2-1 Charge, current and potential difference...................................................................................30 2-2 Resistance .................................................................................................................................30 2-3 Ohms law and measuring resistance........................................................................................30

2-3-1 Graph of V against I and I against V for the ideal resistor ............................................30 2-4 Resistivity .................................................................................................................................30 2.5 Circuit components symbols: ....................................................................................................30

2-5-1 IV and VI graphs for different conductors.............................................................30 2-6 Circuits......................................................................................................................................30 2-7 Potential divider........................................................................................................................30 2-8 Electromotive force and internal resistance..............................................................................30

2-9 38 Worked examples......................................................................................................................30 3 Nature of light ......................................................................................................................................30

3-1 Particles, antiparticles and photons...........................................................................................30 3-2 The photoelectric effect ............................................................................................................30 3-3 Collisions of electrons with atoms............................................................................................30 3-4 Wave-particle duality ................................................................................................................30

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3-5 46 Worked examples......................................................................................................................30

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AS-Unit-2-Physics-at-work 1 Waves Chapter 1 Mechanical waves 1.1 Mechanical waves A mechanical wave is a physical disturbance in an elastic medium. Note: (i) A mechanical wave depends upon a mechanical source and a material medium. (ii) It is important to recognize that all disturbances are not necessarily mechanical. For example, light waves, radio waves, and heat radiation propagate their energy by means of electric and magnetic disturbances. No physical medium is necessary for the transmission of electromagnetic waves. However, many of the basic ideas presented in this chapter for mechanical waves are also applicable to electromagnetic waves. 1.2 Types of waves There are two main types of progressive waves: Transverse waves: The oscillations are at right-angles to the direction of travel: Longitudinal waves: The oscillations are in line with the direction of travel, so that a compression is followed by a rarefaction, and so on. 1.3 Calculating wave speed It can be shown that the speed of a transverse pulse in a string is given by

F Fv ml

Where F is the tension in the string; l is the length of the string; m is the mass of the string; m l is called mass per unit length that is usually referred to as the linear density of the string. And if F is expressed in Newton and in kilograms per meter, the speed is in meters per second.

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1.4 Periodic wave features So far we have been considering single, nonrepeated disturbances called pulses. What happens if similar disturbances are repeated periodically? The wave below (Fig. 1.1) is a propagation of a periodic wave.

Wave features:

(i) Wavelength

The wavelength of a periodic wave train is the distance between any

hase if they have the same

is the magnitude of the oscillation.

d to cover a wavelength is therefore equal to the period

H n be related to the wavelength

two adjacent particles which are in phase. Note: two particles are said to be in pdisplacement and if they are moving in the same direction. (A and B; C and D are in phase). (ii) Amplitude Amplitude: this(iii) Period time The time requireT of the vibrating source.

ence, the wave speed v ca and period T by the equation

vT

(iv) Frequency of a wave is the number of waves that pass a particular The frequency f

point in a unit of time. It is the same as the frequency of the vibrating

source and is therefore equal to the reciprocal of the period ( 1f ) T

The SI unit of frequency is the hertz (Hz). For example, a frequency of 5Hz means that 5 waves are being emitted per second.

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11 1 /Hz cycle ss

(v) The relationship between the frequency, wavelength, and speed of a periodic wave:

fvT

speed freNote: the speed of waves d

uency wavelength e

q

pends on the properties of the medium

.5 Energy of a periodic wave e found that the maximum velocity of a

(material) through which they are traveling. 1In chap. 9 on harmonic motion, wparticle oscillating with frequency f and amplitude A is given by

max 2v fA When a par icle hast this speed, it is passing through its equilibrium position, where its potential energy is zero and its kinetic energy is a maximum. Thus the total energy of the particle is

