Pra-U STPM Physics Penggal 3 2018 CB039248c

Physics Term 3 STPM Chapter 23 Wave Optics 195 23 Slit Light source Collimator ∗ Converging lens Diffraction grating f Turntable Telescope Zeroth order θ 1 Eye Figure 23.33 Optical Spectrometer 12. Diffracting gratings are widely used in experiment to measure the wavelength of light from a light source or light from a distant star. 13. Figure 23.33 shows the main components of the optical spectrometer which consist of: (a) collimator to produce parallel beam. It consists of a converging lens arranged at a distance equal to its focal length f from a narrow slit. (b) a piece of diffraction grating arranged vertically on a turntable. The position of the grating is adjusted so that parallel beam from the collimator is at normal incident with the gratings. (c) a telescope which can be rotated about the vertical axis of the turntable. A circular vernier scale is fi xed to the telescope to measure the angle θ1 between the fi rst order line and the zeroth order line. Using d sin θ1 = λ and d = 1 N Wavelength λ = sin θ1 N N = number of lines per unit length of the diffraction grating. Example 13 Light from a lamp is incident normally on a plane diffraction grating ruled with 4 000 lines per cm. The light consists of two wavelengths 656 nm and 410 nm. Calculate the angular separation of the second order diffracted beams corresponding to these two wavelengths. Solution: Using d sin θm = mλ m = 2 and d = 1 N = 1 cm 4 000 = 1 × 10–2 4 000 m when λ = 656 nm, sin θ2 = 2(656 × 10–9) × 4 000 1 × 10–2 θ2 = 31°39′ when λ′ = 410 nm sin θ2 ′ = 2(410 × 10–9) × 4 000 1 × 10–2 θ2 ′ = 19°9′ Angular separation = 31°39′ – 19°9′ = 12°30′

Physics Term 3 STPM Chapter 23 Wave Optics 196 23 Example 14 A diffraction grating which is 2.0 cm long and has 104 lines is illuminated normally with light of wavelength 495 nm. How many bright lines can be observed? Solution: Using d sin θm = mλ and d = 1 N m = sin θm Nλ m fi 1 Nλ since sin θm fi 1 fi 2.0 × 10–2 104 × (495 × 10–9) N = 2.0 cm 104 fi 4.04 Since m is an integer, the highest order diffraction, m = 4. Total number of bright lines = 9 which consists of one zeroth order line, and 2 each for the 1st, 2nd, 3rd and 4th order lines. Example 15 Light from a source that emits wavelengths 400 nm and 600 nm is incident normally on a plane diffraction grating. It is observed that the 400 nm line in one order is diffracted at the same angle, 30°, as the 600 nm light in the adjacent order. Find the number of lines per cm of the grating. Solution: Using d sin θm = mλ If θm = 30°, the same for the two wavelengths m for the wavelength 600 nm must be smaller compared to the order for the 400 nm line. Hence d sin 30° = m(600 nm) = (m + 1) (400 nm) m + 1 m = 600 400 m = 2 Substitute m = 2 in d sin 30° = m(600 nm) d = 2(600 nm) sin 30° Number of lines per cm = (1.0 × 10–2) d = (1.0 × 10–2) sin 30° 2(600 × 10–9) = 4 200

Physics Term 3 STPM Chapter 23 Wave Optics 197 23 Example 16 A piece of diffraction grating has 40 lines per mm. A beam of parallel monochromatic light of wavelength 5.9 × 10–7 m is incident normally on the grating. The diffraction pattern is focussed on a screen using a converging lens of focal length 100 cm. Calculate the distance between the zeroth order line and the fi rst order line on the screen. Solution: f x Screen Converging lens Zeroth order First order Grating θ In order that the diffraction pattern be focussed on the screen, the screen must be at a distance f, which is the focal length of the lens, from the converging lens. Using d sin θ = λ m = 1 sin θ = (5.9 × 10–7) × 40 1.0 × 10–3 = 0.0236 Hence the angle θ is small, and if θ is small θ = sin θ = tan θ = 0.0236 rad = x f Distance x = 0.0236 × (100) cm = 2.36 cm Example 17 P Q 30° 30° P Q β α α β Figure (a) Figure (b) Figure (a) shows the diffraction of a parallel beam of monochromatic light which is incident normally on a plane diffraction grating in air. Figure (b) shows the arrangement in water which has a refractive index 1.33. In water, the angle α < 30°, and the angle β > 30°. (a) Explain the change caused by water.

Physics Term 3 STPM Chapter 23 Wave Optics 198 23 (b) If the diffraction grating has 8.0 × 105 lines per metre, calculate (i) the wavelength of light in air, (ii) the wavelength of light in water, (iii) the angle α, (iv) the angle β. Solution: (a) If λ = wavelength in air then wavelength in water = λ µ = λ 1.33 Hence for the fi rst order diffraction in water, d sin α = λ 1.33 sin α = λ 1.33d compared to in air, sin 30° = λ d > λ 1.33d Hence α < 30° In water, when m = 2, d sin β = 2 ( λ 1.33) = 1.50 λ Comparing with: d sin 30° = λ sin β > sin 30° β > 30° (b) (i) From d sin 30° = λ d = 1 8 × 105 λ = sin 30° 8 × 105 = 6.25 × 10–7 m (ii) Wavelength in water = λ 1.33 = 4.70 × 10–7 m (iii) In water, when m = 1, d sin α = 4.70 × 10–7 sin α = (4.70 × 10–7) (8.0 × 105 ) α = 22.1° (iv) In water, when m = 2, d sin β = 2 × (4.70 × 10–7) sin β = 2 × (4.70 × 10–7) (8.0 × 105 ) β = 48.8°

Physics Term 3 STPM Chapter 23 Wave Optics 199 23 Example 18 (a) A diffraction grating is used to produce spectrum from a source that emits red and violet light. It is found that the fourth line from the centre is a mixture of red and violet. Explain this phenomenon. (b) The grating has 500 lines per mm, and the combined line is diffracted through 43.6°. Find the wavelength of red and violet light. What is the colour of the fi fth line and calculate its angle of diffraction. Solution: (a) The wavelength of violet light λV < λR, wavelength of red light From the equation, d sin θm = mλ for each order of diffraction, θV < θR Hence the fi rst line is violet (V) m = 1 2nd line is red (R) m = 1 3rd line is violet (V) m = 2 4th line is (red (R) m = 2) + (violet (V) m = 3) (b) Fourth line, red (R) m = 2, θ2 = 43.6°, d sin 43.6° = 2 λR λR = (1.0 × 10–3) 500 × sin 43.6° 2 m = 690 nm Fourth line, violet (V) m = 3, θ3 = 43.6°, d sin 43.6° = 3 λV λV = (1.0 × 10–3) 500 × sin 43.6° 3 m = 460 nm If the 5th line is red, m = 3 d sin θ3 = 3 λR sin θ3 = 3 × (6.90 × 10–7) × 500 (1.0 × 10–3) = 1.035 (not possible for sine > 1) Hence 5th line is not red. If the 5th line is violet, m = 4 d sin θ4 = 4 λV sin θ4 = 4 × (460 × 10–9) × 500 (1.0 × 10–3) θ4 = 66.9° Hence the 5th line is violet. V 5 V 1 V V = Violet V V V 3 V + R 4 V + R R 2 R R = Red 0

Physics Term 3 STPM Chapter 23 Wave Optics 200 23 Example 19 A parallel beam of light from a sodium vapour lamp is incident normally on a grating having N lines per unit length. The fi rst order spectrum has two closely-spaced yellow lines which are observed at angles θ and θ + δθ. The corresponding wavelengths are λ and λ + δλ. (a) Show that δθ = ( δλ λ ) tan θ (b) Calculate θ and δθ given that λ = 589 nm δλ = 0.6 nm N = 5.0 × 105 m–1 Solution: (a) Using d sin θm = mλ, m = 1 λ = d sin θ .............................. ① Differentiating with respect to θ, dλ dθ = d cos θ δλ = d cos θ δθ ..............................② ① ② : δλ λ = cos θ sin θ δθ δθ = ( δλ λ ) tan θ (b) From d sin θ = λ sin θ = (589 × 10–9) × (5.0 × 105 ) 1 θ = 17.1° δθ = ( δλ λ ) tan θ = ( 0.6 589 ) tan 17.1° = 3.13 × 10–4 rad. Quick Check 7 1. Monochromatic light of wavelength λ is incident normally on a diffraction grating which consists of opaque lines of width a alternate with transparent lines of width b. The angle between the fi rst order spectrum and zeroth order spectrum depends on A a, b, and λ C b and λ B a and λ D a and b 2. A parallel beam of white light (wavelengths from 4.5 × 10–7 m to 7.5 × 10–7 m) is incident normally on a diffraction grating. The most deviated wavelength in the second order spectrum is diffracted through an angle of 60° from the direction of the incident beam. How many lines per metre are there on the grating?

Physics Term 3 STPM Chapter 23 Wave Optics 201 23 A 5.8 × 105 C 11.6 × 105 B 9.6 × 105 D 19.2 × 105 3. A parallel beam of monochromatic light of wavelength λ is incident normally on a diffraction grating G. The angle between the directions of the two second order diffracted beams at P1 and P2 is α, as shown below. P2 P1 G What is separation between successive lines on the grating? A 2λ sin α C 2λ sin ( α 2 ) B λ sin α D λ sin ( α 2 ) 4. R B B R White light Diffraction grating Second order spectrum First order spectrum The wavelength of white light is in the range 400 nm to 700 nm. A beam of white light is incident normally on a diffraction grating which has 6 × 105 lines per metre. The first order and second order spectrum produced is as shown in the figure. The angular separation between the red end and blue end of the first order spectrum is α, and that for the second order is β. Which of the following is true? A α < 1 2 β C α = 2β B α = 1 2 β D α > 2β 5. Monochromatic light of wavelength 600 nm is used in a spectrometer to illuminate normally a diffraction grating which has 3 × 105 lines per metre. A telescope is used to scan the region to the left of the zeroth order line. What is the maximum order line that can be observed? A 2 C 8 B 5 D 10 6. (a) What is meant by diffraction? (b) Monochromatic light of frequency 5.50 × 1014 Hz is incident normally on a diffraction grating which has 500 lines per mm. (i) What is the angle of diffraction for the second order line? (ii) Determine the maximum number of maximum produced. 7. Light from a mercury lamp is incident normally on a plane diffraction grating ruled with 6 000 lines per cm. The spectrum contains two strong yellow lines of wavelengths 577 nm and 579 nm. What is the angular separation of the second order diffracted beams corresponding to these wo wavelengths? 8. In the spectrum of white light obtained by using a certain diffraction grating, the second and third orders partially overlap. What wavelength in the third order spectrum will appear at the angle corresponding to a wavelength of 650 nm in the second order spectrum? 9. (a) Discuss how diffraction and interference contribute to the action of the diffraction grating. (b) Draw a wave diagram to show how a grating produces a second order beam from a plane monochromatic light. (c) Parallel light is incident normally on a grating having N lines per unit length. The spectrum consists of two close lines of wavelength λ and λ + Δλ. If θ is the mean angle of deviation, show that Δθ, the angular separation between the two emergent beams, is given by

Physics Term 3 STPM Chapter 23 Wave Optics 202 23 Δθ = ( Δλ λ ) tan θ Calculate Δθ from the data below wavelength λ = 589.0 nm λ + Δλ = 589.6 nm N = 4.0 × 105 m–1 Order of spectrum employed, m = 2 10. Grating Laser beam 1.5 m 0.55 m 1.44 m O The figure shows a diffraction grating of 5.5 × 105 lines per metre set with its plane normal to a beam of laser. A large screen is at 1.50 m from the grating. Bright spot are observed at 0.55 m and 1.44 m from the central bright spot. (a) Calculate the wavelength of the laser light from the measurements of the first order diffracted light. (b) Suggest one advantage and one disadvantage of obtaining the wavelength by using observation of the second order diffracted light rather that the first order diffracted light. 11. (a) Red light is incident normally on a diffraction grating. The light consists of two wavelengths of 695 nm and 700 nm. The maximum angle of deviation for the first order spectrum is 30°. Calculate the angular separation between the two beams in the first order spectrum. (b) Give two advantages of a diffraction grating over a prism for usage in a spectrometer. 12. A beam of monochromatic light is incident normally on a diffraction grating which has n lines per unit length. What is the wavelength of the light is the mth order spectrum is just visible? 13. (a) What are the factors that determined the number of order of diffracted spectrum which are visible when light is incident normally on a diffraction grating? (b) A diffraction grating has 5 000 lines per cm. Light from a sodium vapour lamp is incident normally on the grating. A converging lens of focal length 40.0 cm is used to focus the spectrum produced on a screen. If the wavelengths for the D-lines of sodium are 589.0 nm and 589.6 nm, calculate the linear separation between the two second order lines on the screen. 14. A parallel beam of monochromatic light of wavelength 580 nm is incident normally on a diffraction grating. The separation between successive lines on the grating is 0.70 μm as shown in the figure. Calculations show that the first, second, and third order spectrum are expected for the angle θ between the directions of the incident beam and the diffracted beams at 16°, 34° and 56°. (a) At what value of θ is light diffracted by a single slit with width 0.70 μm has its first minimum. (b) Sketch a graph to show the variation of light intensity against the angle θ for light diffracted by the grating. Mark clearly values of θ at 16°, 34°, and 56°.

