A-LevelPhysicsSuperposition9702/23 · May/June 2017 · Q510 Marks
A diffraction grating is used to determine the wavelength of light. A diffraction grating is used with different wavelengths of light. The angle θ of the second order maximum is measured for each wavelength. The variation with wavelength λ of sin θ is shown in [Figure 5.1].
- (a)(i)Describe the diffraction of light at a diffraction grating.[2]
- (a)(ii)By reference to interference, explain 1. the zero order maximum, 2. the first order maximum.[3]
- (b)(i)Determine the gradient of the line shown in Fig. 5.1.[2]
- (b)(ii)Use the gradient determined in (i) to calculate the slit separation d of the diffraction grating.[2]
- (b)(iii)On Fig. 5.1, sketch a line to show the results that would be obtained for the first order maxima.[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- 5(a)(i) waves at the elements/slits
- 5(a)(i) waves spread (into the geometric shadow)
- 5(a)(ii) 1. waves (from each element/slit) overlap/meet/superpose
- 5(a)(ii) with a phase difference/path difference of zero
- 5(a)(ii) 2. phase difference is 360°/path difference of λ
- 5(b)(i) e.g. gradient = (0.40 – 0.32) / [(500 – 400) × 10⁻⁹]
- 5(b)(i) = 8(.0) × 10⁵
- 5(b)(ii) d sinθ = nλ
- 5(b)(ii) d = n/gradient
- 5(b)(ii) = 2/8.0 × 10⁵ = 2.5 × 10⁻⁶ m
- 5(b)(iii) straight line drawn with lower gradient (about ½) and all points lower
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