Question
(a) State what is meant by diffraction of a wave. (2 marks)
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(b) A beam of vertically polarised light of wavelength \(540\,\mathrm{nm}\) is incident normally on a diffraction grating, as shown in Fig. 4.1.
Fig. 4.1 shows the diffraction grating, screen, central bright fringe at point \(O\), and point \(P\) at a variable angle \(\theta\) to the line \(XO\).
The diffraction grating has line spacing of \(5.0\times10^{-6}\,\mathrm{m}\).
The light transmitted by the diffraction grating illuminates a circular screen. The diffraction grating is at the centre \(X\) of the circle.
The central bright fringe is formed at point \(O\) on the screen and has intensity \(I_0\).
\(P\) is a point on the screen where the line \(XP\) is at a variable angle \(\theta\) to the line \(XO\). The intensity of light on the screen at \(P\) varies with \(\theta\).
(i) Show that the angle \(\theta\) at which the first-order bright fringe is formed is \(6.2^\circ\). (2 marks)
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(ii) Determine the value of \(\theta\) at which the second-order bright fringe is formed. (1 mark)
\(\theta=\) __________________________ \(^{\circ}\)
(iii) On Fig. 4.2, sketch the variation of the intensity \(I\) with \(\theta\) for values of \(\theta\) from \(-15^\circ\) to \(+15^\circ\). (3 marks)
(c) A polarising filter is placed in the path of the light beam that is incident on the diffraction grating in Fig. 4.1. The transmission axis of the filter is at \(45^\circ\) to the vertical.
Suggest how the variation of intensity with \(\theta\) for the light on the screen compares with the answer in (b)(iii). (2 marks)
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 8.4: The diffraction grating — parts (b)(i), (b)(ii) and (b)(iii)
• 7.5: Polarisation — part (c)
▶️ Answer/Explanation
(a) Diffraction [2 marks]
Diffraction is the spreading out of a wave as it passes through a gap or around an obstacle.
Answer: \( \boxed{\text{spreading out of a wave through a gap or around an obstacle}} \)
(b)(i) First-order bright fringe [2 marks]
For a diffraction grating, bright fringes occur when
\(n\lambda=d\sin\theta\)
For the first order, \(n=1\), so
\(\sin\theta=\dfrac{\lambda}{d}\)
\(\sin\theta=\dfrac{540\times10^{-9}}{5.0\times10^{-6}}\)
\(\theta=\sin^{-1}(0.108)\)
\(\theta=6.2^\circ\)
Answer: \( \boxed{6.2^\circ} \)
(b)(ii) Second-order bright fringe [1 mark]
For the second order, \(n=2\).
\(2\lambda=d\sin\theta\)
\(\sin\theta=\dfrac{2(540\times10^{-9})}{5.0\times10^{-6}}\)
\(\theta=\sin^{-1}(0.216)\)
\(\theta=12^\circ\)
Answer: \( \boxed{12^\circ} \)
(b)(iii) Intensity variation [3 marks]
The bright fringes occur symmetrically about \(\theta=0^\circ\).
The first-order maxima occur at \(\theta=\pm6.2^\circ\), approximately \(\pm6^\circ\), and the second-order maxima occur at \(\theta=\pm12^\circ\).
Therefore, the graph should show peaks at approximately \(\theta=0^\circ\), \(\pm6^\circ\) and \(\pm12^\circ\), with zero intensity between the maxima.
The central maximum has intensity \(I_0\), while the other maxima have intensities equal to or less than \(I_0\).
Answer: \( \boxed{\text{Maxima at }0^\circ,\ \pm6^\circ,\ \pm12^\circ\text{ with zero intensity between them}} \)
(c) Effect of the polarising filter [2 marks]
The positions of the diffraction maxima are determined by the diffraction-grating equation, so the maxima remain at the same angles as in (b)(iii).
The filter has its transmission axis at \(45^\circ\) to the original vertical polarisation. By Malus’ law, the transmitted intensity is
\(I=I_0\cos^2 45^\circ\)
\(I=\dfrac{I_0}{2}\)
Thus, the intensities of all the diffraction maxima are halved, while their angular positions remain unchanged.
Answer: \( \boxed{\text{The maxima occur at the same angles, but all intensities are halved.}} \)