Physics Term 3 STPM Chapter 23 Wave Optics 203 23 23.7 Polarisation Learning Outcomes Students should be able to: • state that polarisation is a property of transverse waves • explain the polarisation of light obtained by refl ection or using a polariser • use the Brewster’s law tan θB = n • use the Malus’s law I = I0 cos2 θ 1. Plane-polarisation of waves means limiting the oscillations of a wave to one plane. 2. Only transverse waves can be plane-polarised. Longitudinal waves, such as sound waves, cannot be plane-polarised. Since light is transverse waves it can be plane-polarised. 3. Light wave is emitted when an electron from a higher energy level in an atom drops to a lower energy level. 4. If the transition of the electron occurs when the electron is in a horizontal orbit round the nucleus, the wave emitted has its electric fi eld oscillations in the horizontal plane. The wave is said to be horizontally plane polarised. (Figure 23.34) 5. On the other hand, if the transition occurs when the electrons orbit is vertical, then the wave emitted in vertically plane-polarised. (a) (b) Nucleus Electron Horizontally plane - polarised wave Electron Nucleus Vertically plane - polarised wave Figure 23.34 6. Light from sources such as the sun, or a lamp are not plane-polarised because the oscillations of the electric fi eld are in all possible planes. This is because transitions of electrons in atoms of the source occur at all possible planes. Figure 23.35 shows the planes of oscillation of the electric fi eld of an unpolarised beam of light. Direction of propagation of light Planes of oscillation Figure 23.35 7. Since the transitions of electron in atoms are random in nature, the amplitudes of oscillation in the various planes are equal. 2012/P1/Q41, 2016/P3/Q9, 2017/P3/Q9 VIDEO Polarisation

Physics Term 3 STPM Chapter 23 Wave Optics 204 23 Polarising direction vertical Polaroid Incident ray Unpolarised light 1_ 2 Io Io Eye Vertically plane - polarised light Transimitted light Figure 23.36 8. Figure 23.36 shows a ray of unpolarised light incident on a polarising sheet, such as a piece of Polaroid with its polarising direction vertical. The light transmitted has its electric field oscillation in the vertical plane only. Hence it is vertically plane-polarised. 9. If I0 is the intensity of the incident unpolarised ray, the electric field oscillations can be resolved into two mutually perpendicular directions, one parallel to the polarising direction of the Polaroid and the other perpendicular to it. The sum of the components in the parallel direction equals to the sum of components in the perpendicular direction. 10. The perpendicular component is absorbed, and the parallel component is transmitted. Hence the intensity of the transmitted light is half that of the incident ray, i.e. 1 2 I0 . A cos I I1 Unpolarised light A Polaroid Q Polaroid P Plane - polarised light Polarising direction θ θ θ Figure 23.37 11. Figure 23.37 shows the light transmitted by Polaroid P is plane-polarised of amplitude A. The intensity of this plane-polarised ray I1 ∝ A2 ..............................① 12. This plane-polarised light is then incident on a second Polaroid Q which has its polarising direciton inclined an angle θ to that of Polaroid P.

Physics Term 3 STPM Chapter 23 Wave Optics 205 23 13. The amplitude of the light transmitted by Polaroid Q is A cos θ. Since intensity ∝ (amplitude)2 Intensity ∝ (A cos θ)2 ..............................② Comparing ② and ①, intensity, I = I1 cos2 θ which is known as Malus’s law. 14. When the Polaroid Q in Figure 23.37 is slowly rotated about the line of vision, the intensity I of the light transmitted decreases. When θ = 90°, the intensity = 0. The intensity increases when θ increases from 90° to 180°. The intensity is maximum again when θ = 180°. The intensity is again zero when θ = 270°, and maximum when θ = 360° (Figure 23.38). 0 Intensity 90 180 270 360 I1 θ /° Figure 23.38 15. A piece of Polaroid may be used to determined whether a beam of light is plane-polarised. The beam is directed towards a piece of Polaroid, and the Polaroid is slowly rotated about the axis of vision. Eye Unpolarised light incident light Intensity remains constant Polaroid Figure 23.39 16. (a) If the intensity of the transmitted light remains constant when the Polaroid is rotated (Figure 23.39), then the incident light is unpolarised. This is because the amplitude of the transmitted light remains constant. Eye Amplitude, A Incident light Amplitude, A cos Transmitted light Polaroid θ θ Figure 23.40

Physics Term 3 STPM Chapter 23 Wave Optics 206 23 (b) If the intensity of the transmitted light changes as the Polaroid is rotated, the incident light is plane-polarised. This is because the amplitude of the transmitted light is A cos θ, which changes when θ is increased. The intensity of the transmitted light is maximum when the polarising direction of the Polaroid is parallel to the plane of Polarisation of the incident light, and zero when perpendicular. Example 20 30° Direction of polarisation P1 P2 P3 Vertical Unpolarised beam Transmitted beam A beam of unpolarised light is directed towards three Polaroids, P1 , P2 and P3 arranged coaxially as shown in the fi gure. The polarising directions of P1 and P3 are vertical and that of P3 inclined at 30° to the vertical. Through what angle in the clockwise direction Polaroid P3 need to be rotated so that the intensity of the beam emerging from P3 is minimum? Solution: The plane of Polarisation of light transmitted by P2 and incident of P3 is inclined at 30° to the vertical. The light transmitted by P3 would be minimum (zero) when the polarising direction of P3 is perpendicular to the plane of Polarisation of the incident light. Hence P3 has to be rotated through (30° + 90°) = 120° Plane of polarisation of incident light Polarising direction of polaroid P3 30°

Physics Term 3 STPM Chapter 23 Wave Optics 207 23 Example 21 A beam of unpolarised light of intensity I0 is along the common axis of two Polaroids P and Q as shown in the fi gure. Unpolarised beam I0 P Q Initially the polarising directions of P and Q are vertical. (a) State in terms of I0 the intensity of light transmitted by P. (b) Find the angle Q need to be rotated so that the intensity of light transmitted by Q is 1 5 I0 . Solution: (a) Intensity of unpolarised light incident on P = I0 Hence intensity of light transmitted by P = 1 2 I0 (b) Intensity of light incident on Q, I1 = 1 2 I0 Intensity of light transmitted, I = 1 5 I0 . Using I = I1 cos2 θ 1 5 I0 = ( I0 2 ) cos2 θ cos2 θ = 2 5 θ = 50.8° Quick Check 8 1. A beam of plane-polarised light of intensity I falls normally on a piece of Polaroid. The intensity of the transmitted light is 1 4 I, what is the angle between the plane of Polarisation of the incident beam and the polarising direction of the Polaroid? A 22.5° C 45° B 30° D 60° 2. The fi gure shows a beam of unpolarised light directed along the axis of through Polaroids P1 , P2 and P3 . The polarising axis of each polaroid is as shown by an arrow. Polaroids P1 and P2 are fi xed, with their polarising axes at 30° to each other, and P3 is set with its polarising axis at a variable angle θ to that of P1 . P1 P2 P3 Incident beam 30° θ For what values of θ do intensity of the emergent beam from P3 minimum? A 30°, 120° C 90°, 270° B 60°, 240° D 120°, 300°

Physics Term 3 STPM Chapter 23 Wave Optics 208 23 3. A beam of plane-polarised light with its electric oscillation in the vertical plane and of amplitude a is incident on a Polaroid P. The polarising axis of P is inclined at 30° to the vertical, and that of Q is horizontal. 30° P Q a What is the amplitude of the electric oscillation of the beam emerging from Q? A 1 2 a C 3 4 AB a B 3 4 a D 3 2 AB a 4. Two polarisers are arranged coaxially and the angle between their polarising axes is 30o . An unpolarised beam is directed along the common axis. What is the percentage of the intensity of the beam is absorbed by the polarisers? A 30% C 63% B 38% D 70% 5. P A Polarising axis Unpolarised light θ A beam of unpolarised light is incident on a Polaroid P whose polarising axis is vertical. The emergent polarised light is then incident on another Polaroid A. Polaroid A is arranged such that the intensity of the emergent beam is maximum. Polaroid A is then rotated through 360° about the direction of the incident light. Sketch a graph to show the variation of intensity with the angle of rotation θ of the emergent beam. 6. Two pieces of Polaroid are arranged so that initially their axes of Polarisation are parallel. One of the polaroids is rotated until the intensity of the emergent beam from the two Polaroids is 1 3 its initial intensity. Calculate the angle turned through by the Polaroid. Polarisation Obtained by Reflection 2009/P2/Q12, 2010/P1/Q39 1. When light from the sun is reflected by the surface of water or a piece of glass, and is observed using a piece of Polaroid that is rotated about the direction of propagation of the light, the intensity of light changes. 2. This is because the reflected light is partially plane-polarised. Electric field oscillation Beam Beam Electric field oscillation Beam Electric field oscillation Beam Electric field oscillation (i) (ii) (a) Unpolarised light (b) Partially plane-polarised light (c) Vertically polarised light (i) vertically (ii) horizontally Figure 23.41

Physics Term 3 STPM Chapter 23 Wave Optics 209 23 3. Figure 23.41 (a) shows that for an unpolarised beam the amplitude of the electric field oscillation is the same in all directions. Figure 23.41 (b) shows a partially polarised beam. Figure 23.41 (b) (i) represents a partially vertically polarised beam. The amplitude of the electric field oscillation in the vertical plane is the greatest. In Figure 23.41 (b) (ii) the amplitude of the electric field oscillation in the horizontal plane is greatest. Hence the beam is partially horizontally polarised. Figure 23.41 (c) shows that for a fully vertically polarised beam, the oscillation of the electric field is in the vertical plane only. 4. Light reflected from a surface would be fully polarised if the angle of incident on the surface satisfies Brewster’s law: tan θB = n, refractive index of the reflecting material. The angle θB that satisfies the equation tan θB = n, is knows as Brewster’s angle. 5. From Brewster’s law:tan θB = sin θB cos i = n from Snell’s law : sin θB sin r = n Hence sin θB sin r = sin θB cos θB sin r = cos θB = sin (90° – θB) r = 90° – θB (θB + r) = 90° That is when a ray is incident at Brewster angle, the reflected ray and the refracted ray are at right angles. (Figure 23.42). Reflected ray Refracted ray (Vertically plane-polarised) Air Glass r Incident ray (Horizontally (Unpolarised) plane-polarised) θB Figure 23.43 6. Figure 23.43 shows an unpolarised ray incident at Brewster angle on a glass surface. The oscillations of the electric field can be resolved into two mutually perpendicular components, one parallel to the plane of incidence and the other perpendicular to it. 7. The component parallel to the plane of incident is refracted into glass and the component that is perpendicular to the plane of incident is reflected. 8. Hence the refracted ray is vertically plane-polarised and the reflected ray is horizontally planepolarised. 9. If the angle of incident is not equal to Brewter angle then reflected ray, and the refracted ray are partially plane-polarised. θB θB r Incident ray Reflected ray Air Glass Refracted ray Figure 23.42