2max1 12 2 2mv m fA 2 2 2 22 f A m

As a periodic wave passes through a medium, each element of the

maxp k kE E E E

medium is continuously doing work on adjacent elements. Therefore, the energy transmitted along the length of a vibrating string is not confined to one position. Let us apply the result obtained for a single particle to the entire length of a vibrating string. The energy content of the entire string is the sum of the individual energies of its constituent particles. If we let m refer to the entire mass of the string instead of the mass of an individual particle, equation 2 2 22 f A mE represents the total wave energy in the string. In a strin , the wave energy per unit length is given by

g of length l

2 2 22 mf AE l lubstituting S for the mass per unit length, we can write

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2 2 22 f AlE

Suppose that a wave travels down a length l of a given string with a speed v. the time t required for the wave to travel this length is

ltv

If the energy in this length of string is represented by E, the power P of the wave is given by

E E EP vlt lv

This represents the rate at which energy is propagated down the string.

Substitution from Eq. 2 2 22 f AlE yields

2 2 22P v f A vlE

The wave power is directly proportional to the energy per unit length and to the wave speed. Note: The dependence of wave energy and wave power on 2f and , as

found in Eqs.

2A

2 2 22 f AlE and 2 2 22P v f A v

lE , is a general

conclusion for all kinds of waves. 1.6 The superposition principle Let us consider transverse waves in a vibrating string. The speed of a transverse wave is determined by the tension of the string and its linear density. Since these parameters are a function of the medium and not the source, any transverse wave will have the same speed for a given string under constant tension. However, the frequency and amplitude may vary considerably. When two or more wave trains exist simultaneously in the same medium, each wave travels through the medium as though the other were not present. The superposition principle:

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When two or more waves exist simultaneously in the same medium, the resultant displacement at any point and time is the algebraic sum of the displacements of each wave. Note: (i) The superposition principle applies only for linear media. (ii) The sum of displacements is algebraic only if the waves have the same plane of polarization. The application of this principle is shown graphically in Fig. 1.2. In Fig. 1.2a the superposition results in a wave of larger amplitude. These waves are said to interfere constructively. Destructive interference occurs when the resulting amplitude is smaller, as in Fig. 1.2b.

1.7 standing waves (i) Standing waves Standing waves are produced by the superposition of two sets of progressive waves (of equal amplitude and frequency) traveling in opposite directions. For example, when a stretched string is vibrated, waves travel along the string, reflect from the ends, and are superimposed on waves traveling the other way (Fig. 1.3).

For the standing waves shown in Fig. 1.3, the positions which the amplitude of the oscillation are always zero are called nodes, are labeled N in the figure. The points of maximum amplitude occur midway between the nodes and are called antinodes, are labeled A in the figure.

(ii) Characteristic frequencies The simplest possible standing wave occurs when the wavelengths of incident and reflected waves are equal to twice the length of the string.

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The standing wave consists of a single loop with nodal points at each end, as shown in Fig. 1.3a. This pattern of vibration is referred to as the fundamental mode of oscillation. The higher modes of oscillation occur for shorter and shorter wavelengths. From the figure, it noted that the allowable wavelengths are 2 2 2 2, , , , ..1 2 3 4l l l l .

Or, in equation form, 2 1, 2, 3, ...nl n

n The corresponding frequencies of vibration are, from v f ,

2 21,2, 3,...n

nv vf n nl l

Where v is the speed of the transverse waves. This speed is the same for all wavelengths because it depends only upon the characteristics of the

vibrating medium. The frequencies given by Eq. 2 2nnv vf n

l

l are

called the characteristic frequencies of vibration. In terms of string tension F and linear density , the characteristic frequencies are

1,2,3,...2nn Ff nl

The lowest possible frequency ( 2

vl ) is called the fundamental

frequency 1f . The others, which are integral multiples of the fundamental,

are known as the overtones, the entire series,

1 1,2,3,...n n ff n Consisting of the fundamental and its overtones, is known as the harmonic series. The fundamental is the first harmonic, the first overtone ( 12 2 ff ) is the second harmonic, the second overtone ( 13 3 ff ) is the third harmonic, and so on.

