HCI 10 Superposition Solutions (Lecture Examples)

Superposition Lecture Examples Worked Solution Example 10.3.1 Two pulses are travelling toward each other, at 10 cm s-1 on a long string, as shown in the figure below. Sketch the shape of the string after 0.6 s. Solution: Dotted lines show the shape of individual waves at t = 0.6 s. Solid line shows the shape of the wave after interference using principle of superposition to add up the displacements of individual waves at a particular point.

Example 10.4.1: Two Source Interference The figure (Fig. 10.4.5) shows two sources X and Y which emit identical sound waves of wavelength 2.0 m. The two sources emit in phase, and the waves emitted have equal amplitudes, each A. What is the amplitude of the sound wave (a) at R, (b) at Q? Suppose the source X is 180o out of phase with source Y. What does an observer hear (c) at R, (d) at Q? Fig. 10.4.5 Solutions: When two sources are in Phase: a) At R, Path difference,  = XR – YR = 0 Therefore, constructive interference occurs  Aresultant = A + A = 2A b) At Q, using Pythagoras theorem, XQ = √6.02 + (3.5 − 1.0)2 = 6.5 m YQ = √6.02 + (3.5 + 1.0)2 = 7.5 m Path difference,  = YQ – XQ = 1.0 m = 0.5  Therefore, the waves meet in antiphase  destructive interference occurs at Q  Resultant amplitude = A - A = 0 If sources X & Y are in Antiphase: c) At R, Path difference,  = XR – YR = 0 Therefore, destructive interference occurs at R  A resultant = A - A = 0  An observer will hear no sound. d) At Q, The waves at Q will meet in phase instead, Thus constructive interference occurs at Q.  A resultant = A + A = 2A  An observer will hear a loud sound.

Example 10.5.1 : Determination of wavelength of light using Young’s Double Slit A screen is separated from the double-slit source by 1.2 m. The distance between the two slits is 0.03 cm. The second-order bright fringe (m = 2) is measured to be 4.5 mm from the centre line (i) What is the fringe separation between 2 neighbouring bright fringes formed on the screen? (ii) Determine the wavelength of light. Solution: (i) Since fringes are formed at small  the fringe separation is constant. 4.5 2.25 mm2y = = (ii) 3 2 7 1.22.25 10 0.03 10 5.63 10 m Ly d    − − − = =  =

Example 10.6.1 : Diffraction Grating White light from a source passes through a filter which transmits only wavelengths of 400 nm to 600 nm. When the filtered light falls normally on a diffraction grating, light of wavelength 600 nm in one order of the spectrum is diffracted at the same angle, 30o, as the 400 nm light in the adjacent order. Find the number of lines per mm for the grating. Solution: 9 61 2(600 10 ) 2.4 10sin sin(30 )   − −= = =   ndm 51 4.17 10N d= =  lines per m = 417 lines per mm Let 600 nm light be the nth order. Then (n+1)th order is the