(a) State the principle of superposition. (2 marks)
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(b) Light of wavelength \(7.2\times10^{-7}\,\mathrm{m}\) is incident normally on a double slit, as shown in Fig. 4.1.
A screen is at a distance \(D\) from the double slit. The double slit and the screen are parallel.
The separation of the slits in the double-slit arrangement is \(0.16\,\mathrm{mm}\). The resulting interference pattern on the screen contains nine dark fringes, as shown in Fig. 4.2.
The distance between the centres of the first and ninth dark fringes is \(3.2\,\mathrm{cm}\).
(i) Calculate \(D\). (3 marks)
\(D=\) ____________________ \(\mathrm{m}\)
(ii) The slit separation is now gradually decreased from \(0.16\,\mathrm{mm}\) to \(0.04\,\mathrm{mm}\). The distance between the centres of adjacent dark fringes is \(x\).
On Fig. 4.3, sketch the variation of \(x\) with slit separation. (3 marks)
Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
(a) Principle of superposition [2 marks]
When two or more waves meet or overlap at a point, the resultant displacement is equal to the sum of the individual displacements.
(b)(i) Distance \(D\) [3 marks]
There are nine dark fringes, so the distance between the centres of the first and ninth dark fringes contains \(8\) fringe spacings.
Therefore, the fringe width is
\(x=\dfrac{3.2\times10^{-2}}{8}\)
\(x=4.0\times10^{-3}\,\mathrm{m}\)
For double-slit interference,
\(x=\dfrac{\lambda D}{a}\)
Rearranging,
\(D=\dfrac{ax}{\lambda}\)
The slit separation is
\(a=0.16\,\mathrm{mm}=0.16\times10^{-3}\,\mathrm{m}\)
Hence,
\(D=\dfrac{(0.16\times10^{-3})(4.0\times10^{-3})}{7.2\times10^{-7}}\)
\(D=0.889\,\mathrm{m}\)
Answer: \( \boxed{0.89\,\mathrm{m}} \)
(b)(ii) Variation of fringe spacing with slit separation [3 marks]
From
\(x=\dfrac{\lambda D}{a}\)
the fringe spacing \(x\) is inversely proportional to the slit separation \(a\).
Therefore, the graph should be a curved line with a negative gradient, with the magnitude of the gradient decreasing as the slit separation increases.
The curve should pass through the points approximately
\((0.04,\,1.6)\)
and
\((0.16,\,0.4)\)
Graph: a decreasing inverse-type curve from approximately \((0.04,1.6)\) to \((0.16,0.4)\).