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1221ll

222ll

3 23 23ll

4 224

l l

Note: Stationary waves that vibrate freely do not transfer energy. The amplitude of a vibrating particle in a stationary wave pattern varies with position from zero at a node to maximum amplitude at an antinode. The phase difference between two vibrating particles is: (i) Zero if the two particles are between adjacent nodes or separated by an even number of nodes. (ii) 180 radians if the two particles are separated by an odd number of nodes. For a string of length L, the resonant wavelengths must be consistent with

2L n , where n = 1, 2, 3, 1.7 11 Worked examples 1. A string has a length and mass of 0.3 g. calculate the speed of a transverse pulse in the string if it is under a tension of 20 N.

2l m

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Solution: We first compute the linear density of the string.

340.3 10 1.5 10 /2

kgm kg ml m

The equation Fv yields

4

20 365 /1.5 10 /

F Nv mkg m s

2. A man sits near the end of a fishing dock and counts the water waves as they strike a supporting post. In 1 min he counts 80 waves. If a particular crest travels 20 m in 8 s, what is the wavelength of the waves? Solution: The frequency and velocity of the waves are calculated as follows:

80 1.3360wavesf Hz

s

20 2.5 /8

mv ms

s

Thus, the wavelength is given by 2.5 / 1.881.33

m s mHz

vf

3. A steel piano wire 50 cm long has a mass of 5 g and is under a tension of 400 N. What are the frequencies of its fundamental mode of vibration and the first two overtones? Solution:

From Eq. 1,2,3, ...2nn Ff nl , the fundamental is found by

setting n = 1:

1 2002 0.5 0.005

0.5

1 1 1 4002 2 Hzm m kgl

m

F F Nf l l

The first and second overtones are

12 4002 Hzff

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13 6003 Hzff 4. A string that is fixed at both ends is made to vibrate in the fundamental (first harmonic) mode.

The fixed ends of the string are at x = 0 and x =L. Each point on the string oscillates in simple harmonic motion. The displacement y of the string at a point x at time t is given by the equation

cos(500 )y A t Where 12sin

2xA

In these formulae x is in metres and t is in seconds. Using this equation, (a) Explain why the amplitude of the standing wave is not constant. Solution:

The amplitude is 12sin2xA that depends on the value of x.

(b) Calculate the frequency of the standing wave. Solution: From cos(500 )y A t , 500 2 f , gives

500 2502

f Hz (c) Show that L= 2.0 m. Solution: At x = L, the amplitude is always equal to zero, thus

12sin 12sin 02 2x LA , gives L = 2.0 m

5. What is the wave speed of a guitar string whose fundamental frequency is 330 Hz if the length of string free to vibrate is 0.651m? Solution:

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The wave speed may be determined from equation v f , which relates wave speed, frequency, and wavelength the wavelength of the

ngth of the string. Thus, s

fundamental frequency is twice the fr e lee(2 ) 330 2 0.651 430 /f L f mv

6. The stationary wave pattern shown in Fig. 6.1 is set up on a rope of length 3.0m.

Fig. 6.1

a, calculate the wavelength of these waves. b. state the phase difference between the particle vibrating at A,

djacent nodes distance is equal to

B, C. Solution:

12The a .

So 13 32

L m a. .0 2m b.

. Between O and A there is one node, so the phase difference is 180 . Between O and B is equal to the sum phase difference of OA and BA,

se there is an even number of nodes, o Phase difference OC = 0.

and Wave optics nd refraction

angle of reflection is equal to the angle of incidence (Fig. 2-1), or

So phase difference (OB) = 180+ 45= 225. . Phase difference OC, becauS Chapter 2 Geometrical optics2.1 Reflection a1.1 Reflection Law of reflection: the

r i 10

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The normal, the incident ray, and the reflected ray all lie in a plane perpendicular to the reflecting surface, known as the plane of incidence. The light ray does not turn out of the plane of incidence as it is reflected.

