(a) Define dffraction.
(b)(i) Explain critical angle. The diagram here illustrates a ray of light passing through a rectangular transparent plastic block \(\alpha\) Determine the value of the critical angle. \(\beta\) Calculate the refractive index of the block.
(c) A pipe closed at one end has fundamental frequency of 200Hz. The frequency of the first overtone of the closed pipe is equal to the frequency of the first overtone of an open pipe. Calculate the:
(iii) length of the open pipe. [Speed of sound in air = 330 ms\(^{-1}\)]
(a) Diffraction
Diffraction is the spreading of a wave into the region behind an obstacle, or the bending of a wave round the edges of an obstacle or through an aperture, as the wave passes them. It is most noticeable when the width of the aperture or obstacle is comparable with the wavelength of the wave.
(b)(i) Critical angle
The critical angle is the angle of incidence in the optically denser medium for which the angle of refraction in the less dense medium is exactly \(90^\circ\). For any angle of incidence greater than the critical angle the light undergoes total internal reflection.
(\(\alpha\)) Value of the critical angle
From the diagram the ray travels along (grazes) the top surface of the block, so its angle of incidence in air is \(90^\circ\). It is refracted into the block, and the refracted ray makes \(44^\circ\) with the surface. Its angle measured from the normal is therefore
\[ r = 90^\circ - 44^\circ = 46^\circ. \]
At grazing incidence the angle of refraction is equal to the critical angle of the block, hence
\[ c = 46^\circ. \]
(\(\beta\)) Refractive index of the block
\[ n = \frac{1}{\sin c} = \frac{1}{\sin 46^\circ} = \frac{1}{0.7193} = 1.39. \]
(c) Pipes
A pipe closed at one end sounds only the odd harmonics. Its fundamental is \(f_c = 200\,\text{Hz}\), so its first overtone is the third harmonic:
\[ 3 \times 200 = 600\,\text{Hz}. \]
(i) Fundamental frequency of the open pipe
An open pipe sounds all harmonics, so its first overtone is the second harmonic, \(2f_o\). Given that the two first overtones are equal:
\[ 2f_o = 600 \Rightarrow f_o = 300\,\text{Hz}. \]
(ii) Length of the closed pipe
For a closed pipe \(f_c = \dfrac{v}{4L_c}\), so
\[ L_c = \frac{v}{4f_c} = \frac{330}{4 \times 200} = 0.41\,\text{m}. \]
(iii) Length of the open pipe
For an open pipe \(f_o = \dfrac{v}{2L_o}\), so
\[ L_o = \frac{v}{2f_o} = \frac{330}{2 \times 300} = 0.55\,\text{m}. \]