Physics Term 3 STPM Chapter 23 Wave Optics 210 23 Reflected ray Incident ray Glass plates Transmitted light almost fully polarised Figure 23.44 10. Figure 23.44 shows a method of producing plane-polarised light. The light transmitted by each glass plate is partially polarised. The emergent light from a stack of glass plates is almost fully plane-polarised. Uses of Plane-polarised Light 1. Polaroid sheets whose polarising direction which is vertical are used in sunglasses (figrure 23.45) for reducing the intensity of direct sunlight, and reduce glare of reflected light. 2. Most reflected light are partially horizontal plane-polarised. Hence a major component of reflected light is absorbed and the transmitted light is of reduced intensity. 3. When polarised light passes through a sugar solution, the plane of polarisation is rotated. The sugar solution is said to be optically active. The angle of rotation increases as the concentration of the solution increases. (Figure 23.46) Eye Polariod B Polaroid A Unpolarised light Sugar solution Vertical polarised light Plane of polarisation rotated through θ θ 1θ Figure 23.46 4. Optical activity is also exhibited by glass or plastic under stress. Example of glass under stress is the curved windscreen of a car. When a piece of strain glass is placed in between two pieces of polaroid with their directions of polarisation at right angles, a stress pattern can be observed. Reflected line which is partially horizontally polarised after falling on the windscreen has its plane of polarisation rotated. A person wearing Polaroid sunglasses with its direction of polarisation vertical is able to observe the stress pattern which consists of parallel bands. 5. In three-dimensional movies, two pictures of the same scene are projected slightly displaced from each other on a curved screen. The scene on the screen looks blurred to the naked eye. When viewed with a pair of glasses made of Polaroids, one with vertical polarising direction and the other horizontal polarising direction, each eye sees a separate scene. The brain interprets the two pictures as a single picture with depth. Figure 23.45

Physics Term 3 STPM Chapter 23 Wave Optics 211 23 Polaroid with horizontal polarising direction Polaroid with vertical polarising direction Dipole aerial Reflectors Figure 23.47 Figure 23.48 6. Television antennas in Malaysia are orientated horizontal because the video signals are carried by horizontal polarised microwaves. If λ is the wavelength, the length of the dipole aerial is λ 2 . Refl ectors are spaced so that constructive superposition between the refl ected waves occurs at the dipole aerial. Quick Check 9 1. A beam of unpolarised light is incident at Brewster angle to a piece of glass. Which one of the following statements is true? A The refl ected and refracted rays are both polarised in the same plane as the plane of the incident beam. B The refl ected ray is polarised in the same plane as the plane of incidence and the refracted ray polarised in the plane normal to the plane of incidence. C The refl ected and refracted rays are both polarised in the plane perpendicular to the plane of the incidence beam. D The refracted ray is polarised in the plane of the incident beam, and the refl ected ray polarised in the plane normal to the plane of the incident beam. 2. Figure (a) shows a beam of unpolarised light incident on a glass surface. The light refl ected is fully plane-polarised. Figure (b) shows a beam of polarised light with plane in the plane of the paper. Which of the following pairs shows correctly the polarising plane of the refl ected beam (if any)? i Unpolarized i Figure (a) Figure (b) A No reflected ray B i i

Physics Term 3 STPM Chapter 23 Wave Optics 212 23 C i No reflected ray i D i i 3. In which one of the following are the waves plane-polarised? A Light from a mercury vapour lamp B Infrared radiation from a hot electric iron C Radiowaves from a dipole aerial D Sound waves from a loud speaker 4. A half-wave dipole aerial emits radio waves of frequency 100 MHz. To reinforce the emission, a refl ector is fi xed behind the dipole so that waves travelling to the back is refl ected in phase with waves emitted in the forward direction. What is the minimum distance of the refl ector from the aerial? A 50 cm C 100 cm B 75 cm D 150 cm 23.8 Optical Waveguides Learning Outcomes Students should be able to: • explain the basic principles of fi bre optics and waveguides • state the applications of fi bre optics and waveguides 1. An optical fi bre is a fl exible, transparent fi ber made of glass or plastic, slightly thicker than human hair. 2. It is used as waveguide or ‘light pipe’ to transit light from one end to the other. 3. The principle used is total internal refl ection. Repeated total internal refl ection bends the light signal around obstacles or corners to get it just where it needs to be. θ1 θ1 θ2 n Medium 2 2 Medium 1 n1 θ3 θ3 θ2 = 90° n Medium 2 2 Medium 1 n1 θ3 θ3 θ2 = 90° n Medium 2 2 Medium 1 n1 (a) Partially refl ected internally (b) Critical angle θc (c) Total internal refl ection Figure 23.49 4. In Figure 23.49 (a), a ray in medium 1 of refractive index n1 is incident on the interface with medium 2 of refractive index n2 (n1 > n2 ). A part of the ray is refracted into medium 2, and part of it is refl ected (θ2 > θ1 ). The angles θ1 and θ2 are related by Snell’s law: n1 sin θ1 = n2 sin θ2 2017/P3/Q10

Physics Term 3 STPM Chapter 23 Wave Optics 213 23 5. When the angle of incidence θ1 is increased, the angle of refraction θ2 increases. When the angle of refraction θ2 = 90o , the angle of incidence θ1 = θc which is known as the critical angle. Applying Snell’s law, n1 sin θc = n2 sin 90o The critical angle at the interface is given by sin θc = n2 n1 6. In figure 23.49(c) the angle θ3 > θc . No ray is refracted and total internal reflection occurs. The two conditions necessary for total internal reflection to occur at an interface are • the incident ray is in the medium of greater refractive index, and • the angle of incidence is greater than the critical angle of the interface. Core, n1 Cladding, n2 Buffer coating Core, n1 Ray Cladding, n2 Figure 23.50 Optical fibre 7. An optical fibre consists of a core made of very pure glass of refractive index n1 and diameter ranging from 10 µm to 200 µm. The core is surrounded by a cladding of refractive index n2 which is smaller than n1 . This is further protected by a buffer coating. Uses of optical fibre as waveguides 1. Telecommunication Fibre optic cables are used to transmit signals. Advantages of fibre optic cables. • Flexible and durable Relatively little loss of intensity, less than 10 percents per kilometer. Loss of intensity due to transmission is known as attenuation. This allows fibre optic cable to transport information over long distances with few repeaters. • Fibre optic cables can carry large number of differentsignalssimultaneously using a technique called wavelength division multiplexing. • Immune to electromagnetic interference. 2. Medical Application The endoscope consists of multiple fibre optic enables medical examination of the interior of the body. The endoscope is passed through the mouth to examine the esophagus, stomach and duodenum. Fibre optic is flexible, hence it is able to reach the interior easily. Since the fibre optic cables are inert, they is no risk of infection.

Physics Term 3 STPM Chapter 23 Wave Optics 214 23 Example 22 A multimode fi bre optic is able to carry difference signals simultaneously. The performance specifi cation of a multimode fi bre optic cable is indicated by the numerical aperture (NA) which is the sine of the maximum angle θmax for light that will be transmitted through the cable. Core, n1 = 1.62 max r c c c c θ Cladding, n2 Cone of acceptance = 1.53 The core of a fi bre optic has refractive index 1.62 and the refractive index of the cladding is 1.53. All light rays within the cone of acceptance will be transmitted through the fi bre optic. Calculate (a) the numerical aperture of the fi bre optic, (b) the apex angle of the cone of acceptance. Solution: (a) When the angle of incidence in air is maximum θmax, the angle of incidence at the corecladding interface is the critical angle, c. Apply Snell’s law, n1 sin c = n2 sin 90o sin c = n2 n1 = 1.53 1.62 Critical angle, c = 70.8o Angle of refraction, r = (90.0 – 70.8)o = 19.2o Apply Snell’s law at the air-core interface, (1.00) sin θmax = n1 sin 19.2o Hence, numerical aperture = sin θmax = (1.62) sin 19.2o = 0.533 θmax = 32.2o (b) Apex angle of the cone of acceptance = 2θmax = 64.4o Quick Check 10 1. The refractive indices of water and glass is 1.33 and 1.50 respectively. (a) Calculate the critical angle for the glass-water interface. (b) A ray in glass is incident on the glassair interface at angles of incidence of (i) 20o , (ii) 62.5o and (iii) 70o . Draw diagram to show the path of the ray. 2. The core of a fibre optic has refractive index of n1 , and the refractive index of the cladding is n2 . For a ray to be transmitted from one end of the fi bre optic to the other end, the angle of incidence of a ray at the air-core interface must be less than θmax. Show that sin θmax = n2 1 – n2 2 3. State the advantages of using fi bre optic cables in telecommunication compared to the use of copper cables.

Physics Term 3 STPM Chapter 23 Wave Optics 215 23 Important Formulae 1. For a geometrical distance of l in medium of refractive index n, the optical path = nl 2. For constructive interference, optical path difference = mλ m = 0, 1, 2, … For destructive interference, optical path difference = (m – 1 2 ) λ m = 1, 2, … 3. Young’s double-slit: fringe width, x = λD a 4. Interference produced by air wedge For dark fringe, 2t = mλ m = 0, 1, 2, … Bright fringe, 2t = (m – 1 2 ) λ m = 1, 2, 3… 5. Blooming of a lens, if refractive index of fi lm µ < refractive index of glass of lens Minimum thickness of fi lm, t = λ 4μ 6. Single slit diffraction: sin θ = λ d 7. Diffraction grating: d sin θm = mλ 8. Intensity of light beam transmitted by a Polaroid, I = 1 2 I0 where I0 = intensity of unpolarised beam 9. Intensity of transmitted polarised beam I = I1 cos2 θ where I1 = intensity of incident polarised beam 10. Brewster’s law: tan θB = n, refractive index STPM PRACTICE 23 1. A ray of monochromatic light of wavelength λ is at near normal incidence to a fi lm of thickness d and refractive index n as shown in the fi gure. R d P Q S Air Air Film What is the path difference between the refl ected rays PQ and RS? A 2d C 2nd – λ 2 B 2nd D 2nd + λ 2. In a Young’s two-slit experiment, when the slit separation is a, and the distance of the screen from the two-slit is D, the fringe separation is x. Changes were made of the slit separation, and distance of the screen from the two-slit. Which of the followings correctly gives the fringe separation? Slit Distance of Fringe separation screen from separation two-slit A a 2D 1 2 x B 2a D 2x C 2a 2D x D 2a 1 2 D 4x 3. Two pieces of polaroids which are arranged such that the polarisation axes make an angle θ to one another are shown in the fi gure. Unpolarised light I 0 θ I If the intensity I of light which emerges from the set-up is 30% of the intensity of the unpolarised light I0, what is the angle θ? A 33.2° C 53.1° B 39.2° D 56.8°

Physics Term 3 STPM Chapter 23 Wave Optics 216 23 4. In a Young’s two-slit experiment, a source of green light is used. The separation of fringes in the interference pattern produced can be increased by increasing. A the distance between the screen and the slit B the distance between the two slits C the size of every slit D the intensity of the green light 5. Two wavetrains are coherent if they have the same A phase C amplitude B speed D wavelength 6. Two Polaroids with parallel planes are centred on a common axis. Their axes of polarization make an angle of θ with each other. When an unpolarised beam of light of intensity I0 is directed towards the arrangement, the intensity of the light emerging from the second Polaroid is I0 3 . What is the value of θ? A 35° C 55° B 48° D 71° 7. Coherent light waves emerge from two fine parallel slits S1 and S2 as shown in the figure. Q P S1 S2 P is the position of the first dark fringe, and Q is the first bright fringe from the central maximum. The phase differences between waves from S1 and S2 at P and Q are respectively A π 2 and π C π and 2π B π 2 and 3π 2 π D 2π and 3π 8. Which of the following monochromatic light waves are not coherent? A Light waves that emerge from the slits of a diffraction grating. B Light waves reflected from a lens surface coated with a thin film. C Light waves from two helium vapour lamps. D Light waves that are pass through a soap film. 9. Water waves of wavelengths 20 cm are produced by two generators S1 and S2 as shown. Each generator S1 and S2 produces waves of amplitude A and 2A respectively at P, which is 140 cm and 90 cm from S1 and S2 respectively. P 90 cm 140 cm S1 S2 When the generator are operated in phase, what is the amplitude of oscillation at P? A 0 C 2A B A D 3A 10. Light of wavelength 600 nm, falls on a double slit, forming fringes 3.00 mm apart on a screen. What would the fringe separation become if the wavelength was 500 nm? A 2.50 mm C 5.00 mm B 3.60 mm D 7.20 mm 11. Which one of the following properties of light shows that it is a transverse rather than a longitudinal wave? A Diffraction B Interference C Reflection D Plane-polarisation 12. A beam of light which consists of two different wavelength λ1 and λ2 (λ1 > λ2 ) is incident normally on a piece of diffraction grating. The third order line for λ2 and the second order line for λ1 overlap. What is the ratio of λ1 λ2 ? A 1.5 C 2.5 B 2.0 D 3.0 13. Monochromatic light of wavelength λ in incident normally on a single slit as shown in the figure.