i r

ri

1.2 Refraction The law of refraction, also . 2-2), can be written called Snells law Fig (

1 1 2 2sin sinn n Where 1n depends only on the optical properties of medium 1 and 2n depends only on the optical properties of medium 2. The constant n is called the index of refraction of the medium. We define the index of refraction n of a medium to be the ratio of the speed of light in vacuum c to the speed of light in that medium v,

cn v

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1 r

2

2.2 Total internal reflection When light travels from the higher refractive index ( ) medium to a

lower refractive index medium ( ), take the light travels from glass

( ) into air ( ) as an example. The Fig. 2-3 below shows that the ray refracts away from the normal. If the angle of incidence is increased to a certain value known as the critical angle, the light ray refracts along the boundary. Fig. 2-4 shows this effect. If the angel of incidence is increased further, the light ray undergoes total internal reflection at the boundary (Fig. 2-5), the same as if the boundary were replaced by a plane mirror.

2n

1n

2 1.5n 1 1n

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i

r

Normal

Incident ray

Refracted ray

Air Glass

Fig. 2-3 critical anglei

i

Normal

Incident ray

Refracted ray

Air Glass

Fig. 2-4 ccritical angle( )i

900

i

ci>critical angle( )

r

In the Fig. 2-4, from Snells law 1 1 2 2sin sinn n That is 0sin sin 90n n and n 01, sin 90 1 2 1i 1So

2

1 , 1.5n fosin r glassi n ,

Then

2

1arcsin 421.5i

So for the air-glass boundary, the critical angle

2.4 Progressive waves & Wave features 2.5 Polarization & Intensity 2.6 Superposition and interference 2.7 Diffraction 7.1 Diffraction by a single silt

42c 2.3 Optical fibers

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The angular locations of the minima are given by sinm b , m = 1, 2, 3, (single-slit minima). Where is the wavelength of the light; b is the silt width.

a is a maximum. The brightest rs right in the center, and the other maxima get

Note: between each pair of minimmaximum occusuccessively dimmer (Fig. 2-16).

intensity

1.0

0.5

Lb 2 L

b 3 L

bL

b2 LbL3b

0

Position along screenFig. 2-16 light intensity along the screen for diffraction by a single slit of width

b located a distance L from the screen. By far the brightest spot occurs at the center line of the slit

7.2 Diffraction grating

1

1

A diffraction grating, as in Fig. 2-17, has many slits (typically, 500 per mm). Constructive interference produces sharp lines of maximum

intensity at set angles either side of a sharp, central maximum. In between,

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destructive interference gives zero or near-zero intensity. To identify the lines, they are each given an order number (0, 1, 2 etc.).

A close-up of part of the grating is shown in Fig. 2-18. d is the grating spacing. 1 is the angle of the first order maximum. In this case, the path difference for any two adjacent slits is one wavelength, . From the triangles:

1sin d

So 1sind For higher orders, the path differences are 2 , 3 etc. the following

ation gives values of

equ for all orders: sind n here n is the order number (0, 1, 2 etc.). W

Note: If d 1 and n are known, can be calculated. A longer wavelength gives a larger angle for each order. If the incoming light is a mixture of wavelengths, each order above zero becomes a spectrum.

examples

the irection of t

tive index of water:

41 Worked

air to water and (b

2.8 1. A light ray strikes an air/water surface at an angle of 46with respect to the normal. Find the angle of refraction when d he ray is (a) from ) from water to air. Refractive index of air: 1 1n

2 1.33n RefracSolution: (a) The incident ray is in air, so 01 146 1.00and n . The refracted ray is in water, so 2 1.33n . Snells law can be used to find the angle of refraction:

01 1

22 1.33n

sin (1.00)sin 46in 0.54n s1 0

2 sin (0.54) 33

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(b) Now the incident ray propagates in water ), and

tes i air

( 01 146 1.00and n ( 2 1.00n ). Snells law yields the refracted ray propaga n

01 1sin (1.33)sin 46sin n 2 n2 0.961.00

hows a rectangular glass fish tank containing water. Three light rays, P, Q and R from the same point on a small object O at the bottom of the tank are shown.

