10 exam-style questions on progressive waves, superposition, standing waves, diffraction and refraction. Tap Show answer for full working.
A progressive wave has frequency \(250\,\mathrm{Hz}\) and wavelength \(1.36\,\mathrm{m}\). Calculate its speed and the phase difference between two points on the wave that are \(0.34\,\mathrm{m}\) apart.
\(v = f\lambda = 250\times1.36 = 340\,\mathrm{m\,s}^{-1}\).
Phase difference \(\Delta\phi = \dfrac{2\pi}{\lambda}\times\Delta x = \dfrac{2\pi}{1.36}\times0.34 = \dfrac{\pi}{2}\,\mathrm{rad}\) (or \(90^\circ\)).
State the principle of superposition. Two coherent sources of sound of wavelength \(0.80\,\mathrm{m}\) produce a minimum of intensity at a point where the path difference is \(1.20\,\mathrm{m}\). Explain why this is a minimum.
When two or more waves meet, the resultant displacement is the vector sum of the individual displacements.
Path difference \(1.20\,\mathrm{m} = 1.5\lambda\), which is an odd multiple of \(\lambda/2\). The waves arrive in antiphase and interfere destructively → minimum intensity.
Light of wavelength \(590\,\mathrm{nm}\) is incident normally on a diffraction grating with \(500\) lines per mm. Calculate the angle of the first-order maximum.
Slit spacing \(d = 1/(500\times10^{3}) = 2.00\times10^{-6}\,\mathrm{m}\).
\(\sin\theta = n\lambda/d = 1\times590\times10^{-9}/2.00\times10^{-6} = 0.295\).
\(\theta = \sin^{-1}(0.295) = 17.2^\circ\).
A string of length \(1.20\,\mathrm{m}\) is fixed at both ends and vibrates in its third harmonic at \(180\,\mathrm{Hz}\). Calculate the wave speed on the string and the wavelength of the progressive waves that form the standing wave.
Third harmonic: \(L = 3\lambda/2 \Rightarrow \lambda = 2L/3 = 2\times1.20/3 = 0.80\,\mathrm{m}\).
\(v = f\lambda = 180\times0.80 = 144\,\mathrm{m\,s}^{-1}\).
A ray of light travels from air into glass. The angle of incidence is \(40^\circ\) and the refractive index of the glass is \(1.52\). Calculate the angle of refraction and the critical angle for light travelling from this glass into air.
\(n\sin i = \sin r\) (air \(n=1\)): \(1.52\sin r = \sin40^\circ \Rightarrow \sin r = 0.423 \Rightarrow r = 25.0^\circ\).
Critical angle: \(\sin c = 1/1.52 = 0.658 \Rightarrow c = 41.1^\circ\).
Two coherent sources \(S_1\) and \(S_2\) emit sound of wavelength \(0.60\,\mathrm{m}\). A detector moves along a line parallel to \(S_1S_2\) and records successive maxima separated by \(0.45\,\mathrm{m}\). The perpendicular distance from the midpoint of \(S_1S_2\) to the line is \(2.0\,\mathrm{m}\). Calculate the separation of the two sources.
Fringe spacing \(w = \lambda D / a\), where \(a\) is source separation, \(D\) is distance to screen/line.
\(0.45 = (0.60\times2.0)/a \Rightarrow a = 1.20/0.45 = 2.67\,\mathrm{m}\).
Explain what is meant by path difference and phase difference for two waves. A standing wave is formed on a string. At a node the amplitude is zero. Explain, in terms of path difference, why a node occurs at the fixed ends of the string.
Path difference: the difference in the distances travelled by two waves from their sources to a point. Phase difference: the corresponding difference in phase (\(\Delta\phi = 2\pi\times\text{path difference}/\lambda\)).
At a fixed end the reflected wave undergoes a \(\pi\) phase change. The incident and reflected waves therefore meet in antiphase (path difference effectively \(\lambda/2\) plus the phase change), producing a permanent node.
Monochromatic light of wavelength \(450\,\mathrm{nm}\) is incident normally on a diffraction grating. The second-order maximum is observed at \(32^\circ\) to the normal. Calculate the number of lines per millimetre on the grating and the maximum order that can be observed.
\(d\sin\theta = n\lambda \Rightarrow d = n\lambda/\sin\theta = 2\times450\times10^{-9}/\sin32^\circ = 1.70\times10^{-6}\,\mathrm{m}\).
Lines per mm: \(1/(d\times10^{3}) = 589\,\mathrm{mm}^{-1}\).
Maximum order: \(n_{\max} = d/\lambda = 1.70\times10^{-6}/450\times10^{-9} = 3.78\), so \(n_{\max}=3\).
A progressive transverse wave on a string is described by \(y = 0.05\sin(4\pi x - 20\pi t)\) where \(x\) and \(y\) are in metres and \(t\) in seconds. Determine the amplitude, wavelength, frequency and speed of the wave. State the direction of propagation.
Amplitude \(A = 0.05\,\mathrm{m}\).
\(k = 4\pi \Rightarrow \lambda = 2\pi/k = 0.50\,\mathrm{m}\).
\(\omega = 20\pi \Rightarrow f = \omega/2\pi = 10\,\mathrm{Hz}\).
\(v = f\lambda = 5.0\,\mathrm{m\,s}^{-1}\).
The form is \(f(kx - \omega t)\), so the wave travels in the positive \(x\)-direction.
Ultrasound of frequency \(2.5\,\mathrm{MHz}\) travels from soft tissue into bone. The speed in soft tissue is \(1500\,\mathrm{m\,s}^{-1}\) and in bone is \(4000\,\mathrm{m\,s}^{-1}\). Calculate the wavelength in each medium and the percentage of intensity reflected at the interface if the intensity reflection coefficient is given by \(\left(\dfrac{Z_2-Z_1}{Z_2+Z_1}\right)^2\) where the acoustic impedances are \(Z_{\mathrm{tissue}}=1.6\times10^{6}\) and \(Z_{\mathrm{bone}}=6.5\times10^{6}\,\mathrm{kg\,m}^{-2}\mathrm{s}^{-1}\).
\(\lambda_{\mathrm{tissue}} = v/f = 1500/(2.5\times10^{6}) = 6.0\times10^{-4}\,\mathrm{m}\).
\(\lambda_{\mathrm{bone}} = 4000/(2.5\times10^{6}) = 1.6\times10^{-3}\,\mathrm{m}\).
Reflection coefficient \(= \left(\dfrac{6.5-1.6}{6.5+1.6}\right)^2 = (4.9/8.1)^2 = 0.366\) (36.6% of intensity reflected).