Explain why the intensity of light at θ=0 is 16I₀.

The width of each slit is 1.0μm. Use the graph to

(i) estimate the wavelength of light.

(ii) determine the separation of two consecutive slits.

The arrangement is modified so that the number of slits becomes very large. Their separation and width stay the same.

(i) State two changes to the graph on page 20 as a result of these modifications.

(ii) A diffraction grating is used to resolve two lines in the spectrum of sodium in the second order. The two lines have wavelengths 588.995nm and 589.592nm.

Determine the minimum number of slits in the grating that will enable the two lines to be resolved.

constructive interference

amplitude/amount of light from 4 slits is 4 × amplitude «from one slit»

intensity is proportional to amplitude² OR shows 4² = 16 in context of intensity

(i)
«diffraction minimum at» θ=0.43rad
\(\lambda  =  \ll b\theta  = 1.0 \times {10^{ - 6}} \times 0.43 =  \gg 4.3 \times {10^{ - 7}}{\rm{m}}\)

Accept θ in range 0.41 to 0.45 rad.
Allow λ=bsinθ but do not allow nλ=dsinθ.
Award [1 max] for solution using factor of 1.22.
Award [0] if use of \(s = \frac{{\lambda D}}{d}\) seen.

(ii)
«first secondary maximum at» θ=0.125rad

\(d = \frac{{1 \times {\rm{value from (b)(i)}}}}{{\sin 0.125}} = 3.4 \times {10^{ - 6}}{\rm{m}}\)

Accept q in range 0.123 to 0.127 rad.
Sine must be seen to award MP2.
Allow ECF from (b)(i).
Allow use of 2nd or 3rd maxima (0.25 rad and 3.46 μm or 0.375 rad and 3.5 μm with appropriate n).

(i)
primary maxima/fringes become brighter/more intense

primary maxima become narrower/sharper

secondary maxima become unimportant/less intense/disappear

Insist on “secondary” for MP3.

(ii)
\(N =  \ll \frac{{\bar \lambda }}{{m\Delta \lambda }} =  \gg \frac{{589.2935}}{{2 \times 0.5970}}\)
N=494 or 500

Allow use of 588.995 nm or 589.592 nm for \({\bar \lambda }\).