1 02 sin (0.96) 74

2. Fig. 2-1 s

(a) (i) Light ray Q is refracted along the water-air surface. The angle oincidence of light ray Q at the w

f ater surface is 49.0. Calculate the

refractive index of the water. Give your answer to an appropriate number of significant figures. Solution: From figure 6, the critical angle is , thus 049.0

0 1sin 49 airwater water

nn n

, gives

0

1 1.33sin 49water

n (a) (ii) Draw on Fig. 2-1 the path of light ray P from the water-air surface. Solution:

al From (i), ray P is refracted away from the normal. Ray R is internreflected. And ray P is also partially reflected.

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(b) In Fig. 2-1, the angle of incidence of light ray R at the water-air 60.0.

nally reflected at the water surface. Solution:

to a certain value known as the critical

flection at the boundary

nd side of the tank. e refractive index of the glass = 1.50

greater than that of water, ray R enters the glass and

at glass equals to 30, thus by the Snells law, i , it can be get that

hrough the glass to strike the glass-gel

surface is (b) (i) Explain why this light ray is y inter totall

If the angle of incidence is increasedangle, the light ray refracts along the boundary.

If the angel of incidence is increased further, the light ray undergoes total internal re(b) (ii) Draw the path of light ray R from the water surface and explain whether or not R enters the glass at the right-hathSolution: Index of glass isrefracts towards the normal (partially reflected). And the angle of incidence

01sinnsin 30watern glass

3. Fig. 3-1 shows a 45 right angled glass prism in air, c

01 26.2i

oated on one side with a film of transparent gel. A ray of light strikes the prism, at an angle of incidence of 45, and continues tinterface at the critical angle.

Fig. 3-1 (a) Calculate the refractive index of (a) (i) the glass,

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(a) (ii) the gel. Solution:

(i) From Snells law: 21 1 2sin sinn n Thus,

, 0 0sin 45 sin 29air glassn n 1airn Therefore,

0

0sin 45 1.46sin 29glass

n (ii) The same way to calculate the geln :

, 0 0sin 74 sin 90n nglass gel glass 1.46n Therefore,

0

0sin 7

gel glassn n 4 1.40n90

draw, of the ray of light with the oved. Mark the values of all relevant angles.

si

Fig. 3-2(b) On gel rem

using a ruler. The path

Solution: Let the critical angle of glass-air , then

1 1sin 1.46glassn

Thus, So, the light ray undergoes total internal reflection at the boundary. The ray then refract away from the normal from the side of the prism. (c) A ray of light passes through a straight, 5.00 m long, optical fiber of

043

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refractive index 1.59. Calculate the time taken for a ray of light to travel along the axis of the fiber. Solution:

Time taken stv

and s = 5 m, so we firstly calculate the speed of the ray in the optical fiber. And

c , we can get airfibren v n8

83.00 10 / 1.89 10 /1.59

air

fibre fibre

n c c m sv mn n

s

Therefore, 8

8 2.65 101.89 10 /m

v m s

5.00st s 4. A beam of light is propagating through diamond ( 1 2.42n ) and strikes

ond-air interface at an angle of incidence of ill part of

water

028 . (a) Wa diamthe beam enter the air or will the beam be totally reflected at the interface? (b) Repeat part (a), assuming the diamond is surrounded by( 2 1.33n ). Solution: (a) The critical angle c for total internal reflection at the diamond-air interface is given by sin airc

diamondn , thus n

1 1 1sin sinairn 024.42

Because th is greater than the critical angle,

, rather than air, surrounds the diamond, the critical angle for total internal reflection become ger:

