Question
(a) State the principle of superposition. (2 marks)
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(b) Coherent light is incident normally on two identical slits X and Y. The diffracted light emerging from the slits superposes to produce an interference pattern on a screen positioned at a distance of \(1.9\,\mathrm{m}\) from the slits.
Fig. 4.1 shows the arrangement and the central part of the interference pattern of bright and dark fringes formed on the screen.
The separation of the slits is \(0.65\,\mathrm{mm}\). The distance between the centres of adjacent bright fringes is \(1.7\,\mathrm{mm}\).
Calculate the wavelength \(\lambda\) of the light. (3 marks)
\(\lambda\) = __________________________________________ \(\mathrm{m}\)
(c) Light waves from slits X and Y in (b) arrive at a point between adjacent bright fringes on the screen. Fig. 4.2 shows the variation of displacement with time for the waves arriving at the point where they meet.
A student makes two statements about the waves at this point:
Statement 1: “The phase difference between the waves is \(90^\circ\).”
Statement 2: “The amplitude of the resultant wave is zero.”
(i) Explain how statement 1 is correct. (1 mark)
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(ii) State and explain whether statement 2 is correct. (1 mark)
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(d) The width of each slit in (b) is decreased by the same amount. There is no change to the separation of the slits.
Describe and explain the effect, if any, of this change on the appearance of the interference pattern. (2 marks)
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Syllabus Topic Codes (Cambridge International AS & A Level Physics 9702):
• 8.3: Interference — parts (b), (c)(i) and (c)(ii)
• 8.2: Diffraction — part (d)
▶️ Answer/Explanation
(a) [2 marks]
The principle of superposition states that when two or more waves meet or overlap at a point, the resultant displacement is the sum of the individual displacements.
Answer: \( \boxed{\text{resultant displacement}=\text{sum of individual displacements}} \)
(b) Wavelength of the light [3 marks]
For double-slit interference, the fringe spacing is given by
\( x=\frac{\lambda D}{a} \)
Therefore,
\( \lambda=\frac{ax}{D} \)
where \(a=0.65\times10^{-3}\,\mathrm{m}\), \(x=1.7\times10^{-3}\,\mathrm{m}\) and \(D=1.9\,\mathrm{m}\).
\( \lambda=\frac{(0.65\times10^{-3})(1.7\times10^{-3})}{1.9} \)
\( \lambda=5.8\times10^{-7}\,\mathrm{m} \)
Answer: \( \boxed{5.8\times10^{-7}\,\mathrm{m}} \)
(c)(i) Phase difference [1 mark]
The waves are displaced in phase by one quarter of a cycle, or one quarter of a period.
Since one complete cycle corresponds to \(360^\circ\),
\( \text{phase difference}=\frac{360^\circ}{4}=90^\circ \)
Answer: \( \boxed{90^\circ} \)
(c)(ii) Resultant amplitude [1 mark]
Statement 2 is not correct.
The waves are not \(180^\circ\) out of phase, so they are not in antiphase. One wave has some displacement when the other has zero displacement.
Therefore, the displacements are not always equal and opposite, so the resultant amplitude is not zero.
Answer: \( \boxed{\text{Statement 2 is incorrect}} \)
(d) Effect of decreasing slit width [2 marks]
Decreasing the width of each slit causes more diffraction, so the light from each slit spreads out more.
The light from the slits therefore has a lower intensity, so the bright fringes are less bright or dimmer.
The separation of the interference fringes is unchanged because the slit separation has not changed.
Answer: \( \boxed{\text{more diffraction and dimmer fringes; fringe spacing unchanged}} \)