Physics Term 3 STPM Chapter 23 Wave Optics 217 23 S Q P R x In the direction of the first minimum, the path difference x between the rays PQ and RS is A λ 4 C λ B λ 2 D 3 2 λ 14. A diffraction grating has 6000 lines per cm. When monochromatic light is incident normally on the grating, the first order diffracted beam makes an angle of 15° with the direction of incident beam. What is the frequency of the light? A 4.7 × 1013 Hz C 3.6 × 1014 Hz B 1.9 × 1014 Hz D 7.0 × 1014 Hz 15. A beam of monochromatic light falls at normal incidence on a diffraction grating. Second order diffracted beams are formed at angle of 45° to the original direction. What is the highest order of diffracted beam produced by this grating? A 2 C 4 B 3 D 5 16. (a) (i) Explain why light can be planepolarised, but not sound? (ii) With the aid of a labelled diagram explain the plane-polarisation of light by reflection. (iii) The refractive index of water is 1.33. At what angle of incidence is the light reflected from the surface of a calm pool planepolarised? What is the plane of polarization of the reflected light? (b) Two pieces of Polaroid P and Q are arranged as shown in the figure. The axes of polarization of the Polaroids are initially parallel and vertical. A beam of unpolarised light of intensity I0 is directed towards the Polaroid. Polaroid Q is then slowly rotated about a horizontal axis. Polarisation axis Polaroid Q Polaroid P Unpolarised beam (i) What is the intensity of the beam that emerges from Polaroid P? Explain your answer. (ii) Sketch a graph to show how the intensity of the beam that imerges from Polariod Q varies with the angle of rotation θ from a complete rotation. (iii) Find two values of θ when the intensity of the beam that emerges from Polaroid Q is 25% of the intensity of the unpolarised beam. 17. (a) In Young’s two-slit experiment using monochromatic light, light waves that emerge from the two slits are coherent. With the aid of a diagram, use Huygens’s principle to explain why the waves are coherent. (b) The distance of a point P from two coherent sources S1 and S2 are x and (x + δ) respectively. The displacements at P due to the waves from S1 and S2 are given by y1 = A sin (wt – 2π l x) and y2 = A sin [wt – 2π l (x + δ)] (i) Use the principle of superposition of waves to deduce the resultant displacement Y at P. (ii) Write the expression for the amplitude of the resultant displacement. Hence deduce the values of the path difference δ in terms of the wavelength l for constructive interference, and destructive interference to occur at P. (c) Two coherent sources separated by a distance of 2.00 m emit radio waves of frequency 100 MHz in phase. Calculate the angular separation between the first minimum and the central maximum.

Physics Term 3 STPM Chapter 23 Wave Optics 218 23 18. S1 S2 x O y d In the figure shown, S1 and S2 are two coherent sources. The distance between S1 and S2 is 4.0 m. Electromagnetic waves of wavelength 1.0 m is emitted from S1 and S2 with the same power. (a) Find the position of the first maximum (nearest), second maximum, and third maximum along the line Ox. (b) Discuss whether the intensity of the first (nearest) minimum is zero. 19. (a) What is meant by two coherent sources? (b) A condition required for sharp interference fringes is that the amplitude of the two coherent sources must be the same of approximately equal. Explain what would happen to the interference pattern if there is a big difference in the amplitudes of the two sources. (c) In a Young’s two-slit experiment, the angular separation between two fringes is 0.20° when the wavelength of light is 589 nm. What is the angular separation if the arrangement is in water of refractive index 1.33? 20. (a) (i) With the aid of a diagram, explain the formation of bright and dark fringes in a Young’s double slit experiment. (ii) If λ is the wavelength of light and a is the separation between the slit, deduce the expression for the fringe separation, y on a screen which is at a distance D from the two-slit. (iii) A beam which consists of two different wavelengths λ1 and λ2 is used in a Young’s two-slit experiment. The third bright fringe from the axis of fringe pattern for λ1 overlaps the fourth bright fringe for λ2 . Determine the value of the ratio λ1 λ2 . (b) (i) A soap film of thickness x is illuminated with light of wavelength λ. The refractive index of the soap film is n. Deduce in terms of λ and n, the expression for the thickness x when destructive interference occurs for the reflected light. (ii) Explain the use of thin film in non-reflecting lenses. 21. (a) An extended light source of wavelength 680 nm is used to illuminate normally two glass plates of length 12.0 cm. The plates are in contact at one end and separated by a wire of diameter 0.048 mm on the other end. How many bright fringes can be seen along the length of 12.0 cm? (b) The surface of a piece of glass of refractive index 1.50 is coated with a transparent film of refractive index 1.25. Destructive interference occurs when light of wavelength 600 nm is reflected from the glass surface. Calculate the thickness of the transparent film. 22. (a) Parallel monochromatic light is incident normally on a diffraction grating of 3.0 × 105 lines per metre. A half metre rule is placed 1.00 m from the grating and parallel to its plane as shown in the figure. 1.00 cm 43.0 cm 25.0 cm 7.0 cm Bright lines are observed on the rule at the 7.0 cm, 25.0 cm, and 43.0 cm marks. Calculate the wavelength of light. (b) Light from a source incident normally on a difraction grating consist of two wavelengths 420 nm and 600 nm. Show that the third order line of one of the wavelength is diffracted through a bigger angle than the fourth order line of the other wavelength.

Physics Term 3 STPM Chapter 23 Wave Optics 219 23 23. A diffraction grating which has 5 000 lines cm–1 is used to determine the wavelength of light from a source. When light from the source is incident normally on the grating, the angle between the third order line and the normal to the grating is 60°19’. (a) Calculate the wavelength of light. (b) Explain why the value of the wavelength obtained using the third order line may be more accurate than the value obtained using the first order line. 24. A light source that emits 6.30 × 10–7 m and 6.00 × 10–7 m is used in a spectrometer. Light from the source is incident normally on a diffration grating with 5 000 lines per cm. The first order spectrum is observed using telescope which has an objective lens of focal length 15.0 cm and an eyepiece of focal length 3.0 cm. Calculate (a) the width of the first order spectrum formed on the focal plane of the objective lens. (b) the angular separation between the two lines in the first order spectrum when observed through the eyepiece. 25. A beam of light that consists of two wavelengths is incident normally on a piece of diffraction grating. When the telescope of a spectrometer is rotated through 20° from the direction of the incident light, the second order line of one wavelength is observed to overlap with the third order line of the other wavelength. The shorter wavelength is 400 nm. Calculate (a) the longer wavelength, (b) the number of lines per metre of the grating, (c) the other angles, if any, where overlapping of two lines of different order also occur. 26. (a) The refractive index of diamond is 2.42, find the angle of incidence of a beam of light so that the light reflected is fully plane-polarised. (b) Describe how you would prove experimentally that the reflected light is fully plane-polarised. What is the direction of the plane of polarisation of the reflected light? 27. (a) (i) State Huygens’s principle. (ii) What is meant by constructive interference. (iii) Two waves have the same frequency and amplitude. The phase difference between them is θ, and one of the waves can be represented by the equation y = A sin ωt Determine the range of value of θ so that the amplitude of the resultant wave is less than A. (b) In a Young’s double slit experiment, the slits separation is d and a screen is at a distance D from the double slit (d << D). The intensity of light from each of the slit at the centre of the fringe pattern is I0 . Find the combined intensity of light at the central bright fringe in terms of I0 . Hence explain how your answer is consistent with the principle of conservation of energy. 28. (a) Draw a labelled diagram to show the experiment to determine the wavelength of light using a double slit. (b) A beam of microwave of wavelength 3.1 cm is incident normally on a double slit. Interference fringes are detected along a plane parallel to the double slit at a distance of 40.0 cm from the double slit. The distance between the centre of the maximum on each side of the central maximum is 70.0 cm. Estimate the separation between the slits. Explain why the equation you used to find the slit separation is only an approximation. (c) Division of wavefront is used in obtained two coherent sources in the double slit experiment. (i) Explain what is meant by division of wavefront. (ii) With the aid of a diagram explain another method of obtain two coherent sources of light. 29. An air wedge is formed between two glass plates separated by a thin foil at one end, so that the two plates subtend an angle θ at the point of contact. When the air wedge is illuminated normally by monochromatic light interference fringes can be observed.

Physics Term 3 STPM Chapter 23 Wave Optics 220 23 (a) Explain how the fringes are formed. (b) Draw a diagram to show the fringe pattern. (c) Derive an expression for the fringe separation y in terms of λ and θ. (d) How many dark fringes can be seen if the thickness of the foil is 55 μm and the wavelength of light used is 550 nm? (e) Describe qualitatively, the change to the fringe pattern when the separation between the two plates at the point of contact is increased until the plates are parallel. 30. (a) Explain what is meant of coherent sources. The figure below shows the arrangement used in a Young’s two-slit experiment. Monochromatic light source S1 S2 S ∗ Screen O d (b) (i) Explain why the slit S should be narrow. (ii) Explain why the separation d between the slits S1 and S2 should be small. (iii) Explain how the slits S1 and S2 act as coherent sources. (c) The intensity of light on the screen varies with the distance from the centre O of the interference pattern. In the graph below, the intensity at O is shown. (i) Complete the graph below. (ii) Draw immediately below the graph the interference pattern so as to correspond with the graph. 0 Intensity Distance 31. (a) A beam of red light from a laser is incident normally on a diffraction grating. Bright light is seen emerging at certain angles as shown in the figure. Diffraction grating Red light Use the principle of superposition to explain for the observation. (b) A diffraction grating with a grating spacing of 2.20 × 10–6 m is used to examine light from a source. It is found that the first order violet light emerges at an angle of 11.8° and the first order red light at an angle of 15.8° as shown in the figure. 11.8° 11.8° 15.8° 15.8° 1st order red 1st order violet 1st order violet 1st order red (i) Calculate the wavelengths of the two colours. (ii) Describe and explain what will be observed at an angle of 54.8°. (iii) Without any further calculations, draw a label sketch to show the whole pattern observed. 32. (a) Describe briefly how you would show experimentally that light waves are transverse waves. (b) A beam of unpolarised light is directed towards two pieces of Polaroid P and Q arranged with their polarising direction mutually perpendicular as shown in the figure. I0 P Q Unpolarised light (i) Give in terms of the intensity of the unpolarised light, the intensity of light emerging from Polaroid P, emerging from Polaroid Q.

Physics Term 3 STPM Chapter 23 Wave Optics 221 23 (ii) By using a third piece of Polaroid, explain how you can ensure that light emerges from Polaroid Q. (c) A beam of unpolarised light of intensity I0 passes through three pieces of Polaroid A, B, and C as shown in the figure below. The direction of polarising of the Polaroids are as shown. Unpolarised light I0 20° 60° Direction of polarising A B C Deduce the intensity of the emergent light from Polaroid C in terms of I0 . 33. (a) With the aid of labelled diagrams, explain the difference between unpolarised light and a plane-polarised light beam. (b) Explain why sound waves cannot be plane-polarised. (c) When a beam of unpolarised light is incident at an angle of 30° to a piece of glass, the refl ected light is partially plane-polarised. (i) What is meant by partially planepolarised? (ii) Describe how you would show experimentally that the refl ected light is partially plane-polarised. 34. (a) Explain why light emitted from a fi lament lamp is unpolarised. (b) Two pieces of Polaroid are initially arranged such that their directions of polarising are parallel. Calculate the angle rotated by one of the Polaroid so that the intensity of light transmitted by the two Polaroid is 1 4 the initial intensity. (c) A beam of light, consists of an unpolarised component and a plane-polarised component is incident on a piece of Polaroid. When the Polaroid is rotated, the intensity of the emergent beam varies from a maximum value I m to a minimum value 1 3 Im . Calculate the ratio of the intensity of the unpolarised component to that of the planepolarised component. 1 1. (a) t = 0 t = T t = 2T t = 3T (b) A B ANSWERS

