Question

A two-dimensional square plate (20 mm sides) contains a homogeneous circular inclusion of 5 mm diameter in it. A parallel beam of X-rays (beam width 30 mm) is used in a tomography system to determine the location of the inclusion. What is the minimum number of views required to approximately determine the location of the inclusion?

Hint:

- For symmetric shapes (disk), one projection gives the distance \(s\) of the center along the view normal; two non-collinear views are enough to triangulate \((x_0,y_0)\). - Beam width and object size matter for coverage, not for the number of views required to infer position. - Orthogonal views are a simple choice to make the two equations independent.

Answer 1

Correct Answer: 2

Step 1: Geometry of a disk under parallel-beam projection. In parallel-beam CT, the Radon projection at view angle \(\theta\) of a circular inclusion (disk) is a \emph{rectangular} profile of constant height whose center lies at the signed distance \[ s = x_0 \cos\theta + y_0 \sin\theta \] from the rotation origin, where \((x_0,y_0)\) is the center of the inclusion. The width of this rectangle equals the disk diameter (here \(5\ \mathrm{mm}\)) and is independent of \(\theta\). Step 2: Number of views needed to solve for the center. Each single view gives one linear equation in the unknowns \((x_0,y_0)\) via \(s\). Two distinct view angles \(\theta_1\) and \(\theta_2\) yield \[ \begin{cases} s_1 = x_0 \cos\theta_1 + y_0 \sin\theta_1,
[2pt] s_2 = x_0 \cos\theta_2 + y_0 \sin\theta_2, \end{cases} \] which can be solved uniquely for \((x_0,y_0)\) so long as \(\theta_2 \not\equiv \theta_1 \ (\mathrm{mod}\ \pi)\).
Thus, the minimum number of distinct projection views needed to locate the inclusion is \(\boxed{2}\) (e.g., orthogonal views at \(0^\circ\) and \(90^\circ\)). The given beam width (30 mm) exceeds the object size (20 mm), ensuring full coverage but not affecting this minimum.