RI Y6 Remedial Superpositions Soln

Raffles Institution Year 5-6 Physics Department 1 2025 Year 6 H2 Physics Remedial Chapter 12: Superposition Suggested Solutions 1 Rayleigh criterion for 2 images to be just resolved: limiting angle of resolution min b λθ ≈ Approximate minimum separation, minsr r b λθ≈≈ Wavelength of visible light region is in the range 400 nm to 700 nm. Subst. 500 nmλ = , ( ) 9 15 10500 1020 9.5 10 3.2 10 m3.0sr b λ −×≈=×× =× Option C gives the correct order of magnitude. NOTE: The exact wavelength of the visible light used is not crucial as the options given are all in different orders of magnitude. 2 Single slit first order minima: sin ----- (1)b θλ= Double slit maxima sin ----- (2)an θλ= Where the first minima of the single slit diffraction pattern coincide with the maxima of the double- slit pattern, θ is the same. 0.18(2) 6(1) 0.030 a nnb⇒ =⇒= = The 6th order bright fringe will coincide with the first minima of the single slit diffraction pattern. So there will be 5 bright fringes on either side of the central axis and a bright fringe at the centre. Number of bright fringes between the two first order minima of the single slit diffraction pattern is 5 + 5 + 1 = 11. 3 (a) A stationary wave is formed when 2 identical waves with same amplitude, frequency and wavelength (or speed) travelling in opposite directions towards each other superpose. (i) 0 (ii) 3 (iii) A solid straight line across the centre of the wave (ie at the dotted line) 4 (a) (i) Microwaves. The degree of diffraction of a wave depends on its wavelength relative to the opening/obstacle size. (ii) 8 6 3.00 10 3.315 m90.5 10 c fλ ×= = = × 400 nodes correspond to 399 λ /2 segments. Hence, distance is 3.315399 661 m2×= . (b) (i) ( ) ( ) ( ) 9 3 690 10 1.20 0.002486 2.49 mm 0.333 10 Dx a λ − − × = = = = ×

Raffles Institution Year 5-6 Physics Department 2 (ii) Since the screen is 2.40 m wide and a distance 1.20 m away, the maximum diffraction angle is 112.40 2tan tan 1 451.20θ −−= = = ° . ( ) 9 sin sin 1 sin45 690 10 3.4300000 dn dn θλ θ λ − = = = ° ×=  The highest visible order is 3. 5 (a) Waves must be of the same (transverse/longitudinal) type. Transverse waves must be unpolarised or polarised in the same plane. (b) (i) To produce coherent sources at the two slits (s1 and s2). (ii) intensity = k(amplitude)2  amplitude of each wave = I k resultant intensity = k(resultant amplitude) 2 = 2 4IIkI kk  += (iii) π rad OR 180° (iv) Light from S1 and S2 interferes destructively. This leaves light from S3 at point P. Point P will not be dark. 6 (a) Interference is the superposition of two or more coherent waves to give a resultant wave whose resultant amplitude is given by the principle of superposition. (b) To produce 2 water waves with circular wavefronts, 2 (ball-typed) dippers are used. To ensure that they are coherent, the dippers are connected to the same vibrator source / motor. [Note: Also accept 1 dipper as source and 2 slits] For observations of interference patterns, a lamp is set-up above the tank and a screen below the tank. (c) (i) 320 400 0.80 m vf λ λ λ = = × = (ii) It is a maximum. Distance SPX is 7.0 m and distance SX is 5.0 m, hence the path difference is 2.0 m which is equivalent to 2.5 λ. After considering the π radians phase change upon reflection, the 2 waves meet in phase.