| Date | November 2017 | Marks available | 3 | Reference code | 17N.2.HL.TZ0.6 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
| Command term | Determine | Question number | 6 | Adapted from | N/A |
Question
Yellow light from a sodium lamp of wavelength 590 nm is incident at normal incidence on a double slit. The resulting interference pattern is observed on a screen. The intensity of the pattern on the screen is shown.
The double slit is replaced by a diffraction grating that has 600 lines per millimetre. The resulting pattern on the screen is shown.
Explain why zero intensity is observed at position A.
The distance from the centre of the pattern to A is 4.1 x 10–2 m. The distance from the screen to the slits is 7.0 m.
Calculate the width of each slit.
Calculate the separation of the two slits.
State and explain the differences between the pattern on the screen due to the grating and the pattern due to the double slit.
The yellow light is made from two very similar wavelengths that produce two lines in the spectrum of sodium. The wavelengths are 588.995 nm and 589.592 nm. These two lines can just be resolved in the second-order spectrum of this diffraction grating. Determine the beam width of the light incident on the diffraction grating.
Markscheme
the diagram shows the combined effect of «single slit» diffraction and «double slit» interference
recognition that there is a minimum of the single slit pattern
OR
a missing maximum of the double slit pattern at A
waves «from the single slit» are in antiphase/cancel/have a path difference of (n + \(\frac{1}{2}\))λ/destructive interference at A
θ = \(\frac{{4.1 \times {{10}^{ - 2}}}}{{7.0}}\) OR b = \(\frac{\lambda }{\theta }\) «= \(\frac{{7.0 \times 5.9 \times {{10}^{ - 7}}}}{{4.1 \times {{10}^{ - 2}}}}\)»
1.0 × 10–4 «m»
Award [0] for use of double slit formula (which gives the correct answer so do not award BCA)
Allow use of sin or tan for small angles
use of s = \(\frac{{\lambda D}}{d}\) with 3 fringes «\(\frac{{590 \times {{10}^{ - 9}} \times 7.0}}{{4.1 \times {{10}^{ - 2}}}}\)»
3.0 x 10–4 «m»
Allow ECF.
fringes are further apart because the separation of slits is «much» less
intensity does not change «significantly» across the pattern or diffraction envelope is broader because slits are «much» narrower
the fringes are narrower/sharper because the region/area of constructive interference is smaller/there are more slits
intensity of peaks has increased because more light can pass through
Award [1 max] for stating one or more differences with no explanation
Award [2 max] for stating one difference with its explanation
Award [MP3] for a second difference with its explanation
Allow “peaks” for “fringes”
Δλ = 589.592 – 588.995
OR
Δλ = 0.597 «nm»
N = «\(\frac{\lambda }{{m\Delta \lambda }}\) =» \(\frac{{589}}{{2 \times 0.597}}\) «493»
beam width = «\(\frac{{493}}{{600}}\) =» 8.2 x 10–4 «m» or 0.82 «mm»