2.4nd c diamon

e angle of incidence of 028 the light is totally reflected. (b) If water

s lar

1 1 1.33sin sinwatern 033.3

2.42c diamondn

(Now a ray of light that has an angle of incidence of less than the critical angle) at the diamond-water interface is refracted into the water. 5. Show that a 0 0 045 45 90 glass prism can deflect a light beam 090

028

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b ntey total i rnal reflection. The index of refraction of the glass is 1.52. ent ray strikes the short side of the prism normally Strategy: if the incid

(that is, perpendicularly; Fig. 5-1), there is no refraction at the interface

1I and the transmitted beam hits the back surface 2I , making an angle of e is le than , all of the

light is reflected from the back of the prism. So we n to find the critical angle

incidence of 045 . Thus, if the critical angl ss 045eed

c and compare it with the angle of incidence of 045 .

Fig. 5-1

450

450

I2I1

450

900 450

I3

Solution:

To calculate c , we use the equation 21

sin cnn

with 1 1.52n for glass

2 1.00n for air. The critical angle is and 1 1 02

1

1.00sin sin 41.11.52c

nn

0The angle of incidence of exceeds the critical angle. Thus, the light

striking 45

2I is indeed totally reflected. It strikes the third interface, 3I , normally and emer from the initial diPeriscopes make use of two such prisms to enable the viewer to see around corners or other obstacles. 6. An optical fiber used for communications has a core of refractive index 1.55 which is surrounded by cladding of refractive index 1.45.

ges at an angle of 090 rection.

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Fig. 6-1 (a) Fig. 6-1 shows a light ray P inside the core of the fiber. The light ray

e- at an ang idence-cladding boundary.

Solution:

strikes the cor cladding boundary at Q le of inc of 60.0. (a) (i) Calculate the critical angle of the core

1.45sin1.55c coren

Gives

claddingn 1 01.45sin 70

1.55c

(a) (ii) State why the light ray enters the cladding at Q. Solution:

he angle of incidence is less than the critical angle. T(a) (iii) Calculate the angle of refraction, , at Q. Solution: By Snells law: 2i , gives 1 1 2sin sinn i n

sin 60 sincore claddingn n sin 60 1.55 0.866sin 0.9257

1.45core

cladding

nn

Therefore, (b) Explain why optical fibers used for communications need to have cladding. Solution:

067.8

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7. A glass cube is held in contact with a liquid and a light ray is directed

at a vertical face of the cube. The angle of incidence at the vertical face isthen decreased to 42as shown in Fig. 7-1. At this point the angle of

e first refraction is 27and the ray is totally internally reflected at P for thtime.

(a) Complete Fi e path og. 7-1 to show th f the ray beyond P until it returns to air. (b) Show that the refractive index of the glass is about 1.5. Solution: From the Snells law:

1 1 2 2sin i , gives 0 0sin 42airn

sinn i nsin 27glassn

Therefore,

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www.gcephysics.com 0 0

0 0

sin 42 1 sin 42 1.5sin 27 sin 27

airglass

nn (c) Calculate the critical angle for the glass-liquid boundary. Solution: In Fig. 7-1, the light is totally internally reflected at point P. Thus the critical angle for the glass-liquid is 63 (d) Calculate the refractive index of the liquid. Solution:

sin liquidcglass

nn

Therefore, 0sin 1.5sin 63 1.34liquid glass cn n

8. Fig. 8-1 shows a pool of water of depth 1.0 m which has a lamp set into the bottom corner as shown. The angle c marked on the diagram is the critical angle for a water-air boundary. Refractive index of water = 1.33

(a) Calculate

itical angle(i) The speed of light in water, (ii) The cr .

edium 1 to medium 2:

cSolution: For a ray passing

1 1 2 2sin sinn i n i and 1 1 2 2n c n cfrom m

Thus (i)

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www.gcephysics.com 81 3 10 /w water air airn c n c m s

here ater

T fore8 8

81 3 10 / 1 3 10 / 2.26 10 /m s m sc m s water 1.33watern(ii) For the water: 1sin c n , 1.33watern . waterThus

1(1.3c

Fig.

