Question 17N.2.HL.TZ0.6

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 = «$\frac{\lambda }{m\mathrm{\Delta }\lambda }$ =» $\frac{589}{2\times 0.597}$ «493»

beam width = «$\frac{493}{600}$ =» 8.2 x 10^–4 «m» or 0.82 «mm»