Physics Term 3 STPM Chapter 23 Wave Optics 222 23 2 (a) Ray Wavefront (b) (i) Amplitude ∝ 1 r , Intensity = energy area , A = 4πr2 Intensity ∝ amplitude2 (ii) Amplitude ∝ 1 ABr , A = 2πrd 2 1. Q Q P x d P (a) Change in optical path = [nd + (x – d)] – x = d(n – 1) (b) Change in optical path of λ = 2π, change in phase Change in phase = d(n – 1) λ (2p) rad. 2. (a) Not coherent. Phase difference is not constant if the frequencies are different. (b) Yes. Phase difference is constant = p rad. (c) Yes. Phase difference is constant = π 2 rad 3. (a) S1 and S2 in phase For the point X: S2 X = 5 cm Path difference = S2 X – S1 X = (5 – 3) cm = 2 cm = λ Hence constructive interference For the point Y: S2 Y = S2 Y Path difference = S2 Y – S1 Y = 0 Hence constructive interference For the point Z: S1 Z = 5 cm Path difference = S1 Z – S2 Z = (5 – 3) cm = 2 cm = λ Hence constructive interference (b) S1 and S2 in antiphase (Phase difference = π rad) For the point X: S2 X = 5 cm Path difference = S2 X – S1 X = (5 – 3) cm = 2 cm = λ ≡ 2p rad Total phase difference = (π + 2π) = 3π rad Hence destructive interference For the point Y: S2 Y = S2 Y Path difference = S2 Y – S1 Y = 0 Total phase difference = (0 + π) rad Hence destructive interference For the point Z: S1 Z = 5 cm Path difference = S1 Z – S2 Z = (5 – 3) cm = 2 cm = λ ≡ 2π rad Total phase difference = (π + 2π) = 3π rad Hence destructive interference 3 1. B: x = λD a , D = g, a = f 2. D: Dark fringe (destructive interference) Phase difference = (n + 1 2 )(2π) = (2n + 1)π rad 3. C: Destructive interference, totalpathdifference=(l 1 + l3 ) – (l 2 + l4 ) = (m + 1 2 )λ (l 1 + l3 ) – (l 2 + l4 ) = (2m + 1) λ 2 4. C 5. D: Constructive interference at centre of fringe pattern. Resultant amplitude = 2A , intensity I ∝ (2A)2 When one slit is covered, amplitude = A, intensity = I 4 6. C: PS2 = 5 cm, path difference = PS2 – PS1 = (5 – 3) cm = 2 cm = λ Constructive interference. Resultant amplitude = A + A = 2A 7. A: x = λD a , 0.4 mm ∝ 600 nm, 3.3 10 mm ∝ λ λ = 0.33 0.40 (600 nm) = 495 nm 8. D 9. D: x2 = 2lD a = 2( x2 D ) = 4l a = 8.0 × 10–3 rad 4 1. D: x = λD a , x ∝ D = l 2. D: x = λD a = (3)(1 000) 300 mm = 1.0 cm, A: 3000 mm; B & C: 100 mm (too large). 3. B: Use x = λD a 4. D: New path difference, S2 O – S1 O > 0, O not central maximum 5. C: From x = λD a , when a is large x is very small. Individual fringes are not visible.

Physics Term 3 STPM Chapter 23 Wave Optics 223 23 6. A: New path difference, S2 O – S1 O > 0, O not central maximum. Intensity of light from S1 is lower. 7. 6.6 × 10–6 m Use (µ – 1)d = mλ 8. (a) • Two coherent sources. • Same amplitude. • Small path difference. (c) λ = 5.82 × 10–7 m 9. (a) 0.200 mm (b) 0.300 mm (c) 0.600 mm M M R + B B R B R + B B R B M: magenta R: red B: blue 10. (b) (i) s = λD d (ii) d = (0.2 mm → 0.8 mm) D = (50 cm → 100 cm) (c) (i) Fringe pattern dimmer. – light through double slit decreases. (ii) Intensity of bright fringes decreases. Intensity of ‘dark’ fringes ≠ 0 Fringe pattern less sharp. (iii) Fringe pattern displaced upwards. (iv) When axes of polaroid parallel, fringe pattern sharp, when B rotated, difference of intensity between bright to dark fringe decreases. When axes perpendicular – no interference. 11. (c) (i) Constructive interference Path difference = (XP – YP) = 0 (ii) x = λD a Distance = 4x = 3.84 mm (c) Constructive interference because total path difference = (SX + XP) – (SY + YP) = λ 5 1. C: Colourful pattern is due to interference of light reflected from the oil film. 2. B: Use minimum thickness of coating, t = λ 4 3. B: Path difference of λ 2 occurs when light is reflected from the top surface and the interface. Hence no path difference due to reflection at the two surfaces. 2n1 t = λ 2 (destructive interference) t = 600 4(2.00) nm = 75 nm 4. B: Refer to 3 above, for n1 > n2 , destructive interference 2t – λ 2 = λ 2 , t = λ 2 5. Factors: • refractive index of film • wavelength selected for destructive interference. Thickness = 8.9 × 10–8 m, t = λ 4µ 6. 4.80 × 10–7 m 2µt = (n – 1 2 ) λ, try n = 1, 2, 3, ... 6 1. C: Maximum intensity ∝ (slit width)2 sin θ = λ slit width 2. B: sin θ = λ slit width 3. D: Refer to graph for B, question 1 above. 4. D: YP – XP = λ 2 + λ 2 , sin θ = λ a 5. B: sin θ = λ d , for small θ , θ = λ d = x f x = λf d 6. λ = 7.2 × 10–7 m θ = λ a = x f 7. (b) slit width a  0.2 mm to 0.5 mm. Distance of screen from slit  20 cm 8. (b) Width of central maximum halved. Intensity of central maximum 4 times. 9. Resolving power, θmin = 1.22 λ a = 1.22 ( 500 × 10–9 10.4 ) rad = 5.9 × 10–8 rad 10. θmin = 1.22 λ a = 1.22 ( 450 × 10–9 5 × 10–3 ) rad = 1.1 × 10-4 rad 7 1. A: d sin θ = (1) λ, d = (a + b) 2. A: d sin θ2 = (2)λ, d = 1 N, sin 60o = 2(7.5 × 10–7)N, N = 5.8 × 105 m–1 3. C: d sin θ2 = (2)λ, θ2 = a 2 , d = 2λ sin ( a 2 ) 4. A: m = 1, sin θ1 = Nλ1 = (6 × 105 )(7.00 × 10–7), θ1 = 24.8o sin θ1 ' = Nλ2 = (6 × 105 )(4.00 × 10–7), θ1 ’ = 13.9o a = (24.8 – 13.9)o = 10.9o m = 2, sin θ2 = 2Nλ1 = 2(6 × 105 )(7.00 × 10–7), θ2 = 57.1o sin θ2 ’ = Nλ2 = 2(6 × 105 )(4.00 × 10–7), θ2 ’ = 28.7o b = (57.1 – 28.7)o = 28.4o , a = 10.9o < 1 2 b

Physics Term 3 STPM Chapter 23 Wave Optics 224 23 5. A: d sin θ2 = mλ, d = 1 N, m < 1 Nλ sin 90o = 2.4, maximum order = 2 6. (b) (i) d sin θ2 = 2λ, d = 1 N and λ = c f θ2 = 33° (ii) d sin θ = nλ n < d λ sin 90° gives n =3 Number of maximum = 2n +1 = 7 7. 0.19° Use d sin θ = 2λ twice 8. 433 nm d sin θ = 2 (650 nm) = 3λ′ 9. Δθ = 5.44 × 10–4 rad. 10. (a) 6.26 × 10–7 m tan θ = 0.55 1.50 , d sin θ = λ (b) θ2 bigger % error smaller. 11. Δθ = 14′ use d sin θ = λ twice 12. λ = 1 mn θ = 90° for nth order 13. (a) n = d λ = 1 Nλ Factors : N = number of lines per m λ = wavelength of light (b) 0.04 cm θ2 = 36.09°, x2 = 29.16 cm θ2 ′ = 36.13′, x2 ′ = 29.20 cm 14. (a) 56° sin θ = λ a (b) Intensity –56° –34° –16° 0 16° 34° 56° sin θ = ––d λ θ 8 1. D: I’ = I cos2 θ = I 4 , θ = 60o 2. D: θ = (90 + 30)o = 120o and (180 + 90 + 30)o = 300o 3. C: I ∞ a2 , I p = I cos2 30°, I Q ∞ A2 = I p cos2 (90° – 30°) A = a cos 30° cos 60° = 3 4 a 4. C: I1 = 1 2 I0 I2 = I1 cos2 30o = 0.375 I0 Absorbed = (100 - 37.5)% 5. Intensity 0 90° 180° 270° 360° θ 6. 54.7° I = I0 cos2 θ 9 1. D 2. A: Refer to D from question 1 above. 3. C 4. B: λ = c f = 3 × 108 100 × 106 m = 3.0 m. Reflection at reflector causes a path change = λ 2 For constructive interference between reflected wave and incident wave x + x + λ 2 = λ, x = λ 4 Distance between reflector, x = λ 4 = 75 cm 10 1. (a) Snell’s law: ng sin c = nw sin 90o sin c = nw ng = 1.33 1.50 Critical angle, c = 62.5o (b) (i) 1.33 sin r = 1.50 sin 20o r = 22.7o (ii) 62.5o = critical angle (iii) Total internal reflection occurs. 20° Glass Water r Glass 90° Water Glass 70° Water 2. At the core-cladding interface, n1 sin c = n2 sin 90o sin c = n2 n1 If r is the angle of refraction in the core when the incidence angle at the core-cladding interface is c, then (r + c) = 90o r = (90 – c) At the air-core interface, (1.00) sin θmax = n1 sin r sin θmax = n1 sin (90 – c) = n1 cos c sin c = n2 n1 cos c = n2 1 – n2 2 n1 sin θmax = n1 ( n2 1 – n2 2 n1 ) = n2 1 – n2 2 r c n1 n2

Physics Term 3 STPM Chapter 23 Wave Optics 225 23 STPM Practice 23 1. C: Reflection at upper surface increases path of PQ by λ 2 . Path of RS increases by 2nd. Net path difference = 2nd – λ 2 . 2. C: x = λD a , λ(2D) (2a) = x 3. B: I = 0.30I0 = ( 1 2 I0) cos2 θ, θ = cos-1( 0.60 ) = 39.2o 4. A: x = λD a , x increases when D increases. 5. A: Waves having the same phase are coherent, phase difference = 0 (constant) 6. A: Intensity after first Polaroid, l 1 = l 0 2 . Intensity after second Polaroid, l 2 = l 1 cos2 θ = l 0 2 cos2 θ = l 0 3 . cos θ = 2 3 , θ = 35° 7. A 8. C 9. B: Path difference = (140 – 90) cm = 50 cm = 2 1 2 λ Destructive interference occurs. Resultant amplitude = (2A – A) = A 10. A: x = λD a , x’ = (λ' λ )x = 500 600(3.00 mm) = 2.50 mm 11. D: Longitudinal wave cannot be planepolarised. 12. A: d sin θ = 3λ2 = (2)λ1 . λ1 λ2 = 1.5 13. C 14. D: d sin θ = λ, d = 1 N, λ = sin 15° 6000 × 102 m = 431 nm, f = 3.00 × 108 431 × 10–9 Hz = 7.0 × 1014 Hz 15. A: d sin 45o = 2λ, d sin 90o > mλ, m < 2 sin 45° = 2.8 Highest order is 2 16. (a) (i) Light wave is transverse wave. Sound wave is longitudinal wave. Only transverse wave can be planepolarised. (ii) i Unpolarised incident ray Reflected ray is polarised Refracted ray is polarised (iii) Use Brewster’s law, tan i = n angle of incidence, i = tan-1 1.33 = 53.1o Plane of polarization of reflected ray is perpendicular to the plane of incidence. (b) (i) Intensity = 1 2 I0 Explanation: Vibration of the electric field of the unpolarised beam is in all possible planes. These vibration can be resolved into two mutually perpendicular components – one parallel to the axis of polarization of the Polaroid, and the other perpendicular to it. The perpendicular component is absorbed, and the parallel component is transmitted. (ii) I 1 2 I 0 0 0 90 180 270 360 θ° (iii) I = ( 1 2 I 0) cos2 θ = 0.25I0 θ = cos-1( 0.50 ) = 45.0o and 135o 17. (b) (i) Y = y1 + y2 = (2A cos 2p l δ)[sin (wt – 2p l (x + δ)] (ii) Amplitude, A’ = (2A cos p l δ) Constructive interference, A’ = 2A when cos (p l δ) = 1 or p l δ = 0, p, 2 p, … δ = 0, l , 2l, … = m l where m = 0, 1, 2, 3,… Constructive interference, A’ = 0 when cos (p l δ) = 0 or p l δ = p 2 , 3p 2 , … δ = l 2 , 3l 2 , … = (m+ 1 2 )l where m = 0, 1, 2, 3,… 18. (a) x1 = 1.2 m x2 = 3.0 m x3 = 7.5 m xn S1 S2 P 4 m (xn + nλ) Use Pythagoras Theorem, n = 1, 2, 3, ... (b) Not zero Amplitude of waves from S1 and S2 at the first minimum are not equal. 19. (a) Coherent : constant phase difference.

Physics Term 3 STPM Chapter 23 Wave Optics 226 23 (b) No interference fringes seen because constructive and destructive interference fringes have almost equal amplitude. (c) 0.15° Angular separation = x D = λ a 20. (a) (i) Refer to page 166 (ii) Refer to page 167 (iii) 3x1 = 4x2 3λ1 D a = 4λ2 D a = λ1 λ2 = 4 3 (b) (i) Derive: 2nx = mλ m = 1, 2, 3, ... (ii) Refer to Section 23.4 21. (a) 141, 2t = nλ (b) 1.2 × 10–7 m, 2µt = λ 2 22. (a) λ = 591 nm tan θ = 18 100 , d sin θ = λ (b) Use d sin θ3 = 3 (600 nm) d sin θ4 = 4 (420 nm) θ3 > θ4 23. (a) λ = 579 nm d sin θ3 = 3λ θ3 bigger, % error smaller 24. (a) 2.6 mm (b) 1.38° (a) sin θ1 = λ d , x1 = f 0 tan θ1 sin θ 1 ′ = λ’ d , x2 = f 0 tan θ 2 x = x2 – x1 (b) tan α1 = x1 f e , tan α1 ′ = x2 f e Δα = α1 ′ – α 25. (a) 600 nm d sin θ = 2λ1 , d sin θ = 3 λ2 (b) 2.85 × 105 lines m–1 (c) 43.2° 26. (a) 67.5° tan i = n (b) Observe reflected light through polaroid. Rotate Polaroid. Minimum intensity = 0. 27. (a) (ii) Amplitude of resultant wave bigger than amplitude of either waves. (iii) 120° < θ < 240°. Find resultant amplitude. (b) Average intensity of a bright fringe and a dark fringe = 1 2 (4I 0 + 0) = 2I0 = I0 + I 0 28. (b) 3.5 cm x = λD d f0 fe x2 x1 θ1 θ1 ' α' 1 α 1 Objective Eyepiece (c) (ii) (Draw diagram of interference in thin film) – Division of amplitude 29. (d) 200 2t = nλ (e) Fringe separation decreases until not visible. 30. (a) Coherent sources: constant phase difference (b) (i) S should be narrow to act a a point source. (ii) small d: – separation between interference fringes is large. – can be resolved by the eye. (iii) – at any instant, the same circular wavefront arrives at S1 and S2 . – circular wavefronts from S1 and S2 are in phase. Hence coherent. (c) 31. (b) (i) λ1= 4.50 × 10–7 m, λ2= 5.99 × 10–7 m (ii) 4th order of λ1 and 3rd order of λ2 overlap (iii) n = 3 n = 2 n = 1 n = 0 n = 4 n = 1 n = 2 n = 3 n = 4 λ1 λ2 λ1 λ2 λ1 λ2 λ1 λ1 λ2 λ1 λ2 λ1 λ2 λ1 λ2 32. (a) Show the light can be plane-polarized. (b) (i) 1 2 I0, 0 (ii) Insert 3rd Polaroid between P and Q with its polarizing direction at an angle θ (≠ 90°) to the vertical. (c) 0.259 I0 I = ( 1 2 I0) (cos2 20°) (cos2 40°) 33. (b) Sound waves – longitudinal waves – longitudinal waves cannot be plane-polarised (c) (ii) Observe reflected light through a piece of Polaroid. Rotate Polaroid-minimum intensity not zero. 34. (b) θ = 60° I = 1 4 I0 = I0 cos2 θ (c) Im = 1 2 I1 + I2 1 3 Im = 1 2 I1 I1 I2 = 1.0 0 (i) (ii) Intensity Bright Distance Dark

Physics Term 3 STPM Chapter 24 Quantum Physics CHAPTER QUANTUM PHYSICS 24 Concept Map Photons E = hf Photoelectric Effect hf = W + 1 2 mv2 max Wave-particle Duality de Broglie‘s wavelength λ = h p Nanoscience Atomic Structure Bohr’s Hydrogen Atom En = – Z2 e4 m 8ε2 0 h2 n2 X-ray λmin = hc eV Bragg’s Law 2d sin θ = mλ Quantum Physics 227 Bilingual Keywords Absorption spectrum: Spektrum penyerapan Duality: Kedualan Electroscope: Elektroskop Emission: Pancaran Excited state: Keadaan teruja Fluorescence: Berpendarfl uor Ground state: Keadaan dasar Instantaneous: Seketika Line spectrum: Spektrum garis Penetrating power: Kuasa penembusan Photocell: Sel foto Photoelectric: Fotoelektrik Postulate: Postulat Stopping potential: Keupayaan menahan Threshold frequency: Frekuensi ambang Ultraviolet: Ultraungu Wave-particle: Gelombang-zarah Work function: Fungsi kerja INTRODUCTION 1. In 1887 Hertz discovered photoelectric effect. Further research were carried out by Max Planck and Einstein. 2. In 1905 Einstein suggested the idea of photon to explain the photoelectric effect.

Physics Term 3 STPM Chapter 24 Quantum Physics 228 24 24.1 Photons Learning Outcomes Students should be able to: • describe the important observations in photoelectric experiments • recognise the features of the photoelectric effect that cannot be explained by wave theory, and explain these features using the concept of quantisation of light • use the equation E = hf for a photon • explain the meaning of work function and threshold frequency • use Einstein’s equation for the photoelectric effect hf = W + 1 2 mv 2 max • explain the meaning of stopping potential, and use eVs = 1 2 mv 2 max Photoelectric Effect 1. Photoelectric effect is the emission of electrons from a metal surface when irradiated with electromagnetic radiation. 2. Photoelectric effect is identical to thermonic emission of electrons from a hot cathode. In photoelectric effect, free electrons in the metal absorb energy from the incident electromagnetic radiation to enable them to escape from the metal surface. 3. Figure 24.1 shows the arrangement used to show photoelectric effect. Ultraviolet radiation from a mercury vapour lamp is incidence on a freshly cleaned zinc plate placed on a negatively charged gold leaf electroscope. 4. The divergence of the gold leaf decreases immediately when the zinc plate is irradiated by ultraviolet radiation. This shows the emission electrons from the zinc surface. 5. If the experiment is repeated with a piece of glass placed between the mercury vapour lamp and zinc plate, there is no change in the divergence of the gold leaf. This shows that no electrons are emitted. 6. The piece of glass absorbs ultraviolet radiation. Light transmitted by the glass plate and incident on the zinc surface is not able to eject electrons from the zinc surface. 7. If the experiment (without the glass plate) is repeated using a positively charged electroscope, divergence of the gold leaf does not change when ultraviolet radiation falls on the zinc plate. Any photoelectrons emitted by the zinc plate are attracted back immediately to the positively charged electroscope. 8. Photoelectrons are emitted from metals, such as sodium, potassium and caesium when light falls on these metals. Info Physics Photoelectric effect is a major problem for spacecraft exposed to sunlight. The part of the spacecraft that is exposed to sunlight loses electrons and becomes positively charged. Figure 24.1 e Zinc plate – – – – – – Gold leaf electroscope Ultraviolet radiation from mercury vapour lamp 2012/P1/Q42, 2012/P2/Q7, 2013/P3/Q11, 2014/P3/Q10,11 2015/P3/Q10, 2016/P3/Q10, 2017/P3/Q11

Physics Term 3 STPM Chapter 24 Quantum Physics 229 24 Features of Photoelectric Emission 1. Photoelectric effect can be investigated using a photo cell (Figure 24.2). Light Anode Vacuum Cathode + – Figure 24.2 2. A photocell consists of a cathode whose surface is coated with caesium, and an anode sealed in an evacuated glass tube. μA V Monochromatic light Vacuum Anode Cathode 0 Saturation current V V s I (a) (b) Figure 24.3 3. Figure 24.3 (a) shows the circuit used to investigate the characteristics of a photocell. A monochromatic light source is placed at fixed distance from the photocell so that the intensity of light on the cathode is constant. 4. The voltage across the photocell is varied by adjusting the potential divider. 5. Figure 24.3 (b) is the characteristics I-V graph of the photocell. 6. When the intensity of light is increased by placing the light source closer to the photocell, the saturation current increases. 7. The classical wave theory of light is unable to explain various features of photoelectric effect. These features are explained using Einstein’s quantum theory. (a) The existence of a threshold frequency • The threshold frequency f 0 of a metal is the minimum frequency of the incident electromagnetic radiation which would remove electrons from the metal surface. If the frequency of the incident radiation is below the threshold frequency, no photoelectron is emitted. • Visible light does not remove photoelectron from a zinc surface because the frequency of visible light is less than the threshold frequency of zinc. According to Einstein’s quantum theory, energy of an electromagnetic radiation is quantizied, or in discrete quantity or packet. The quantum of electromagnetic radiation is the photon. E = hf where h = Planck constant, 6.63 × 10–34 J s f = frequency of electromagnetic radiation

Physics Term 3 STPM Chapter 24 Quantum Physics 230 24 • The work function W of a metal is the minimum energy required by a free electron in the metal to escape from the metal surface. From the definition of the threshold frequency f 0 , the work function of a metal is given by Work function, W = hf0 • Hence, if the frequency f of the incident radiation is less than the threshold frequency f 0, the energy of a photon E = hf < hf0 = W the work function of the metal. Electrons would not be removed from the metal surface. • The classical wave theory of light fails to explain the existence of the threshold frequency. According to the classical wave theory, light energy is absorbed continuously by the free electrons in the metal. When the sufficient energy has be absorbed, the electrons would escape from the metal. Hence, the frequency of the incident radiation needs not exceed a certain minimum value. This does not happen. (b) Instantaneous emission of photoelectron • When radiation of frequency f, greater than the threshold frequency f 0 , is incident on a metal surface, photoelectrons are emitted instantaneously . If the energy of a photon E = hf absorbed by an electron is enough for it to escape from the metal surface, the electron is immediately emitted as a photoelectron. • According to the classical wave theory, an electron needs time to absorb sufficient energy for it to escape from the metal surface. Hence, there would be a time lapse before photoelectrons are emitted. Experimental observations shows otherwise. (c) The photoelectric current is directly proportional to the intensity of the incident radiation. 0 Intensity Photoelectric current Figure 24.4 • When the intensity of the incident radiation is increased, the number of photons per second incident on the metal surface increases. More photoelectrons are emitted and the photoelectric current in the photocell increases. The increase in the photoelectric current is due to the increase in the rate of emission of photoelectrons, not because the photoelectrons are emitted with greater speeds. • According to the classical wave theory, increase in intensity of radiation means that the amplitude of the incident electromagnetic wave increases and photoelectrons would be emitted with a greater speed which is not correct. (d) The maximum kinetic energy of the photoelectrons increases when the frequency of the incident radiation increases. • If f = frequency of the incident radiation f 0 = threshold frequency of metal (f > f 0 ) Then according to the principle of conservation of energy, energy of incident photon = work function + maximum kinetic energy of photoelectron

Physics Term 3 STPM Chapter 24 Quantum Physics 231 24 Hence, maximum kinetic energy of photoelectron = energy of incident photon – work function Emax = (hf – W) (W = hf0 ) = h (f – f 0 ) or 1 2 mv2 max = h(f – f 0) where vmax = maximum velocity of photoelectrons m = mass of photoelectron The above equation is known as Einstein’s equation. • In Einstein’s equation Emax = (hf – W) Emax represents the maximum kinetic energy of the photoelectrons. There are photoelectrons whose kinetic energy is less than Emax. Electrons that were further from the metal surface require energy more than the work function W to be emitted. Hence, these electrons would be emitted with kinetic energy less than Emax. • According to the classical wave theory, the maximum kinetic energy of the photoelectron would increase if the intensity of the incident radiation is increased. This is incorrect as discuss in (c) above. Example 1 The work function of a metal is 3.63 eV. Ultraviolet radiation of wavelength 300 nm is incident on the metal surface. (a) Calculate (i) the threshold frequency of the metal. (ii) the maximum kinetic energy of the photoelectron. (b) If the metal surface is irradiated by ultraviolet radiation of wavelength 200 nm, discuss whether the maximum kinetic energy of the photoelectron would be greater or smaller. Solution: (a) (i) Using W = hf0 . Threshold frequency, f 0 = 3.63 × (1.60 × 10–19) 6.63 × 10–34 = 8.76 × 1014 Hz (ii) Using Emax = hf – W (f = c λ ) = hc λ – W = (6.63 × 10–34) × (3.0 × 108 ) (300 × 10–9) – 3.63 × (1.60 × 10–19) = 8.22 × 10–20 J (b) When λ′ = 200 nm < λ = 300 nm E′ max = hc λ′ – W Since λ′ < λ, maximum kinetic energy of photoelectron E′ max is greater.