(c) A layer of oil is poured over he su

01sin ) 48.8 (b) On 8-1 draw the continuation of the paths taken by the three rays

rther calculations are required. rface of the water. Without

alculation explain how the critical angle for the water-oil boundary

3

shown. No fut

cdiffers from the critical angle, c , for the water-air boundary. Solution: The refractive index of the oil is greater than that of air. Thus the critical

ger. angle for water-oil boundary is lar

sinSinc ine soil airc cwater water

n nn n

9. (a) Explain the term critical angle when applied to a transparent material in air. Solution: When light travels from the higher refractive index ( ) medium to a

lower refractive index medium ( , if the angle of incidence is increased to a certain value known as the critical angle, the light ray refracts alo

prism, ABC, is surrounded by air as of light enters one side normally.

index of the glass = 1.60

2n

1n )ng

the boundary. (b) Dense glass in the shape of ashown in Fig. 9-1. A rayRefractive

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Fig. 9-1 for this dense glass-air boundary. (i) Calculate the critical angle Solution:

1 1airn sin1.6c glass glass

in n

Gives 1 01sin ( ) 39i

1.c 6

t of incidence on side AB. No calculation(iii) A t type of glass is fixed to the riginal as shown in Fig. 9-2. The ray of light now passes parallel to the

(ii) Complete the path of the ray on Fig. 9-1 showing it emerging into the air. Mark all relevant angles at the poin

s are expected. second prism made from a differen

oboundary with the second prism.

Calculate the refractive index of the glass from which the second prism is

Solution:

Let the refractive index of the second prism

made.

glassn And from the figure 4, the critical angle is given by 60

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Thus

0sin 60 glass

glass

nn

Gives

0 3sin 60 1.6 1.42glass glass

n n 10. Fig. 10-1 shows a glass optical fiber bent through 90 and surrounded

s of the ncident normally on the flat end of the fiber. They then strike

the internal surface of the fiber at the glass-air boundary. Ray Q is incident at the critical angle.

by air. Three light rays PQR, initially travelling parallel to the axifiber, are i

(a) (i) Explain what is meant by critical angle.

(ii) Calculate the critical angle for a boundary between the glass and air.

Refractive index of the glass = 1.54 Solution: (i) If the angle of incidence is increased to a certain value known as the critical angle, the light ray refracts along the boundary.

2

1sin i , for glass, (ii) 2 1.54nc n1aci The rcsin 41

(b) Draw, using a ruler, on Figure 1 the path taken by rays P, Q and R on

1.54

So for the air-glass boundary, the critical angle 41ci

26

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striking the internal surface. Solution:

w Ray Q showing refraction at 90

showing correct refraction Q if the glass shown in Fig. ture with a glass cladding of

g away from the normal. (ii) State one reason why a glass cladding is normally used. Solution: The glass cladding is used to protect the core to avoid the leakage of light. (iii) Calculate the critical angle for a boundary between the glass core and the glass cladding. Refractive index of the glass used for the cladding = 1.46 Solution:

Ray P sho ing

(c) (i) Describe what would happen to ray10-1 had been surrounded during manufac

TIR

Ray R

lower refractive index. You may be awarded additional marks to those shown in the brackets for the quality of written communication in your answer. Solution: The ray Q would leave the core bendin

21 2

1

1.46sin1.54c

ni n n 1 01.46sin ( ) 71.5i

1.54

c

11. gh a step index optical fiber. Fig. 11-1 shows a cross-section throu

(a) (i) Name the parts A and B of the fiber.

light th ugh the fiber. (a) (ii) On Fig. 11-1, draw the path of the ray of roAssume the light ray undergoes total internal reflection at the boundary

