(i)     On the axes below, sketch a graph to show how the intensity of the light on the screen varies with the angle $\theta$ shown in the diagram.

(ii)     The wavelength of the light is 520 nm, the width of the slit is 0.04 mm and the screen is 1.2 m from the slit. Show that the width of the central maximum of intensity on the screen is about 3 cm.

Points P and Q are on the circumference of a planet as shown.

By considering the two points, outline why diffraction limits the ability of an astronomical telescope to resolve the image of the planet as a disc.

(i)     

general correct shape touching axis and symmetric about $\theta  = 0$ (at least onesecondary maxima on each side); (judge by eye)

central maximum wider than secondary maxima;

secondary maxima at most one third intensity of central maximum;

(ii)     $\frac{d}{2} = \frac{{D\lambda }}{b}$;

$d = \frac{{2.0 \times 1.2 \times 5.2 \times {{10}^{ - 7}}}}{{4.0 \times {{10}^{ - 5}}}} = 3.12 \times {10^{ - 2}}{\text{ m}}$;

$\approx 3{\text{ cm}}$

Award [2 max] for a sensible argument.

e.g. light from each point forms a diffraction pattern after being focussed by the eyepiece of the telescope;

if the diffraction patterns are not sufficiently well separated then the points will not be resolved as separate sources;

Award [1 max] for the conclusion.

e.g. if the points cannot be resolved as separate sources the planet cannot be seen as a disc;

Intensity distributions were often drawn well but quite a few candidates did not have their graphs in contact with the $\theta$ axis. Candidates’ working was often difficult to follow in the calculation part of this question.

Very few candidates recognised the role played by diffraction in the resolution of the planet as a disc.