Physics Term 3 STPM Chapter 24 Quantum Physics 232 24 Experiment to Investigate the Variation of the Maximum Kinetic Energy of Photoelectrons from a Metal Surface with the Frequency of Incident Radiation 2008/P1/Q42, 2009/P1/Q43, Q44, 2011/P1/Q41, 42, 2011/P1/Q13 1. To measure the maximum kinetic energy Emax of the photoelectrons, a reverse potential is applied to the cathode of the photocell as shown in Figure 24.5. 2. Initially the slider S of the potential divider XY is at X, so that there is no potential difference across the photocell, and the voltmeter reading is zero. Figure 24.5 µA V + – Cathode Anode S X Y 3. When monochromatic light of frequency f is incident on the cathode, emitted photoelectrons have sufficient energy to reach the anode. Current flows through the photocell and the microammeter shows a reading. 4. The slider S of the potential divider is moved from X towards Y so that the potential of the cathode is positive. This reverse potential opposes the motion of the photoelectrons from the cathode to the anode. The microammeter reading decreases. 5. The reverse potential is increased until the microammeter reading becomes zero. This shows that photoelectrons with the maximum kinetic energy Emax are stopped. The reading Vs of the voltmeter gives the value of the stopping potential. Hence, maximum kinetic energy of photoelectrons Emax = eVs where e = charge of electron 6. The experiment is repeated for monochromatic light of different frequencies and the corresponding stopping potential Vs is noted. 7. The graph of Emax = eVs against f is plotted as shown in Figure 24.6. The shape of the graph verifies Einstein’s equation Emax = hf – W. E max 0 –W f f0 Gradient = h Figure 24.6

Physics Term 3 STPM Chapter 24 Quantum Physics 233 24 8. From Einstein’s equation Emax = hf – W = h (f – f 0 ) (a) gradient of the graph = h, Planck constant, (b) when Emax = 0, f = f0 , the threshold frequency of the cathode. 9. When the experiment is repeated using photocells of different metals as cathode, the graphs as shown in Figure 24.7 are obtained. f f1 f2 f3 –W1 –W2 –W3 E max 1 2 3 Metal 0 Figure 24.7 f1 , f 2 , f 3 respectively are the threshold frequencies of metals 1, 2 and 3. W1 , W2 , W3 respectively are the work functions of metals 1, 2 and 3. Example 2 365 nm I/mA V/V = 365 nm 5 λ –1 1 2 0 Figure (a) Figure (b) Figure (a) shows a photocell connected in a circuit. The cathode is illuminated with monochromatic radiation of wavelength 365 nm and the current I in the circuit is measured for various values of the applied potential difference V between the cathode and anode. The results are as shown in the graph in Figure (b). (a) Find the maximum kinetic energy of the photoelectrons. (b) Deduce the work function of the cathode. (c) If the experiment are repeated using monochromatic radiation of wavelength 313 nm, where would the new graph meet the V-axis? Solution: (a) From Figure (b), when current I = 0, V = –1.0 V. Hence, the stopping potential Vs = 1.0 V. Maximum kinetic energy of photoelectrons. Emax = 1.0 eV Exam Tips hc λ is in J. To change energy in J to eV, divide the value by e =1.60 × 10–19 C

Physics Term 3 STPM Chapter 24 Quantum Physics 234 24 (b) Using Einstein’s equation Emax = hf – W (f = c λ ) Work function, W = hc λ – Emax = (6.63 × 10–34) (3.0 × 108 ) (365 × 10–9) (1.60 × 10–19) – 1.0 eV = 2.41 eV (c) When λ′ = 313 nm E′ max = hc λ′ – W = (6.63 × 10–34) (3.0 × 108 ) (313 × 10–9) (1.60 × 10–19) – 2.41 = 1.56 eV Hence, stopping potential = 1.56 V New graph meets the V-axis at V = –1.56 V Example 3 Photons of energy 3.5 eV is incident on a plane cathode of work function 2.5 eV in an evacuated tube. Close to the cathode is a parallel collector. (a) Calculate the maximum kinetic energy Emax of the photoelectrons emitted from the cathode. (b) Find the reverse potential difference between the cathode and collector that would prevent photoelectrons of energy Emax from reaching the collector if the electrons are emitted (i) normal to the cathode, (ii) at an angle of 60° to the cathode. Solution: (a) Using Einstein’s equation Emax = hf – W = (3.5 – 2.5) eV = 1.0 eV (b) (i) When photoelectrons are emitted normal to the cathode, eVs = Emax Stopping potential difference, Vs = (1.0 eV) e = 1.0 V (ii) When the photoelectrons are emitted with a velocity v at 60° to the cathode, component of velocity normal to cathode = v sin 60°. Hence, 1 2 m (v sin 60°)2 = eVs . Stopping potential difference, Vs = 1 2 mv2 sin2 60° e = Emax e sin2 60° = (1.0 eV) e × sin2 60° = 0.75 V

Physics Term 3 STPM Chapter 24 Quantum Physics 235 24 Example 4 In a photomultiplier tube, incident light ejects electrons from a cathode. The electrons are accelerated and strike a target from which 5 electrons are ejected from every incident electron. These 5 electrons are accelerated, strike a second target and produce 25 electrons and so on. There are a total of 9 such targets in the photomultiplier. (a) The emitted electrons from the fi nal target constituted a current of 0.10 mA. Calculate the rate at which electrons are emitted from the cathode. (b) One in three photons of wavelength 400 nm incident on the cathode succeeds in ejecting an electron from the cathode. Calculate the power of the light incident on the cathode. Solution: Cathode 1 2 9 1e 5e 52 e 53 e 59 e (a) For an electron ejected from the cathode, and strikes the 1st target, 5 electrons are ejected. After the 2nd target, 25 = 52 electrons are ejected, after the 9th target, 59 electrons are ejected. Let rate of electrons ejected from cathode = n per second. Then after the 9th target, number of electron = 59 n. Current = charge per second 0.10 mA = (59 n)e (e = charge of electron) n = 0.10 × 10–3 59 × (1.6 × 10–19) = 3.2 × 108 s–1 (b) Since 1 in 3 photons succeeds in ejecting 1 electron, number of photon s–1 = 3 (3.2 × 108 ) = 9.6 × 108 Power of incident light = (9.6 × 108 ) hc λ = (9.6 × 108 ) (6.63 × 10–34) (3.0 × 108 ) 400 × 10–9 = 4.77 × 10–10 W Example 5 In a photoelectric experiment, ultraviolet radiation of wavelength 254 nm and intensity 210 W m–2, is incident on a silver surface of area 12 mm2 . A photoelectric current of 4.8 × 10–10 A is produced. (a) Calculate (i) the rate of incidence of photons on the silver surface, (ii) the rate of emission of electrons.

Physics Term 3 STPM Chapter 24 Quantum Physics 236 24 (b) The photoelectric quantum yield is defi ned as the ratio number of photoelectrons emitted per second number of photons incident per second (i) Find the quantum yield of the silver surface at the wavelength 254 nm. (ii) Give two reasons why this value might be expected to be much less than one. (c) When the experiment is repeated with ultraviolet radiation of wavelength 313 nm, no photoelectrons were emitted. Explain this observation. Solution: (a) (i) Intensity, I = 210 W m–2 Surface area, A = 12 mm2 = 1.2 × 10–5 m2 If rate of photon = n then n ( hc λ ) = IA n = (210) (1.2 × 10–5) (6.63 × 10–34) (3.0 × 108 ) × (254 × 10–9) = 3.2 × 1015 s–1 (ii) Current, I = Ne Rate of emission of photoelectron, N = I e = 4.8 × 10–10 (1.60 × 10–19) = 3.0 × 109 s–1 (b) (i) Photoelectric quantum yield = Rate of emission of photoelectron Rate of incidence of photon = 3.0 × 109 3.2 × 1015 = 9.4 × 10–7 (ii) The value of the photoelectric quantum yield is much less than one because • not all incident photons succeed in ejecting an electron. • not all emitted photoelectrons manage to reach the anode. (c) When λ = 313 nm, no photoelectrons are emitted because the energy of a photon of wavelength 313 nm is less than the work function of silver. Quick Check 1 1. Monochromatic light is incident on the cathode of a photocell. An electron is emitted with the maximum kinetic energy if A it absorbed the most photons B it is at the surface of the cathode C its temperature is the highest D it gained thermal energy 2. The fi gure shows how the stopping potential of a photocell varies with the frequency of the incident radiation. 0 Frequency Stopping potential The gradient of the graph depends on the

Physics Term 3 STPM Chapter 24 Quantum Physics 237 24 A intensity of the incident radiation. B ratio of the Plancks’s constant to the electronic charge. C work function of the irradiated surface. D wavelength of the incident radiation. 3. In a photoelectric experiment, when the wavelength of the incident radiation on a metal surface is increased, the electrons emitted have A a greater average kinetic energy. B a greater maximum kinetic energy. C a smaller minimum kinetic energy. D a smaller average kinetic energy. 4. Light photons of energy 3.5 × 10–19 J is incident on a cathode of a photocell. When the potential of the cathode is 0.25 V positive with respect to the anode, the current through the photocell falls to zero. The work function of the cathode is A 2.9 × 10–19 J C 3.5 × 10–19 J B 3.1 × 10–19 J D 3.9 × 10–19 J 5. In a photoelectric experiment, electrons are ejected from metals X and Y by light of frequency f. The potential difference V required to stop the electron is measured for various frequencies. If Y has a greater work function than X, which graph illustrates the expected results? A V X Y f C V X Y f B V Y X f D V Y X f 6. In a photoelectric emission experiment, when the incident photon is of frequency f 1, the maximum kinetic energy of the photoelectron is E1. When the photon frequency is f 2, the maximum kinetic energy of the photoelectron is E2 . Which of the following is equal to the Planck’s constant? A 1 2 ( E1 f 1 ) + ( E2 f 2 ) C (E1 + E2 ) (f 1 + f 2 ) B (f 1 + f 2 ) (E1 + E2 ) D (E1 – E2 ) (f 1 – f 2 ) 7. Photoelectrons are ejected from the surface of a metal when the wavelength of the incident light is equal or less then λ. When the wavelength of the incident light is λ 2 , the maximum kinetic energy of the photoelectron is E. What is the maximum kinetic energy of the photoelectrons when the wavelength is λ 4 ? A E 2 C 3E B 2E D 8E 8. Radio waves of frequency 101 MHz are transmitted at a rate of 150 kW from a radio station. What is the number of photons per second transmitted from the station? A 1.12 × 1030 C 2.24 × 1030 B 1.52 × 1030 D 2.29 × 1030 9. When the surface of a metal is illuminated by photons of energy 2.00 eV, emitted electrons have maximum kinetic energy 0.10 eV. What is the maximum wavelength of the incident radiation which would eject electrons from the metal surface? A 562 nm C 654 nm B 592 nm D 684 nm 10. When a metal surface is illuminated with light of wavelength 520 nm, the stopping potential required is 0.80 V. If light of wavelength λ is used, the stopping potential is 1.58 V. What is the value of λ? A 374 nm C 748 nm B 392 nm D 772 nm 11. In a series of photoelectric emission experiments, a number of metals of work function Φ were illuminated with light of different frequencies f, and intensities I. The maximum kinetic energy of the emitted electrons depends on A Φ, but not f or I B Φ and f, but not I C Φ and I, but not f D I and f, but not Φ

Physics Term 3 STPM Chapter 24 Quantum Physics 238 24 12. A beam of monochromatic radiation falls on a metal X and photoelectrons are emitted. The rate of emission of photoelectrons is doubled if A the intensity of the beam is doubled B radiation of double the wavelength is used C radiation of double the frequency is used D the thermodynamic temperature of the metal is doubled 13. A laser pen has a power of 5.0 W and it emits monochromatic light of wavelength 500 nm. How many photons are emitted per second? A 5.0 s–1 C 4.5 s–1 B 1.3 s–1 D 3.3 s–1 14. In a photoelectric experiment, the cathode is illuminated with radiations of different frequencies f. The potential difference required to prevent any photoelectron from reaching the collector is V. The graph obtained as shown. When the frequency of the radiation is f 1, the maximum kinetic energy of the photoelectron is A hf1 C eV1 B h(f 1 – f0 ) D eV1 (f 1 – f0 ) 15. When the surface of copper is illuminated with light, no photoelectrons are emitted because A the threshold frequency of copper is less than the frequency of light B the intensity of light is not enough to eject photoelectron C the energy of the light photon is less than the work function of copper D the kinetic energy of the photoelectron is less than a photon of light 16. When light of wavelength 300 nm falls on a metal surface, photoelectrons of maximum kinetic energy 2.0 eV is ejected from the surface. What is the maximum wavelength of light that would be able to eject electrons from the surface? A 440 nm C 580 nm B 460 nm D 620 nm 17. When a metal surface is illuminated separately by radiations of frequencies f 1 , f 2 , and f 3 photoelectrons of maximum kinetic energies E1, E2, and E3 respectively are emitted. If f1 > f 2 > f3 , what is the maximum kinetic energy of the photoelectrons when the metal surface is illuminated simultaneously by radiations of frequencies f 1 , f 2 , and f3 ? A E1 C E1 + E2 + E3 B E3 D E1 + E2 + E3 3 18. Two beams, P and Q of the same frequency, illuminate the same metal surface causing photoemission of electrons. The photoelectric current produced by P is four times that produced by Q. Which of the following gives the ratio wave amplitude of beam P wave amplitude of beam Q A 1 4 C 4 B 2 D 16 19. Monochromatic light illuminates the cathode of a photocell causing emission of photoelectrons which are collected by the anode. For a given intensity of light, the variation of photocurrent I with potential difference V across the photocell is as shown in the figure. If the experiment is repeated with light of twice the intensity but the same wavelength which of the graphs below best represents the new relation between I and V? A

