Waves of Wavelength - Physics - Solved Paper, Exams of Physics

Download Waves of Wavelength - Physics - Solved Paper and more Exams Physics in PDF only on Docsity! SOLUTION to PHYS1221 FINAL EXAM S82 2010 Total for 3 Questions: 45 Marks Question4. EM Waves (Marks 14) (a) A laser emits sinusoidal electromagnetic (EM) waves that travel in the negative x-direction. The EM waves of wavelength A = 10,600 nm are emitted from the laser into vacuum with E field parallel to the z-axis; the E field amplitude is 1.5x10° Vm", Write vector equations for E and B as a function of time and position. (8 marks) Solution The wave is travelling in the direction -i. The equations for E and B have the form E(x,t) = KE, cos(kx + wt) Bx,t) = jz. cos(kx + cot) mk where the unit vectors k,j give the orientation of E and B. E 15x10% and where Ey, =1.5x10° Vm" and B,,,. = =, = 5.0x10° T c 3.0x10 2. 27 The wave number is k = = = eae Z =5,93x10° radm™ The angular frequency is a =ck = (3x10°)(5.93x10°) rads =1.78x10" rads” We obtain E(xt) = k(15x10*)cos[5.93x10*x +1.78x10"t] and B(x) = j(5.0x10° cos[5.93x10°x +1.78x10"t] (b) InaCD ROM drive light from a semiconductor diode laser having wavelengtha = 780nm travels a distance 125 nm in a polycarbonate layer, Polycarbonate is a transparent medium of refractive index 1.58. Calculate, (i) the optical path length (3 marks) (ii) the wavelength of the light in the transparent medium (3 marks) Solution i) the optical path length (o.p.1) is (i) o.p.l = (physical distance travelled in medium) x (refractive index) = (125x107 m}(1.58) =1.97x107 The wavelength in the polycarbonate medium is A,, given by where A, is the free space (vacuum) wavelength.