13119 11 Diamond has a refractive index of 2.42. (a) Show that the critical angle for diamond is 24°. [2] (b) Pieces of diamond are cut so as to ‘sparkle’. This is partly due to the phenomenon of total internal reflection of light within the stone. A ray of white light incident at 90° to the surface of a cut diamond is shown in Fig. 11.1. A C B 100 40 40 Fig. 11.1 (i) Explain why the ray has taken the path shown after striking boundary AB. [2] 13119 (ii) Continue the path of the ray on Fig. 11.1 to show the path it follows before and just after striking boundary BC. Draw the normal line and include a value for the angle of incidence at the boundary BC. [3] (iii) Diamonds which are ‘shallow cut’, such as the one in Fig. 11.2, would not sparkle in the same manner. Explain why this is the case. You may illustrate your answer by continuing the path of the ray after striking boundary EF on Fig. 11.2. [2] 140 20 E F Fig. 11.2 THIS IS THE END OF THE QUESTION PAPER 13119 BLANK PAGE DO NOT WRITE ON THIS PAGE 13119 BLANK PAGE DO NOT WRITE ON THIS PAGE Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright holders may have been unsuccessful and CCEA will be happy to rectify any omissions of acknowledgement in future if notified. For Examiner’s use only Question Number Marks 1 2 3 4 5 6 7 8 9 10 11 Total Marks Examiner Number SPH21/6 262603 DO NOT WRITE ON THIS PAGE DATA AND FORMULAE SHEET Physics [SPH11/SPH21] ADVANCED SUBSIDIARY General Certifi cate of Education Assessment Units AS 1 and AS 2 11378.02 11378.02 2 Data and Formulae Sheet for AS 1 and AS 2 Values of constants speed of light in a vacuum c = 3.00 × 108 m s–1 elementary charge e = 1.60 × 10–19 C the Planck constant h = 6.63 × 10–34 J s mass of electron me = 9.11 × 10–31 kg mass of proton mp = 1.67 × 10–27 kg acceleration of free fall on the Earth’s surface g = 9.81 m s–2 electron volt 1 eV = 1.60 × 10–19 J the Hubble constant H0 ≈ 2.4 × 10–18 s–1 Useful formulae The following equations may be useful in answering some of the questions in the examination: Mechanics conservation of energy 1 2 mv 2 – 1 2 mu 2 = Fs for a constant force Waves two-source interference λ = ay d diffraction grating d sinθ = nλ 11378.02 3 Light lens equation 1 u 1 v 1 f + = Electricity terminal potential difference V = E – Ir (e.m.f., E; Internal Resistance, r) potential divider Vout = R1Vin R1 + R2 Particles and photons Einstein’s equation 1 2 mv max 2 = hf – hf0 de Broglie equation λ = h p Astronomy red shift z = Δλ λ recession speed z = v c Hubble’s law v = H0 d 11378.02 4