Physics Term 3 STPM Chapter 24 Quantum Physics 239 24 B C D 20. Which of the following phenomena cannot be explained using the classical wave theory of light? A Polarization of light B Interference of light C Photoelectric emission D Diffraction 21. In a series of photoelectric emission experiments, a graph of the quantity y is plotted against the quantity x. The graph is as shown below. y 0 x Which of the following correctly identifies y and x? y x A Photoelectric current Frequency of radiation B Maximum kinetic energy of photoelectron Frequency of radiation C Intensity of radiation Maximum kinetic energy of photoelectron D Photoelectric current Intensity of radiation 22. Which graph correctly shows the variation of the work function W of a metal with the frequency f of the incident light? A 0 W f C 0 W f B 0 W f D 0 W f 23. The energy of a photon is 14.5 eV. What is its wavelength? A 8.57 × 10-8 m B 1.17 × 10-7 m C 1.52 × 10-7 m D 3.22 × 10-6 m 24. Light of wavelength 420 nm is incident on a metal surface. The maximum kinetic energy of the photoelectrons produced is 1.35 eV. What is the maximum wavelength of light that will cause photo emission from the metal? A 288 nm C 772 nm B 522 nm D 921 nm 25. Light of wavelength 435 nm is used to illuminate the surface of a piece of sodium metal in a vacuum. (a) Calculate the energy of a photon of light of this wavelength. (b) What is the maximum kinetic energy of the photoelectrons if the maximum wavelength of light that can produce photoelectrons from sodium is 650 nm. 26. In a photoelectric experiment, a photoelectric current of 200 μA is produced when light of wavelength 550 nm illuminates a metal surface of area 2.0 cm2. Calculate the intensity of the incident light. State any assumptions you made. 27. The maximum kinetic energy of photoelectrons ejected from a tungsten surface by ultraviolet radiation of wavelength 248 nm is 8.6 × 10–20 J. Find the work function of tungsten.

Physics Term 3 STPM Chapter 24 Quantum Physics 240 24 28. In a photoelectric experiment, light of frequency 8.2 × 1014 Hz illuminates the surface of a metal in a vacuum. The maximum kinetic energy of the photoelectrons is 1.2 eV. What is the work function of the metal? Discuss briefly how you would measure the maximum kinetic energy of the photoelectrons. 29. In a photoemission experiment, the surface of a metal in an evacuated tube is illuminated with monochromatic light. If the experiment is repeated with light of the same wavelength but twice the intensity, explain the changes, if any, to (a) the energy of a photon, (b) the maximum kinetic energy of the photoelectrons, (c) the work function of the metal, (d) the photocurrent. 30. A metal surface in an evacuated tube is illuminated with monochromatic light. The photoelectrons emitted are collected by an adjacent electrode. Draw sketch graphs to show how the photoelectric current I depends on (a) f the frequency of the incident light, the potential difference V, and the intensity P of the incident light remaining constant, (b) V, when P and f remaining constant, (c) P, when f and V remaining constant. Explain the main features of these graphs. 31. (a) What do you understand by (i) Planck’s constant, h, (ii) the work function Φ of a metal? (b) Outline an experiment to determine the value of h, and Φ of a metal using photoelectric effect. Explain how the readings obtained in the experiment is processed to give the values of h and Φ. (c) The surface of a metal is illuminated by a beam of monochromatic ultraviolet radiation, and photoelectrons are collected by an electrode. The potential difference V between the metal and the electrode is varied and the corresponding current I measured. The curve A in the graph below shows how I changes when V changes. C B A State and explain for each case below, how the beam of ultraviolet radiation is changed to obtain the curve labelled (i) B (ii) C. 32. (a) Explain what is meant by photoelectric effect. (b) When a photocell is illuminated by monochromatic light of frequency f, photoelectrons are emitted, a photoelectric current flows in the photocell. The photoelectric current can be reduced to zero by applying a suitable potential difference V. (i) Sketch a graph to show the variation of V with the frequency f. (ii) Derive an expression for V in terms of f. Identify any constants in your expression. (c) An evacuated tube has two parallel electrodes Eand C separated by 5.0 mm. When the electrode E is illuminated with ultraviolet radiation of wavelength 275 nm, photoelectrons are produced. When the potential of the electrode E is +2.0 V relative to the electrode C, all photoelectrons are stopped from reaching the electrode C. (i) What is the energy of a photon of the ultraviolet radiation? (ii) What is the work function of the electrode E? (iii) Calculate the maximum speed of photoelectrons emitted from E. (iv) Sketch a graph to show how the electric potential energy of a photoelectron depends on its distance from the electrode E. On the same horizontal axis, sketch labelled graphs to show how the kinetic energy of the fastest electron, and its total energy depends on the distance from E.

Physics Term 3 STPM Chapter 24 Quantum Physics 241 24 24.2 Wave-Particle Duality Students should be able to: • state de Broglie’s hypothesis • use the relation λ = h p to calculate de Broglie wavelength • interpret the electron diffraction pattern as an evidence of the wave nature of electrons • explain the advantages of an electron microscope as compared to an optical microscope Learning Outcomes 1. Wave-particle duality is the phenomenon where under certain circumstances a particle exhibits wave properties, and under other conditions a wave exhibits properties of a particle. 33. (a) Write down Einstein’s equation for photoelectric effect. Explain the physical process represented by each term of the equation. (b) The results of two experiments using a photocathode of work function 2.05 eV is shown in the fi gure below. Only 5% of the incident photons succeeded in ejecting photoelectrons in each of the experiments. Use the results of experiment 1, to determine I/mA 0 –1.10 –0.50 0.28 0.14 Experiment 1 Experiment 2 V/V (i) the maximum kinetic energy of the photoelectrons, (ii) the energy of an incident photon, (iii) the number of photons per second incident on the photocathode, (iv) the threshold frequency of the photocathode. (c) (i) “The graph for experiment 2 shows that the intensity of the incident radiation of experiment 1 is twice that in experiment 2”. Comment whether this statement is correct, give reasons for your answer. (ii) What changes were carried out in experiment 2? 34. (a) State what is meant by the quantum theory of light. What do you understand by the term photon. (b) State four characteristics of photoelectric emission. Explain each of the characteristics. 35. (a) Write Einstein’s equation for photoelectric emission. Identify each term in the equation. Which of the terms is characteristic of the metal. (b) A metal is illuminated with monochromatic light. Explain why the photoelectrons emitted have a range of kinetic energy. (c) A gold surface is illuminated with monochromatic electromagnetic radiation of photon 1.20 × 10–18 J. The maximum kinetic energy of the photoelectrons is 4.2 × 10–19 J. Calculate (i) the wavelength of the electromagnetic radiation and identify the electromagnetic radiation, (ii) the work function of gold, (iii) the threshold frequency of gold. 2009/P1/Q44, 2012/P1/Q43, 2013/P3/Q12, 2015/P3/Q11

Physics Term 3 STPM Chapter 24 Quantum Physics 242 24 2. Particle properties of a wave (a) According to the quantum theory, a photon of electromagnetic radiation of wavelength λ has energy E = hf = hc λ ............................ ① where h = Planck constant c = speed of light in vacuum (b) According to Einstein’s theory of special relativity, the energy equivalent E of a mass m is given by E = mc2 ............................. ② Equating equations ① and ②, hc λ = mc2 mc = h λ That is momentum p of a photon of wavelength λ is p = mc = h λ Hence a photon behaves as a particle having a momentum p ∝ 1 λ 3. Crompton demonstrated the particle property of a photon using hard X-rays to collide with free electrons in graphite. (Figure 24.8) Incident X-rays Graphite Emergent X-rays Figure 24.8 4. X-rays emerging from the graphite have longer wavelengths showing a decrease in momentum. Part of the momentum of the incident X-rays has been transferred to the free electrons in graphite during collisions. 5. The shift in the wavelength of X-rays after colliding with free electrons in graphite is known as Crompton effect. Diffraction of Electron – de Broglie Wavelength 2010/P1/Q45 e Cathode +4 000 V Anode Graphite film Vacuum Fluorescence screen Figure 24.9 1. The wave property of a particle is illustrated using the apparatus shown in Figure 24.9. A beam of electrons in a cathode-ray tube is directed towards a graphite film.

Physics Term 3 STPM Chapter 24 Quantum Physics 243 24 2. A diffraction pattern is observed on the fl uorescence screen. This shows that a beam of fast moving particles (electrons) behaves as a wave, exhibiting diffraction – a wave property. 3. According to de Broglie, the wavelength λ associated with a particle of mass m travelling at a velocity v is given by momentum of particle, p = mv = h λ de Broglie wavelength, λ = h mv Example 6 Estimate the de Broglie wavelength of an electron beam which has been accelerate through a potential difference of 3 600 V. Solution: Kinetic energy gained by electron, 1 2 mv2 = eV v = 2eV m de Broglie wavelength, λ = h mv = h m 2eV m = h 2meV = (6.63 × 10–34) 2 × (9.1 × 10–31) (1.6 × 10–19) (3 600) = 2.0 × 10–11 m Note: From the above calculation, the wavelength of an electron beam is about 1 30 000 the wavelength of visible light. In an electron microscope, a beam of electron is used. Since the wavelength of the electron beam is 1 30 000 that visible light, then the resolving power of the electron microscope is 30 000 times of the optical microscope. Electron Microscope 1. There are two types of electron microscope, the transmission electron microscope and the scanning electron microscope. 2. Both types of electron microscope use a beam of fast electrons from an electron gun. 3. Instead of using lenses to focus light as in the optical microscope, magnetic fi elds are used as lenses to focus the electron beam in electron microscopes. 4. After been focussed, the electron beam passes through a thin sample of the specimen in the transmission electron microscope. An image is then projected onto a fl uorescence screen for viewing.

Physics Term 3 STPM Chapter 24 Quantum Physics 244 24 5. In the scanning electron microscope, the beam of fast electrons is used to scan the specimen, just like the electron beam scanning the screen of a television screen. 6. The bombardment of the fast electrons on the specimen surface causes the emission of electrons from the specimen surface. These electrons are detected by a detector and an image formed on the screen of a cathode-ray tube. 7. The biggest advantage is that electron microscopes have higher resolving power. The resolving power is given by smallest angular separation θ between two points that can be resolved by the microscope. θmin = 1.22 λ a a = diameter of aperture 8. From the above example, the de Broglie’s wavelength of an electron beam is 1 30 000 the wavelength of visible light. This means that the angular separation θmin for electron microscope is ( 1 30 000)θmin for optical microscope. 9. Optical microscope has a maximum magnifying power of 2000. Therefore the magnifying power of electron microscope can reach (30,000) × (2000) = 6 × 107 . 10. The second advantage is that scanning electron microscope is able to produce three dimensional external shape of the specimen. 11. Besides the high cost of electron microscopes, other disadvantages compared to optical microscopes are • the image is only in black and white • Impossible to observe living specimen Example 7 Calculate the de Broglie wavelength of a bullet of mass 0.02 kg travelling at 300 m s–1. Solution: Using the equation λ = h mv = (6.63 × 10–34) 0.02 × 300 = 1.1 × 10–34 m Example 8 (a) Both X-rays and electrons are diffracted strongly in certain directions by a crystalline solid. (i) What can be deduced about the characteristics of the arrangement of atoms in crystalline solid? (ii) What properties of X-rays and electrons that made this possible? (b) The fi gure shows a row of atoms on a crystal surface. The separation between atoms is a. A beam of electron of wavelength λ is incident normally on the crystal and is diffracted in all directions. (i) Find the path difference between the beam PA and QB which are at angles a to the vertical and in the same vertical plane as the row of atoms.