Electromagnetic Sinusoidal Plane - Higher Physics - Solved Past Paper, Exams of Physics

Download Electromagnetic Sinusoidal Plane - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! SOLUTION for PHYS1231 Final Exam S2 2009 Total for 5 Questions: 80 marks For formula sheet: € Emax = cBmax € I = Sav = Emax 2 2µ0c = cBmax 2 2µ0 € I = Sav = cuav € P = 2I c € En = − me4 8ε0 2h2       1 n2 Question 1. EM Waves (Marks 22) (a) An electromagnetic (EM) sinusoidal plane wave with frequency € f = 90.0 MHz propagates in the € +x direction. The electric field of the EM wave has a peak value € E = 2mV/m directed along the € ±y direction. (i) Find the wavelength, period and the maximum value of the magnetic field for this EM wave. (ii) Write down expressions for the space and time variations of the electric and magnetic fields; give the expressions in SI units. Include in your expressions the appropriate unit vectors € ˆ i , ˆ j , ˆ k . (iii) Find the average power per unit area carried by this wave. (iv) Find the average energy density in the radiation. Solution: For this EM wave € f = 90 MHz, Emax = 2.00x10 −3 Vm−1 (i) € λ = c f = 3x108 90x106 = 3.33m € T= 1 f =1.1x10−8 s =11.8 ns € Bmax = Emax c = 2.00x10−3 3x108 = 6.67x10−12 T (ii) The EM wave is represented in the standard form: € E = Emax cos kx −ωt( ) using the substitution € k = 2π λ and € ω = 2πf = 2π T we can express this as € E = Emax cos2π x λ − t T       € B= Bmax cos2π x λ − t T       so that € E = 2.00 mV/m( )cos2π x 3.33 m − t 11.1 ns       ̂ j € B = 6.67 pT( )cos2π x 3.33 m − t 11.1 ns       ̂ k (iii) € I = Sav = Emax 2 2µ0c = 2.00x10−3( )2 2 4πx10−7( ) 3x108( ) = 5.30x10−9 Wm−2 (iv) € I = cuav uav = I c = 5.30x10−9 3x108 =1.77x10−17 Jm−3 (b) Light with free space wavelength € λ = 780nm travels a distance € 2x10−6m in a transparent medium of refractive index 1.6. Calculate, (i) the optical path length (ii) the wavelength of the light in the transparent medium (iii) the phase difference after travelling the distance € 2x10−6m with respect to light travelling the same distance in free space. Solution: (i) the optical path length (o.p.l) is o.p.l = (physical distance travelled in medium) x (refractive index) € = 2x10−6 m( ) 1.6( ) = 3.2x10−6 (ii) The wavelength in the medium is € λm given by € λm = λ0 n = 780 nm 1.6 = 487.5nm where € λ0 is the free space (vacuum) wavelength. (iii) The physical distance travelled is € l = 2.0x10−6m Solution: (i) (ii) Minimum reflection occurs when exiting rays 1. and 2. interfere destructively. Rays 1. and 2. both undergo a 1800 phase change upon reflection, i.e. no net change on reflection. A phase change of ray 2. w.r.t. ray 1. occurs in transmission a distance 2t within the SiO, where t is the SiO film thickness. € ∴ 2nt = λ 2 t = λ 4n = 550x10−9 4 1.45( ) = 9.48x10−8 m ≈ 95 nm Question 3 (Marks 10) A student attends a physics lecture with a camera to record the notes the lecturer has projected on to the theatre’s screen. The camera has a lens diameter 2mm. The lecturer has written the notes in blue ink ( € λblue = 450nm ) and each character (letter or symbol) can be considered to be a 3mm diameter circle on the screen. (i) Will the camera resolve individual characters on the screen if the student sits at the back of the theatre, at a distance of 25 m from the screen? If not, (ii) what is the minimum distance the camera must be from the screen such that individual characters are just resolvable? Note: a plain yes/no answer will obtain no marks; the principle and argument used to show resolvability must be given, with a simple sketch if appropriate, along with all working in your calculation. Solution: (a) The camera has lens diameter D = 2mm . The wavelength of light from the screen is λ = 450nm . Characters on the screen have height h = 3mm . 1800 phase change on reflection Si-SiO 00 phase change on transmission SiO-air Air n=1 Si n=3.5 SiO coating n=1.45 1800 phase change on reflection SiO-air 00 phase change on transmission air-SiO 1. 2. h (character height on screen) s=25m θchar (i) If the student is 25 m from the screen we have: θchar = h s = 3x10−3m 25m =1.2x10−4 rad The Rayleigh Criterion gives θr =1.22 λ D = 1.22 450x10 −9 2x10−3 = 2.7x10−4 rad θr = 2.7x10−4 rad is the angle subtended by the character on the screen at the viewer’s position which allows resolution of individual characters by the camera lens, when resolution is limited by diffraction. At 25 m the character height is too small to allow resolution. (ii) For characters to be just resolvable by the camera lens we must have θchar = θr = 2.7x10 −4 rad so the student must sit at, or closer than, s = h θr = 3x10−3 2.7x10−4 =11.1m Question 4 (Marks 11) Three ideal polarizing sheets are arranged as shown below. Unpolarised light of intensity € Iin is incident upon polarising sheet A. The polarisation axes of the sheets are indicated by the broken arrows. Sheets A and C are arranged as shown with their axes of polarization at 90º, sheet C is rotated such that its polarization axis is at angle θ to the vertical (and therefore also at θº to the polarization axis of sheet A). A B Iin, Ein IA θ C IB IC (i) What is the intensity of light transmitted by sheet A? (ii) If θ = 45º , find the intensity of light transmitted by each of the three sheets and therefore that transmitted by the system of sheets A, B and C. (iii) If θ = 30º , find the intensity of light transmitted by the system. Solution: (i) What is the intensity of light transmitted by sheet A? When polariser A absorbs one of the components (e.g. the z-component, see diagram above) half the incident intensity is removed. Therefore, IA = 1 2 Iin (ii) If θ = 45º , find the intensity of light transmitted by each of the three sheets and therefore that transmitted by the system of sheets A, B and C. IA = Iin 2 and using Malus’ law, IB = IA cos 2 θ = IA cos 2(45º) = IA 2 = Iin 4 , ∴ Ic = (Iin 4).cos2(45º ) = Iin 8 (iii) If θ = 30º , find the intensity of light transmitted by the system of sheets A, B and C. IA = Iin 2 and IB = IA cos 2 θ = IA cos 2(30º) = 3IA 4 = 3 Iin 8 Ic = (3Iin 8).cos 2(60º ) = 3Iin 32 Question 5 (22 Marks) (a) Find the de Broglie wavelength of a 25kV electron. Solution: