Light from a monochromatic source is incident at right angles to two slits. After passing through the slits the light is incident on a distant screen. Point M is the mid-point of the screen.

The separation of the slits is large compared to their width. A pattern of light and dark fringes is observed on the screen.

(i) State the phenomenon that enables light to reach point M on the screen.

(ii) On the axes below, sketch the intensity of light as observed on the screen as a function of the angle θ . (You do not have to put any numbers on the axes.)

(iii) The distance of the screen from the slits is 1.8 m and the slit separation is 0.12 mm. The wavelength of the light is 650 nm. Point Q on the screen shows the position of the first dark fringe.

Calculate the distance MQ.

Suggest why, even though there are dark fringes in the pattern, no energy is lost.

(i) diffraction;

(ii) correct general shape (cos²θ) touching the horizontal axis;
constant amplitude;
equally spaced maxima; 
Diagram must have at least three fringes.
Award [0] for single slit diffraction pattern.

Award [3] for correct graph that shows modulation by single slit diffraction.

(iii) \({\rm{MQ}} = \frac{1}{2}\frac{{\lambda D}}{d}\);
\({\rm{MQ}} = \left( {\frac{{650 \times {{10}^{ - 9}} \times 1.80}}{{2 \times 0.12 \times {{10}^{ - 3}}}} = } \right)4.9{\rm{mm}}\);

the energy gets redistributed/the total energy in the pattern is the same as the total emitted energy;
the energy that would have appeared at minima now appears at the maxima;