27

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between A and B. Solution: (i) A: cladding B: core (ii) As shown in Fig. 11-1 (b) Calculate the critical angle for the boundary between A and B.

he refractive index of part B = 1.48

Give your answer to an appropriate number of significant figures. The refractive index of part A = 1.46 T

Solution: 1.46sin Ac

n 1.48

itical angle

Bn

Gives

The cr 1 01.46sin ( ) 811.48

(c) State and explain one reason why part B of the optical fiber is made as

c

ould cause signal to get weaker to

(d) State one application of optical fibers and explain how this has benefited society

olution:

Benefit: to improve medical or improve transmission of data. 12. AM and FM radio waves are transverse waves that consist of electric and magnetic disturbances. These waves travel at a speed of . A station broadcasts an AM radio wave whose frequency is 1230 kHz and an FM radio wave whose frequen

The distance between adjacent crests is the wavelength

narrow as possible. Solution: Reason: to prevent light loss which wbe less secure.

. SApplication: endoscope or communications.

83.00 10 /m s

cy is 91.9 MHz. find the distancebetween adjacent crests in each wave. Solution:

. Since the 28

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speed of each wave is known to be 83.00 10 /v m s and the frequency established at the b

is roadcasting station in each case, v f can be used elength: to determine the wav

8

3

3.00 10 / 2441230 10

v m s[AM] m f Hz

8

6

3.00 10 / 3.2691.9 10

v m s[FM] m f Hz

sings a 13. What is the wavelength of the sound waves when someone standard A ( 440f Hz )? SolutionAssume he speed of

s . We may use

: the room temperature to be about 15

Eq. v f, so that t

sound is to get: 340 /v m340 / 0.77440

v m s mf Hz

Chapter 3 Sound 3.1 The speed of sound 3.2 vibrating air columns

le sound w3.3 Audib aves 3.4 the Doppler Effect 3.5 9 Worked examples

29

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f V against I and I againstty

omponents symbols: d VI graphs for differen

The photoelectri-3 Collisions of electr

30

difference

g resistance h o V for the ideal resistor

vic

an t conductors -6 Circuits

tive force and internal resistance orked examples

ns 3-2 c effect 3 ns with atoms

-4 Wave-particle duality xamples

2 DC electricity 2-1 Charge, current and potential 2-2 Resistance 2-3 Ohms law and measurin2-3-1 Grap2-4 Resisti2.5 Circuit2-5-1 IV22-7 Potential divider 2-8 Electromo2-9 38 W3 Nature of light 3-1 Particles, antiparticles and photo

o33-5 46 Worked e

AS-Unit-2-Physics-at-work1 WavesChapter 1 Mechanical waves1.1 Mechanical waves1.2 Types of waves1.3 Calculating wave speed1.4 Periodic wave features 1.5 Energy of a periodic wave1.6 The superposition principle1.7 standing waves

1.7 11 Worked examplesChapter 2 Geometrical optics and Wave optics2.1 Reflection and refraction2.2 Total internal reflection2.3 Optical fibers2.4 Progressive waves & Wave features2.5 Polarization & Intensity2.6 Superposition and interference2.7 Diffraction

2.8 41 Worked examplesChapter 3 Sound3.1 The speed of sound3.2 vibrating air columns3.3 Audible sound waves3.4 the Doppler Effect

3.5 9 Worked examples2 DC electricity2-1 Charge, current and potential difference2-2 Resistance2-3 Ohms law and measuring resistance2-3-1 Graph of V against I and I against V for the ideal resistor

2-4 Resistivity2.5 Circuit components symbols:2-5-1 IV and VI graphs for different conductors

2-6 Circuits2-7 Potential divider2-8 Electromotive force and internal resistance

2-9 38 Worked examples3 Nature of light3-1 Particles, antiparticles and photons3-2 The photoelectric effect3-3 Collisions of electrons with atoms3-4 Wave-particle duality

3-5 46 Worked examples