DHS 14 Superposition (Tutorial Solutions)

1 2025 YEAR 6 SUPERPOSITION TUTORIAL SOLUTIONS 1 For a closed pipe, at the fundamental frequency, the length of the tube is 1 4 1 4 1 4 1 330 4 256 0.32 m (to 2sf) L v f       For an open pipe, at the fundamental frequency, the length of the tube is 1 2 1 2 1 2 1 330 2 256 0.64 m (to 2sf) L v f       2 A 3 D 4 (c + 0.46) – (c + 0.14) = λ / 2  λ = 0.64 m 0.14 m 0.46 m c c

2 5 (a) (i) λ / 2 (ii) Phase difference = π rad (b) fλ of a stationary wave represents the speed of the two progressive waves that superpose to form the stationary wave. When two progressive waves of the same type of equal amplitude, equal frequency, equal wavelength, equal speed travelling in opposite directions meet and undergo superposition with each other, a stationary wave is formed. 6 C 7 D 8 (a) Use b  sin since  is small, sin   tan  0.14/2.7 =  /12 x 10-6  = 620 nm (b) More spreading with narrower slit with lower intensity. (c) (i) From the formula b  sin , when a longer wavelength is used with b unchanged,  increases which means the central maxima is broader with the minima further on both sides further away from the central. (ii) According to the formula b  sin , for the same slit width, the different wavelengths in white light diffract at different angles, with the shorter wavelength (violet) nearer to the central and larger wavelength (red) further from the central, hence there is colour separation away from the central. At the central region, all the different colours have constructive interference, hence central is white.

3 9 The wavelength is estimated to be 500 nm. 9 7(500 10 ) 1.57 103.0 x xb     since  is small,   tan  = x D = 1520(9.5 10 ) x  or x = 3.0 x 1010 m (for this problem, any estimate of wavelength in the visible light region is correct). 10 B 11 12 B  earth Double stars D x

4 13 Using a Dx  (i) If a is decreased, fringe separation x increases from the formula. Intensity of fringes unaffected by changes in a. (ii) If D is increased, fringe separation x increases from the formula. Intensity of fringes decreases as the distance from the source increases (I 1/r2). (iii) Contrast between bright and dark fringes is reduced. This is because the bright fringe will be less bright while the dar k fringes will becomes brighter due to incomplete destructive interference. Fringe separation is unchanged. (iv) If one slit is covered with transparent medium, the fringe pattern is shifted. Fringe separation x is unaffected and the intensity of the fringes are reduced. (v) Fringe separation remained the same. Bright fringes less bright now (as the intensity of light passing through the double slit is reduced). Dark fringe remains dark (no change). More fringes would be observed (due to more diffraction). (vi) Fringe separation changes. Smaller D, smaller fringe separation (i.e. fringe separation closer on the screen nearer the slit) and fringes are brighter. Larger D, larger fringe separation and fringes are less bright. 14 (i) From b  sin and for small , sin   tan  thus, 5.90 x 10 -7 / 0.15 x 10-3 = y / 2 (3.2) or y = 0.025 m (ii) Rayleigh criterion states that for the two diffraction patterns to be just distinguishable, the central maximum of one must lie on the first minimum of the other. Thus, the minimum angle  for two patterns to be just resolved is The angular separation s of the two slits is 2.3 x 10 -3 / 3.2 = 7.2 x 10 -4 rad which is less than . Hence the diffraction patterns cannot be resolved and are not seen as separate. (iii) From a Dx  = 5.90 x 10-7 (3.2) / 2.3 x 10-3 = 8.2 x 10-4 m (iv) Number of fringes = 2.5 x 10 -2 / 8.2 x 10-4 = 30 (round down to whole number) 15 D 16 C = 5.90 x 10-7 / 0.15 x 10-3 = 3.9 x 10-3 rad

5 17 d sin θ = nλ First order: sin θV = 1 (450 x 10-9) / (10-2/6000)  θV1 = 15.7° sin θR = 1 (700 x 10-9) / (10-2/6000)  θR1 = 24.8° angle of dispersion = 24.8o – 15.7o = 11.1o Second order: sin θ V = 2 (450 x 10-9) / (10-2/6000)  θV2 = 32.7° sin θR = 2 (700 x 10-9) / (10-2/6000)  θR2 = 57.1° angle of dispersion = 57.1o – 32.7o = 24.4o Third order: sin θ V = 3 (450 x 10-9) / (10-2/6000)  θV3 = 54.1° < θR2 Hence, there is overlap between 2nd order red and 3rd order violet. sin θR = 3 (700 x 10-9) / (10-2/6000) = 1.26  no third order red 18 (i) sin θ < 1  (n λ / d) < 1  n < (d / λ) = [1/(4.00 x 105)] / (589 x 10-9) = 4.24 So, max n = 4. (ii) 1. 4 th order (greatest diffraction) 2. sin θ = 4 λ / d  θ589 = 70.459°, and θ589.6 = 70.624° The 2 wavelengths can be distinguished since their angular separation is 0.165°. 19. (a) d sin θ = nλ Where tan θ = 0.55 / 1.50  θ = 20.14° λ = d sin θ / n = (1/ (5.5 x 105)) (sin 20.14°) /1 = 626 nm (b) Advantage: Higher order diffracted beams are more spaced out, and hence, the percentage error in θ is smaller. Disadvantage: Higher order diffracted beams have a lower intensity due to diffraction, hence observation could be more difficult.